7.2 Calculus of Parametric Curves

Learning Objectives

  • Determine derivatives and equations of tangents for parametric curves.
  • Find the area under a parametric curve.
  • Determine the arc length of a parametric curve.
  • Apply the formula for the surface area of the surface generated by revolving a parametric curve about the x-axis or the y-axis.

Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve?

Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher’s hand. If the position of the baseball is represented by the plane curve \ds \left(x\left(t\right),y\left(t\right)\right), then we should be able to use calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate just how far that ball has traveled as a function of time.

Derivatives of Parametric Equations

We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations

\ds x\left(t\right)=2t+3,\phantom{\rule{1em}{0ex}}y\left(t\right)=3t-4,\phantom{\rule{1em}{0ex}}-2\le t\le 3.

The graph of this curve appears in Figure 1 below. It is a line segment starting at \ds \left(-1,-10\right) and ending at \ds \left(9,5\right).

A straight line from (−1, −10) to (9, 5). The point (−1, −10) is marked t = −2, the point (3, −4) is marked t = 0, and the point (9, 5) is marked t = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t – 4, and −2 ≤ t ≤ 3
Figure 1. Graph of the line segment described by the given parametric equations.

We can eliminate the parameter by first solving the equation \ds x\left(t\right)=2t+3 for t:

\ds \begin{array}{ccc}\hfill x\left(t\right)&\ds =\hfill &\ds 2t+3\hfill \\[5mm]\ds \hfill x-3&\ds =\hfill &\ds 2t\hfill \\[5mm]\ds \hfill t&\ds =\hfill &\ds \frac{x-3}{2}.\hfill \end{array}

Substituting this into \ds y\left(t\right), we obtain

\ds \begin{array}{ccc}\hfill y\left(t\right)&\ds =\hfill &\ds 3t-4\hfill \\[5mm]\ds \hfill y&\ds =\hfill &\ds 3\left(\frac{x-3}{2}\right)-4\hfill \\[5mm]\ds \hfill y&\ds =\hfill &\ds \frac{3x}{2}-\frac{9}{2}-4\hfill \\[5mm]\ds \hfill y&\ds =\hfill &\ds \frac{3x}{2}-\frac{17}{2}.\hfill \end{array}

The slope of this line is given by \ds \frac{dy}{\,dx }=\frac{3}{2}. Next we calculate \ds {x}^{\prime }\left(t\right) and \ds {y}^{\prime }\left(t\right). This gives \ds {x}^{\prime }\left(t\right)=2 and \ds {y}^{\prime }\left(t\right)=3. Notice that \ds \frac{dy}{\,dx }=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}. This is no coincidence, as outlined in the following theorem.

Derivative of Parametric Equations

Consider the plane curve defined by the parametric equations \ds x=x\left(t\right) and \ds y=y\left(t\right). Suppose that \ds {x}^{\prime }\left(t\right) and \ds {y}^{\prime }\left(t\right) exist, and assume that \ds {x}^{\prime }\left(t\right)\ne 0. Then the derivative \ds \frac{dy}{\,dx } is given by

\ds (*)\quad\frac{dy}{\,dx }=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{{y}^{\prime }\left(t\right)}{{x}^{\prime }\left(t\right)}.

Proof

This theorem can be proven using the Chain Rule. In particular, assume that the parameter t can be eliminated, yielding a differentiable function \ds y=F\left(x\right). Then \ds y\left(t\right)=F\left(x\left(t\right)\right). Differentiating both sides of this equation using the Chain Rule yields

\ds {y}^{\prime }\left(t\right)={F}^{\prime }\left(x\left(t\right)\right){x}^{\prime }\left(t\right), so

\ds {F}^{\prime }\left(x\left(t\right)\right)=\frac{{y}^{\prime }\left(t\right)}{{x}^{\prime }\left(t\right)}.

But \ds {F}^{\prime }\left(x\left(t\right)\right)=\frac{dy}{\,dx }, which proves the theorem. □

The formula (*) from the previous theorem can be used to calculate the first derivative for a curve defined parametrically at the given value t_0 of t, and hence the slope of the tangent line to the curve at the point (x(t_0),y(t_0)) corresponding to t=t_0. For the purpose of sketching parametric curves, it is useful to determine, where the tangent to the curve is horizontal and where it is vertical. It has a direct analogy with considering the critical points on the graph of a curve with explicit equation y=f(x), corresponding to the values x_0 of x, such that f'(x_0)=0 or f'(x_0) is undefined. Based on (*), we see that the tangent to the parametric curve x=x(t), y=y(t) is horizontal where \dfrac{dy}{dt}=0 and \dfrac{dx}{dt}\ne0, and the tangent is vertical where \dfrac{dx}{dt}=0 and \dfrac{dy}{dt}\ne0.

Finding the Derivative for a Parametric Curve

For each of the following parametrically defined plane curves, calculate the derivative \ds \frac{dy}{dx } as well as determine the points, where the tangent line is horizontal and the points, where the tangent line is vertical.

  1. \ds x\left(t\right)={t}^{2}-3,\phantom{\rule{1em}{0ex}}y\left(t\right)=2t-1,\phantom{\rule{1em}{0ex}}-3\le t\le 4
  2. \ds x\left(t\right)=2t+1,\phantom{\rule{1em}{0ex}}y\left(t\right)={t}^{3}-3t+4,\phantom{\rule{1em}{0ex}}-2\le t\le 2
  3. \ds x\left(t\right)=5\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}y\left(t\right)=5\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}0\le t\le 2\pi

Solution

  1. To apply (*), first calculate \ds \dfrac{dx}{dt} and \ds \dfrac{dy}{dt}\text{:}

    \ds \begin{array}{c}\dfrac{dx}{dt}=2t,\ \dfrac{dy}{dt}=2.\hfill \end{array}

    Next substitute these into (*):

    \ds \begin{array}{c}\ds\frac{dy}{\,dx }=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{2t}=\frac{1}{t}.\hfill \end{array}

    Since \dfrac{dy}{dt}\ne0, there are no points on the curve, where the tangent line is horizontal. Solving \dfrac{dx}{dt}=2t=0, we find the value of \ds t=0, corresponding to the point on the curve, where the tangent line is vertical. Calculating \ds x\left(0\right) and \ds y\left(0\right) gives \ds x\left(0\right)={\left(0\right)}^{2}-3=-3 and \ds y\left(0\right)=2\left(0\right)-1=-1, yielding the point \ds \left(-3,-1\right) on the curve. Note that, eliminating the parameter, we can determine that this curve is a parabola opening to the right, and the point \ds \left(-3,-1\right) is its vertex as shown below.

    A curved line going from (6, −7) through (−3, −1) to (13, 7) with arrow pointing in that order. The point (6, −7) is marked t = −3, the point (−3, −1) is marked t = 0, and the point (13, 7) is marked t = 4. On the graph there are also written three equations: x(t) = t2 − 3, y(t) = 2t − 1, and −3 ≤ t ≤ 4.
    Figure 2. Graph of the parabola described by parametric equations in part a.
  2. Again, we start by calculating \ds \dfrac{dx}{dt} and \ds \dfrac{dy}{dt}\text{:}

    \ds \begin{array}{c}\dfrac{dx}{dt}=2,\ \dfrac{dy}{dt}=3{t}^{2}-3.\hfill \end{array}

    Next substitute these into (*) to find \ds\frac{dy}{dx}:

    \ds \begin{array}{c}\ds\frac{dy}{dx }=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3{t}^{2}-3}{2}.\hfill \end{array}

    Since \dfrac{dx}{dt}\ne0, there are no points on this curve, where the tangent line is vertical. To determine the points, where the tangent line is horizontal, we solve \dfrac{dy}{dt}=3t^2-3=0, and find that \ds t=\pm1. When \ds t=-1, \ds x\left(-1\right)=2\left(-1\right)+1=-1 and y\left(-1\right)={\left(-1\right)}^{3}-3\left(-1\right)+4=6,  which corresponds to the point \ds \left(-1,6\right) on the curve. When \ds t=1,

    \ds x\left(1\right)=2\left(1\right)+1=3 and \ds y\left(1\right)={\left(1\right)}^{3}-3\left(1\right)+4=2,
    which corresponds to the point \ds \left(3,2\right) on the curve. The following figure provides the sketch of the curve.
    A vaguely sinusoidal curve going from (−3, 2) through (−1, 6) and (3, 2) to (5, 6). The point (−3, 2) is marked t = −2, the point (−1, 6) is marked t = −1, the point (3, 2) is marked t = 1, and the point (5, 6) is marked t = 2. On the graph there are also written three equations: x(t) = 2t + 1, y(t) = t3 – 3t + 4, and −2 ≤ t ≤ 2.
    Figure 3. Graph of the curve described by parametric equations in part b.
  3. Calculating \ds \dfrac{dx}{dt} and \ds \dfrac{dy}{dt}, we obtain

    \ds \begin{array}{c}\dfrac{dx}{dt}=-5\sin(t),\ \dfrac{dy}{dt}=5\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t).\hfill \end{array}

    Therefore, (*) yields

    \ds \begin{array}{c}\ds\frac{dy}{dx }=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{5\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)}{-5\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)}= - \text{cot}\phantom{\rule{0.2em}{0ex}}(t).\hfill \end{array}

    We see that \dfrac{dy}{dt}=5\cos(t)=0 when t=\dfrac{\pi}2, \dfrac{3\pi}{2} in the interval \ds[0,2\pi]. Note that

    \dfrac{dx}{dt}\left(\frac{\pi}{2}\right)=-5\ne0 and \dfrac{dx}{dt}\left(\dfrac{3\pi}{2}\right)=5\ne0.

    Hence, each of these values of t yields a point on the curve, where the tangent line is horizontal. To find the coordinates of these points, we substitute t=\dfrac{\pi}2 and t=\dfrac{3\pi}{2} into x(t) and y(t):

    x\left(\dfrac{\pi}{2}\right)=0, y\left(\dfrac{\pi}{2}\right)=5, yielding the point (0,5), and

    x\left(\dfrac{3\pi}{2}\right)=0, y\left(\dfrac{3\pi}{2}\right)=-5, yielding the point (0,-5).

    Solving \dfrac{dx}{dt}=-5\sin(t)=0, we find t=0, \pi, 2\pi within [0,2\pi]. Since \dfrac{dy}{dt} is non-zero at all these values of t, each of them corresponds to a point on the curve, where the tangent line is vertical. Substituting t=0, \pi, 2\pi into x(t) and y(t), we find the coordinates of these points to be (5,0), (-5,0) and (5,0) respectively.

    The above computations agree with the sketch of the parametric curve, which is a circle of radius 5 with the center at the origin.


    A circle with radius 5 centered at the origin is graphed with arrow going counterclockwise. The point (5, 0) is marked t = 0, the point (0, 5) is marked t = π/2, the point (−5, 0) is marked t = π, and the point (0, −5) is marked t = 3π/2. On the graph there are also written three equations: x(t) = 5 cos(t), y(t) = 5 sin(t), and 0 ≤ t ≤ 2π.
    Figure 4. Graph of the curve described by parametric equations in part c.

Calculate the derivative \ds \frac{dy}{dx} for the curve defined by the parametric equations \ds x\left(t\right)={t}^{2}-4t, y\left(t\right)=2{t}^{3}-6t, -2\le t\le 3 and find all points on the curve, where the tangent line is horizontal or where the tangent line is vertical.

Answer

\ds \frac{dy}{\,dx }=\frac{6{t}^{2}-6}{2t-4}=\frac{3{t}^{2}-3}{t-2}.

The tangent line line is horizontal at (-3,4) and (5,4), corresponding to t=1 and t=-1 respectively. The tangent line is vertical at (-4,4), corresponding to t=2.

A curve going from (12, −4) through the origin and (−4, 0) to (−3, 36) with arrows in that order. The point (12, −4) is marked t = −2 and the point (−3, 36) is marked t = 3. On the graph there are also written three equations: x(t) = t2 – 4t, y(t) = 2t3 – 6t, and −2 ≤ t ≤ 3.

Slope of the Tangent Line in a Special Case

Determine the slope of the tangent line to the hypocycloid

\ds x\left(t\right)=3\cos(t)+\cos(3t),\phantom{\rule{1em}{0ex}}y\left(t\right)=3\sin(t)-\sin(3t)

at the point corresponding to {\rule{0.2em}{0ex}}t=0.

Solution

We first calculate \ds {x}^{\prime }\left(t\right) and \ds {y}^{\prime }\left(t\right)\text{:}

\ds \begin{array}{c}\ds{x}^{\prime }\left(t\right)=-3\sin(t)-3\sin(3t),\ {y}^{\prime }\left(t\right)=3\cos(t)-3\cos(3t).\hfill \end{array}

We see that x'(0)=0, and so (*) cannot be applied to find \dfrac{dy}{dx} when t=0. However, x'(t)\ne0 when t\in\left[-\dfrac{\pi}6,\dfrac{\pi}6\right]\setminus\{0\}, \left(x'(t)>0 \text{ when } t\in\left[-\dfrac{\pi}6,0\right)\text{ and }x'(t)<0 \text{ when } t\in\left(0,\dfrac{\pi}6\right]\right), and so we can consider \lim\limits_{t\to0} \dfrac{dy}{dx}:

\ds \begin{array}{c}\ds\lim\limits_{t\to0} \dfrac{dy}{dx}=\lim\limits_{t\to0} \dfrac{y'(t)}{x'(t)}=\lim\limits_{t\to0} \dfrac{3\cos(t)-3\cos(3t)}{-3\sin(t)-3\sin(3t)}.\hfill \end{array}

Since \ds \lim\limits_{t\to0} \big(3\cos(t)-3\cos(3t)\big)=0=\lim\limits_{t\to0} \big(-3\sin(t)-3\sin(3t)\big), we deal with a \dfrac{0}{0} indeterminate form and can apply L’Hospital’s rule.

\ds \begin{array}{ccc}\hfill \lim\limits_{t\to0} \dfrac{dy}{dx}&\ds =\hfill &\ds \lim\limits_{t\to0} \dfrac{3\cos(t)-3\cos(3t)}{-3\sin(t)-3\sin(3t)}\hfill \\[5mm]\ds \hfill &\ds =\hfill &\ds \lim\limits_{t\to0} \dfrac{-3\sin(t)+9\sin(3t)}{-3\cos(t)-9\cos(3t)}\hfill \\[5mm]\ds \hfill &\ds =\hfill &\ds \dfrac{-0+0}{-3-9}=\frac{0}{-12}=0.\hfill \end{array}

Therefore, when t=0, the slope of the tangent line is zero, and hence the tangent line to the hypocycloid is horizontal at the point (4,0), corresponding to t=0, where the curve has a cusp.

Two circles are drawn both with center at the origin and with radii 3 and 4, respectively; the circle with radius 3 has an arrow pointing in the counterclockwise direction. There is a third circle drawn with center on the circle with radius 3 and touching the circle with radius 4 at one point. That is, this third circle has radius 1. A point is drawn on this third circle, and if it were to roll along the other two circles, it would draw out a four pointed star with points at (4, 0), (0, 4), (−4, 0), and (0, −4). On the graph there are also written two equations: x(t) = 3 cos(t) + cos(3t) and y(t) = 3 sin(t) – sin(3t).
Figure 5. Graph of the hypocycloid described by the given parametric equations.

Finding a Tangent Line

Find the equation of the tangent line to the parametric curve defined by the equations

\ds x\left(t\right)={t}^{2}-3,\phantom{\rule{1em}{0ex}}y\left(t\right)=2t-1,\phantom{\rule{1em}{0ex}}-3\le t\le 4\phantom{\rule{0.2em}{0ex}}

at the point corresponding to {\rule{0.2em}{0ex}}t=2.

Solution

We first calculate \ds {x}^{\prime }\left(t\right) and \ds {y}^{\prime }\left(t\right)\text{:}

\ds \begin{array}{c}\ds{x}^{\prime }\left(t\right)=2t,\ {y}^{\prime }\left(t\right)=2.\hfill \end{array}

Next we substitute these into (*):

\ds \begin{array}{c}\ds\frac{dy}{\,dx }=\frac{y'(t)}{x'(t)}=\frac{2}{2t}=\frac{1}{t}.\hfill \end{array}

When \ds t=2, \ds \frac{dy}{\,dx }=\frac{1}{2}, so this is the slope of the tangent line. Calculating \ds x\left(2\right) and \ds y\left(2\right) gives \ds x\left(2\right)={\left(2\right)}^{2}-3=1 and y\left(2\right)=2\left(2\right)-1=3, which corresponds to the point \ds \left(1,3\right) on the curve, see Figure 5 below. We now use the point-slope form of the equation of a line to find the equation of the tangent line at this point:

\ds \begin{array}{ccc}\hfill y-{y}_{0}&\ds =\hfill &\ds m\left(x-{x}_{0}\right)\hfill \\[5mm]\ds \hfill y-3&\ds =\hfill &\ds \frac{1}{2}\left(x-1\right)\hfill \\[5mm]\ds \hfill y-3&\ds =\hfill &\ds \frac{1}{2}x-\frac{1}{2}\hfill \\[5mm]\ds \hfill y&\ds =\hfill &\ds \frac{1}{2}x+\frac{5}{2}.\hfill \end{array}
A curved line going from (6, −7) through (−3, −1) to (13, 7) with arrow pointing in that order. The point (6, −7) is marked t = −3, the point (−3, −1) is marked t = 0, and the point (13, 7) is marked t = 4. On the graph there are also written three equations: x(t) = t2 − 3, y(t) = 2t − 1, and −3 ≤ t ≤ 4. At the point (1, 3), which is marked t = 2, there is a tangent line with equation y = x/2 + 5/2.
Figure 6. Tangent line to the parabola described by the given parametric equations when \ds t=2.

Find the equation of the tangent line to the curve defined by the equations

\ds x\left(t\right)={t}^{2}-4t,\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{3}-6t,\phantom{\rule{1em}{0ex}}-2\le t\le 3\phantom{\rule{0.2em}{0ex}}

at the pont corresponding to t=5.

Answer

The equation of the tangent line is \ds y=24x+100.

Second-Order Derivatives

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function \ds y=f\left(x\right) is defined to be the derivative of the first derivative; that is,

\ds \frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{\,dx }\left[\frac{dy}{\,dx }\right].

Since \ds \frac{dy}{dx }=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}, we can replace the \ds y on both sides of this equation with \ds \frac{dy}{\,dx }. This gives us

\ds (**)\quad \frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx }\left(\frac{dy}{\,dx }\right)=\frac{\frac d{dt}\left(\frac{dy}{dx} \right)}{\frac{dx}{dt}}.

If we know \ds \frac{dy}{dx} as a function of t, then this formula is straightforward to apply.

Finding a Second Derivative

Calculate the second derivative \ds \frac{{d}^{2}y}{d{x}^{2}} for the plane curve defined by the parametric equations \ds x\left(t\right)={t}^{2}-3,y\left(t\right)=2t-1.

Solution

Using (*), we find that \ds \frac{dy}{\,dx }=\frac{2}{2t}=\frac{1}{t}. Applying (**), we obtain

\ds \frac{{d}^{2}y}{d{x}^{2}}=\frac{\frac d{dt}\left(\frac{dy}{dx} \right)}{\frac{dx}{dt}}=\frac{\frac d{dt}\left(\frac1t\right)}{2t}=\frac{ - {t}^{-2}}{2t}=-\frac{1}{2{t}^{3}}.

Calculate the second derivative \ds \frac{{d}^{2}y}{d{x}^{2}} for the plane curve defined by the equations

\ds x\left(t\right)={t}^{3}+2t,\phantom{\rule{1em}{0ex}}y\left(t\right)=1-t+t^2.

Answer

\ds \frac{{d}^{2}y}{d{x}^{2}}=\frac{4+6t-6t^2}{{\left(3t^2+2\right)}^{3}}.

From before, we know that the second derivative is “responsible” for concavity of the curve with an explicit equation y=f(x): the curve is concave upward where f''(x)=\dfrac{d^2y}{dx^2}>0, and it is concave downward where f''(x)=\dfrac{d^2y}{dx^2}<0. Since, locally, a parametric curve usually admits eliminating the parameter and obtaining an explicit equation, we can still look at the sign of \dfrac{d^2y}{dx^2}, to determine where the curve is concave upward and where it is concave downward. Because, in practice, we won't be finding an explicit equation of the curve, but we will be using (**) to find \dfrac{d^2y}{dx^2} as a function of t, it is the intervals in terms of t that we will be referring to when discussing concavity of parametric curves.

Examining Concavity of a Parametric Curve

Determine where the parametric curve \ds x\left(t\right)=4t-{t}^{2}, y\left(t\right)=t^3+2 is concave upward and where it is concave downward.

Solution

Applying (*), we find that \ds \frac{dy}{dx }=\frac{3t^2}{4-2t}. Using (**) together with the quotient rule, we obtain

\ds \frac{{d}^{2}y}{d{x}^{2}}=\frac{\frac d{dt}\left(\frac{dy}{dx} \right)}{\frac{dx}{dt}}=\frac{\left(\frac{3t^2}{4-2t}\right)'}{4-2t}=\frac{\frac{6t(4-2t)-3t^2(-2)}{(4-2t)^2}}{4-2t}=\frac{24t-6t^2}{(4-2t)^3}.

We now need to determine for which values of t, \ds \frac{{d}^{2}y}{d{x}^{2}} is positive, and for which values of t it is negative. Factoring the numerator and denominator, we rewrite \dfrac{d^2y}{dx^2}:

\ds\frac{{d}^{2}y}{d{x}^{2}}=\frac{24t-6t^2}{(4-2t)^3}=\dfrac{6t(4-t)}{\big(2(2-t)\big)^3}=\dfrac{6t(4-t)}{2^3(2-t)^3}=\dfrac{3t(4-t)}{4(2-t)^3}.

The numerator has zeros t=0 and t=4, while the denominator has a zero t=2 of multiplicity 3. Using sample points or any other appropriate method, we find that \ds \frac{{d}^{2}y}{d{x}^{2}}>0, and hence the parametric curve is concave upward, when t\in(0,2) and t\in(4,\infty), and \ds \frac{{d}^{2}y}{d{x}^{2}}<0, implying that the curve is concave downward, when t\in(-\infty,0) and t\in(2,4).

Determine where the parametric curve \ds x\left(t\right)=t^2+1, y\left(t\right)=t^2+t is concave upward.

Answer

The curve is concave upward when t\in(-\infty,0).

Integrals Involving Parametric Equations

Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations \ds x\left(t\right)=t-\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}y\left(t\right)=1-\text{cos}\phantom{\rule{0.2em}{0ex}}(t). Suppose we want to find the area of the shaded region in the following graph.

A series of half circles drawn above the x axis with x intercepts being multiples of 2π. The half circle between 0 and 2π is highlighted. On the graph there are also written two equations: x(t) = t – sin(t) and y(t) = 1 – cos(t).
Figure 7. Graph of a cycloid with the arch over \ds \left[0,2\pi \right] highlighted.

To derive a formula for the area under the curve defined by the functions

\ds x=x\left(t\right),\phantom{\rule{1em}{0ex}}y=y\left(t\right)\ge0,\phantom{\rule{1em}{0ex}}a\le t\le b,

we assume that \ds x\left(t\right) is increasing and differentiable and start with an equal partition of the interval \ds a\le t\le b. Suppose \ds {t}_{0}=a<{t}_{1}<{t}_{2}<\cdots<{t}_{n}=b and consider the following graph.

A curved line is drawn in the first quadrant. Below it are a series of rectangles marked that begin at the x axis and reach up to the curved line; the rectangle’s height is determined by the location of the curved line at the leftmost point of the rectangle. These lines are noted as x(t0), x(t1), …, x(tn).
Figure 8. Approximating the area under a parametrically defined curve.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is \ds y\left(x\left(t^*_{i}\right)\right)=y\left(t^*_{i}\right) for some value \ds t^*_{i} in the ith subinterval, and the width can be calculated as \ds x\left({t}_{i}\right)-x\left({t}_{i-1}\right). It follows that the area of the ith rectangle is given by

\ds {A}_{i}=y\left(t^*_{i}\right)\phantom{\rule{0.2em}{0ex}}\big(x\left({t}_{i}\right)-x\left({t}_{i-1}\right)\big).

Then a Riemann sum for the area is

\ds {A}_{n}=\sum _{i=1}^{n}y\left(t^*_{i}\right)\phantom{\rule{0.2em}{0ex}}\big(x\left({t}_{i}\right)-x\left({t}_{i-1}\right)\big).

Multiplying and dividing each area by \ds {t}_{i}-{t}_{i-1} gives

\ds {A}_{n}=\sum _{i=1}^{n}y\left(t^*_{i}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{x\left({t}_{i}\right)-x\left({t}_{i-1}\right)}{{t}_{i}-{t}_{i-1}}\right)\left({t}_{i}-{t}_{i-1}\right)=\sum _{i=1}^{n}y\left(t^*_{i}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{x\left({t}_{i}\right)-x\left({t}_{i-1}\right)}{\Delta t}\right)\Delta t.

Taking the limit as \ds n approaches infinity, we obtain

\ds A=\underset{n\to \infty }{\text{lim}}{A}_{n}=\int\limits_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.

Note that if x(t) is decreasing, that is, the curve is traced from left to right, everything in the above derivation stays the same except that the width of a typical rectangle becomes x(t_{i-1})-x(t_i)=-\big(x(t_i)-x(t_{i-1})\big), which results in the formula

\ds A=\underset{n\to \infty }{\text{lim}}{A}_{n}=\int\limits_{a}^{b}y\left(t\right)\big(-{x}^{\prime }\left(t\right)\big)\phantom{\rule{0.2em}{0ex}}dt=-\int\limits_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=\int\limits_{b}^{a}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.

This leads to the following theorem.

Area under a Parametric Curve

Consider the plane curve defined by the parametric equations

\ds x=x\left(t\right),\phantom{\rule{1em}{0ex}}y=y\left(t\right)\ge0,\phantom{\rule{1em}{0ex}}a\le t\le b

and assume that \ds x\left(t\right) is differentiable.

  • If x(t) is increasing, then the area under this curve is given by

    \ds A=\int\limits_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.

  • If x(t) is decreasing, then the area under this curve is given by

    \ds A=\int\limits_{b}^{a}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.

Finding the Area under a Parametric Curve

Find the area under one arc of the cycloid defined by the equations

\ds x\left(t\right)=t-\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}y\left(t\right)=1-\text{cos}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}0\le t\le 2\pi .

Solution

To determine whether x(t) is increasing or decreasing we look at the sign of x'(t). We have that \ds x'(t)=1-\cos(t)\ge0, and hence x(t) is increasing. Applying the above theorem, we have

\ds \begin{array}{cc}\ds \hfill A&\ds =\int\limits_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{2\pi }\left(1-\text{cos}\phantom{\rule{0.2em}{0ex}}(t)\right)\left(1-\text{cos}\phantom{\rule{0.2em}{0ex}}(t)\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{2\pi }\left(1-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)+{\text{cos}}^{2}(t)\right)dt\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{2\pi }\left(1-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)+\frac{1+\text{cos}\phantom{\rule{0.2em}{0ex}}(2t)}{2}\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{2\pi }\left(\frac{3}{2}-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)+\frac{\text{cos}\phantom{\rule{0.2em}{0ex}}(2t)}{2}\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\[5mm]\ds &\ds =\left({\frac{3t}{2}-2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)+\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}(2t)}{4}}\right)\Big|_{0}^{2\pi }\hfill \\[5mm]\ds &\ds =3\pi .\hfill \end{array}

Find the area under the upper half of the hypocycloid defined by the equations

\ds x\left(t\right)=3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)+\text{cos}\phantom{\rule{0.2em}{0ex}}3t,\phantom{\rule{1em}{0ex}}y\left(t\right)=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)-\text{sin}\phantom{\rule{0.2em}{0ex}}3t,\phantom{\rule{1em}{0ex}}0\le t\le \pi .

Answer

\ds A=3\pi

Hint

Use the above theorem, along with the identities \ds \text{sin}\phantom{\rule{0.2em}{0ex}}(\alpha) \phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\beta) =\frac{1}{2}\left[\text{cos}\left(\alpha -\beta \right)-\text{cos}\left(\alpha +\beta \right)\right] and \ds {\text{sin}}^{2}(t)=\frac{1-\text{cos}\phantom{\rule{0.2em}{0ex}}(2t)}{2}. Note that x(t) is decreasing.

Arc Length of a Parametric Curve

The same way we did for a regular curve with explicit equation y=f(x) or x=g(y), to derive a formula for the arc length of a parametric curve, we approximate it by a union of line segments as shown in the following figure.

A curved line in the first quadrant with points marked for x = 1, 2, 3, 4, and 5. These points have values roughly 2.1, 2.7, 3, 2.7, and 2.1, respectively. The points for x = 1 and 5 are marked A and B, respectively.
Figure 9. Approximation of a curve by a union of line segments.

Given a plane curve defined by the parametric equations \ds x=x\left(t\right),y=y\left(t\right),a\le t\le b, we start by partitioning the interval \ds \left[a,b\right] into n equal subintervals: \ds {t}_{0}=a<{t}_{1}<{t}_{2}<\cdots<{t}_{n}=b. The width of each subinterval is \ds \Delta t=\frac{b-a}n. The length of the ith line segment can be found as follows:

\ds \begin{array}{ll}\ds {d}_{i}=\sqrt{{\left(x\left({t}_{i}\right)-x\left({t}_{i-1}\right)\right)}^{2}+{\left(y\left({t}_{i}\right)-y\left({t}_{i-1}\right)\right)}^{2}}.\hfill \end{array}

Adding those from 1\le i\le n, we obatin an approximation of the arc length s of the parametric curve:

\ds (1)\quad s\approx \sum _{i=1}^{n}{d}_{i}=\sum _{i=1}^{n}\sqrt{{\left(x\left({t}_{i}\right)-x\left({t}_{i-1}\right)\right)}^{2}+{\left(y\left({t}_{i}\right)-y\left({t}_{i-1}\right)\right)}^{2}}.

If we assume that \ds x\left(t\right) and \ds y\left(t\right) are differentiable functions of t, then the Mean Value Theorem applies, so in each subinterval \ds \left[{t}_{k-1},{t}_{k}\right] there exist \ds t^*_{k} and \ds t^{**}_{k} such that

\ds \begin{array}{ll}\ds \ds \\[5mm]\ds x\left({t}_{i}\right)-x\left({t}_{i-1}\right)={x}^{\prime }\left(t^*_{i}\right)\left({t}_{i}-{t}_{i-1}\right)={x}^{\prime }\left(t^*_{i}\right)\Delta t\hfill \\[5mm]\ds y\left({t}_{i}\right)-y\left({t}_{i-1}\right)={y}^{\prime }\left(t^{**}_{i}\right)\left({t}_{i}-{t}_{i-1}\right)={y}^{\prime }\left(t^{**}_{i}\right)\Delta t.\hfill \end{array}

With this, (1) becomes

\ds \begin{array}{cc}\ds \hfill s&\ds \approx \sum _{i=1}^{n}{d}_{i}\hfill \\[5mm]\ds &\ds =\sum _{i=1}^{n}\sqrt{{\left({x}^{\prime }\left(t^*_{i}\right)\Delta t\right)}^{2}+{\left({y}^{\prime }\left(t^{**}_{i}\right)\Delta t\right)}^{2}}\hfill \\[5mm]\ds &\ds =\sum _{i=1}^{n}\sqrt{{\left({x}^{\prime }\left(t^*_{i}\right)\right)}^{2}{\left(\Delta t\right)}^{2}+{\left({y}^{\prime }\left(t^{**}_{i}\right)\right)}^{2}{\left(\Delta t\right)}^{2}}\hfill \\[5mm]\ds &\ds =\left(\sum _{i=1}^{n}\sqrt{{\left({x}^{\prime }\left(t^*_{i}\right)\right)}^{2}+{\left({y}^{\prime }\left(t^{**}_{i}\right)\right)}^{2}}\right)\Delta t.\hfill \end{array}

This is a Riemann sum that approximates the arc length over a partition of the interval \ds \left[a,b\right]. If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives

\ds \begin{array}{cc}\ds \hfill s&\ds =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}{d}_{i}\hfill \\[5mm]\ds &\ds =\underset{n\to \infty }{\text{lim}}\left(\sum _{i=1}^{n}\sqrt{{\left({x}^{\prime }\left(t^*_{i}\right)\right)}^{2}+{\left({y}^{\prime }\left(t^{**}_{i}\right)\right)}^{2}}\right)\Delta t\hfill \\[5mm]\ds &\ds =\int\limits_{a}^{b}\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt.\hfill \end{array}

When taking the limit, the values of \ds t^*_{k} and \ds t^{**}_{k} are both contained within the same ever-shrinking interval of width \ds \Delta t, so they must converge to the same value.

We can summarize this method in the following theorem.

Arc Length of a Parametric Curve

Consider the plane curve defined by the parametric equations

\ds x=x\left(t\right),\phantom{\rule{1em}{0ex}}y=y\left(t\right),\phantom{\rule{1em}{0ex}}{t}_{1}\le t\le {t}_{2}

and assume that \ds x\left(t\right) and \ds y\left(t\right) are smooth, that is, their derivatives \ds \frac{dx}{dt} and \ds \frac{dy}{dt} are continuous. Then the arc length of this curve is given by

\ds s=\int\limits_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{\,dx }{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt.

Now suppose that the parameter can be eliminated, leading to a function \ds y=F\left(x\right). We are going to show that the above formula agrees with the formula for the arc length of a regular curve derived in Section 2.4. We have \ds y\left(t\right)=F\left(x\left(t\right)\right) and the Chain Rule gives \ds {y}^{\prime }\left(t\right)={F}^{\prime }\left(x\left(t\right)\right){x}^{\prime }\left(t\right). Substituting this into the above formula gives

\ds \begin{array}{cc}\ds \hfill s&\ds =\int\limits_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{\,dx }{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt\hfill \\[5mm]\ds &\ds =\int\limits_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{\,dx }{dt}\right)}^{2}+{\left({F}^{\prime }\left(x\right)\frac{\,dx }{dt}\right)}^{2}}dt\hfill \\[5mm]\ds &\ds =\int\limits_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{\,dx }{dt}\right)}^{2}\left(1+{\left({F}^{\prime }\left(x\right)\right)}^{2}\right)}dt\hfill \\[5mm]\ds &\ds =\int\limits_{{t}_{1}}^{{t}_{2}}{x}^{\prime }\left(t\right)\sqrt{1+{\left(\frac{dy}{\,dx }\right)}^{2}}dt.\hfill \end{array}

Here we have assumed that \ds {x}^{\prime }\left(t\right)\symbol{"3E}0, and the case when x'(t)<0 is analogous (the extra minus is going to disappear when the limits of integration are interchanged). Using a substitution x=x(t), we have that \ds \,dx ={x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt, and letting \ds a=x\left({t}_{1}\right) and \ds b=x\left({t}_{2}\right) we obtain the formula

\ds s=\int\limits_{a}^{b}\sqrt{1+{\left(\frac{dy}{\,dx }\right)}^{2}}\,dx ,

which is exactly the one we had before.

Finding the Arc Length of a Parametric Curve

Find the arc length of the semicircle defined by the equations

\ds x\left(t\right)=3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}y\left(t\right)=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}0\le t\le \pi .

Solution

The parametric curve is shown in Figure 9 below. To determine its length, we use the formula:

\ds \begin{array}{cc}\ds \hfill s&\ds =\int\limits_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{\,dx }{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{\pi }\sqrt{{\left(-3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)\right)}^{2}+{\left(3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)\right)}^{2}}dt\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{\pi }\sqrt{9\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}(t)+9\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}(t)}\phantom{\rule{0.2em}{0ex}}dt\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{\pi }\sqrt{9\left({\text{sin}}^{2}(t)+{\text{cos}}^{2}(t)\right)}dt\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{\pi }3dt={3t}\Big|_{0}^{\pi }=3\pi .\hfill \end{array}

Note that the formula for the arc length of a semicircle is \ds \pi r and the radius of this circle is 3. This is a great example of using calculus to derive a known geometric formula.

A semicircle is drawn with radius 3. There is an arrow pointing counterclockwise. On the graph there are also written three equations: x(t) = 3 cos(t), y(t) = 3 sin(t), and 0 ≤ t ≤ π.
Figure 10. The arc length of the semicircle is equal to its radius times \pi

Find the arc length of the curve defined by the equations

\ds x\left(t\right)=3{t}^{2},\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{3},\phantom{\rule{1em}{0ex}}1\le t\le 3.

Answer

\ds s=2\left({10}^{3\text{/}2}-{2}^{3\text{/}2}\right)

We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels in a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive x-axis, the parametric equations for this curve can be written as

\ds x\left(t\right)=140t,\phantom{\rule{1em}{0ex}}y\left(t\right)=-16{t}^{2}+2t

where t represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions \ds x\left(t\right) and \ds y\left(t\right) using v as an independent variable, so as to eliminate any confusion with the parameter t:

\ds x\left(v\right)=140v,\phantom{\rule{1em}{0ex}}y\left(v\right)=-16{v}^{2}+2v.

Then we write the arc length formula as follows:

\ds \begin{array}{cc}\ds \hfill s\left(t\right)&\ds =\int\limits_{0}^{t}\sqrt{{\left(\frac{\,dx }{dv}\right)}^{2}+{\left(\frac{dy}{dv}\right)}^{2}}dv\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{t}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv.\hfill \end{array}

The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. To integrate this expression, one needs to make a trigonometric substitution -32v+2=140\tan(\theta), which will lead to a constant multiple of an integral of \sec^3(\theta). After some technical computations, this will result in

\ds \begin{array}{cc}\ds \hfill s\left(t\right)&\ds =-\frac{1}{32}\left[\frac{\left(-32t+2\right)}{2}\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}+\frac{{140}^{2}}{2}\text{ln}|\left(-32t+2\right)+\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}|\right]\hfill \\[5mm]\ds &\ds \phantom{\rule{0.6em}{0ex}}+\frac{1}{32}\left[\sqrt{{140}^{2}+{2}^{2}}+\frac{{140}^{2}}{2}\text{ln}|2+\sqrt{{140}^{2}+{2}^{2}}|\right]\hfill \\[5mm]\ds &\ds =\left(\frac{t}{2}-\frac{1}{32}\right)\sqrt{1024{t}^{2}-128t+19604}-\frac{1225}{4}\text{ln}|\left(-32t+2\right)+\sqrt{1024{t}^{2}-128t+19604}|\hfill \\[5mm]\ds &\ds \phantom{\rule{0.6em}{0ex}}+\frac{\sqrt{19604}}{32}+\frac{1225}{4}\text{ln}\left(2+\sqrt{19604}\right).\hfill \end{array}

This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to t. While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:

\ds \frac{d}{\,dx }\left(\int\limits_{a}^{x}f\left(u\right)\phantom{\rule{0.2em}{0ex}}du\right)=f\left(x\right). Therefore,

\ds \begin{array}{cc}\ds \hfill {s}^{\prime }\left(t\right)&\ds =\frac{d}{dt}\left[s\left(t\right)\right]\hfill \\[5mm]\ds &\ds =\frac{d}{dt}\left[\int\limits_{0}^{t}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv\right]\hfill \\[5mm]\ds &\ds =\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}\hfill \\[5mm]\ds &\ds =\sqrt{1024{t}^{2}-128t+19604}\hfill \\[5mm]\ds &\ds =2\sqrt{256{t}^{2}-32t+4901}.\hfill \end{array}

One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to

\ds \begin{array}{cc}\ds \hfill s\left(\frac{1}{3}\right)&\ds =\left(\frac{1\text{/}3}{2}-\frac{1}{32}\right)\sqrt{1024{\left(\frac{1}{3}\right)}^{2}-128\left(\frac{1}{3}\right)+19604}\hfill \\[5mm]\ds &\ds \phantom{\rule{0.6em}{0ex}}-\frac{1225}{4}\text{ln}|\left(-32\left(\frac{1}{3}\right)+2\right)+\sqrt{1024{\left(\frac{1}{3}\right)}^{2}-128\left(\frac{1}{3}\right)+19604}|\hfill \\[5mm]\ds &\ds \phantom{\rule{0.6em}{0ex}}+\frac{\sqrt{19604}}{32}+\frac{1225}{4}\text{ln}\left(2+\sqrt{19604}\right)\hfill \\[5mm]\ds &\ds \approx 46.69\phantom{\rule{0.2em}{0ex}}\text{feet}.\hfill \end{array}

This value is just over three quarters of the way to home plate. The speed of the ball is

\ds {s}^{\prime }\left(\frac{1}{3}\right)=2\sqrt{256{\left(\frac{1}{3}\right)}^{2}-32\left(\frac{1}{3}\right)+4901}\approx 140.27\phantom{\rule{0.2em}{0ex}}\text{ft/s}.

This speed translates to approximately 95 mph—a major-league fastball.

Surface Area Generated by a Parametric Curve

Recall the problem of finding the surface area of a surface of revolution. In Section 2.4, we derived a formula for the surface area of a surface generated by revolving the curve \ds y=f\left(x\right)\ge0 from \ds x=a to \ds x=b, around the x-axis:

\ds S=2\pi \int\limits_{a}^{b}f\left(x\right)\sqrt{1+{\left({f}^{\prime }\left(x\right)\right)}^{2}}\,dx .

We now consider a surface of revolution generated by revolving a parametrically defined curve \ds x=x\left(t\right),\ y=y\left(t\right),\ a\le t\le b around the x-axis as shown in the following figure.

A curve is drawn in the first quadrant with endpoints marked t = a and t = b. On this curve, there is a point marked (x(t), y(t)). There is a circle with an arrow drawn around the x axis that seems to indicate a rotation about the x axis, and there is a shape that accompanies that curve that seems to be what you would obtain if you rotated the curve about the x axis.
Figure 11. A surface of revolution generated by a parametrically defined curve.

The formula for its surface area is

\ds S=2\pi \int\limits_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt

provided that \ds y\left(t\right) is non-negative on \ds \left[a,b\right].

Finding Surface Area

Find the surface area of a sphere of radius r centered at the origin.

Solution

We start by parametrizing the upper semicircle with center at the origin and radius r:

\ds x\left(t\right)=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}y\left(t\right)=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\phantom{\rule{1em}{0ex}}0\le t\le \pi .
A semicircle is drawn with radius r. On the graph there are also written three equations: x(t) = r cos(t), y(t) = r sin(t), and 0 ≤ t ≤ π.
Figure 11. A semicircle generated by parametric equations.

When this curve is revolved around the x-axis, it generates a sphere of radius r. To calculate the surface area of the sphere, we use the above formula:

\ds \begin{array}{cc}\ds \hfill S&\ds =2\pi \int\limits_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt\hfill \\[5mm]\ds &\ds =2\pi \int\limits_{0}^{\pi }r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)\sqrt{{\left( - r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)\right)}^{2}+{\left(r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)\right)}^{2}}dt\hfill \\[5mm]\ds &\ds =2\pi \int\limits_{0}^{\pi }r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)\sqrt{{r}^{2}{\text{sin}}^{2}(t)+{r}^{2}{\text{cos}}^{2}(t)}\phantom{\rule{0.2em}{0ex}}dt\hfill \\[5mm]\ds &\ds =2\pi \int\limits_{0}^{\pi }r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)\sqrt{{r}^{2}\left({\text{sin}}^{2}(t)+{\text{cos}}^{2}(t)\right)}dt\hfill \\[5mm]\ds &\ds =2\pi \int\limits_{0}^{\pi }{r}^{2}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\[5mm]\ds &\ds =2\pi {r}^{2}\left({ - \text{cos}\phantom{\rule{0.2em}{0ex}}(t)}\Big|_{0}^{\pi }\right)\hfill \\[5mm]\ds &\ds =2\pi {r}^{2}\left( - \text{cos}\phantom{\rule{0.2em}{0ex}}\pi +\text{cos}\phantom{\rule{0.2em}{0ex}}0\right)\hfill \\[5mm]\ds &\ds =4\pi {r}^{2}.\hfill \end{array}

This agrees with the geometric you might have seen before.

Find the area of the surface generated by revolving the plane curve defined by the equations

\ds x\left(t\right)={t}^{3},\phantom{\rule{1em}{0ex}}y\left(t\right)={t}^{2},\phantom{\rule{1em}{0ex}}0\le t\le 1

around the x-axis.

Answer

\ds A=\frac{\pi \left(494\sqrt{13}+128\right)}{1215}

Hint

When evaluating the integral, use a u-substitution.

Key Concepts

  • The derivative of the parametrically defined curve \ds x=x\left(t\right) and \ds y=y\left(t\right) can be calculated using the formula \ds \frac{dy}{\,dx }=\frac{{y}^{\prime }\left(t\right)}{{x}^{\prime }\left(t\right)}. Using the derivative, we can find the equation of a tangent line to a parametric curve.
  • If y(t)\ge0, the area under the parametric curve can be determined by using the formula \ds A=\pm\int\limits_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt, where the choice of sign depends on whether x(t) is increasing or decreasing over [a,b].
  • The arc length of a parametric curve can be calculated by using the formula \ds s=\int\limits_{a}^{b}\sqrt{{\left(\frac{\,dx }{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt.
  • The area of a surface obtained by revolving a parametric curve around the x-axis is given by \ds S=2\pi \int\limits_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt, provided y(t)\ge0 when t\in[a,b]. If the curve is revolved around the y-axis, then the formula is \ds S=2\pi \int\limits_{a}^{b}x\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt,
    provided x(t)\ge0 when t\in[a,b].

Key Equations

  • Derivative of parametric equations
    \ds \frac{dy}{dx }=\frac{dy\text{/}dt}{dx \text{/}dt}=\frac{{y}^{\prime }\left(t\right)}{{x}^{\prime }\left(t\right)}
  • Second-order derivative of parametric equations
    \ds \frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx }\left(\frac{dy}{dx }\right)=\frac{\frac d{dt}\left(\frac{dy}{dx} \right)}{dx \text{/}dt}
  • Area under a parametric curve
    \ds A=\pm\int\limits_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt, where the sign depends on the sign of x'(t)
  • Arc length of a parametric curve
    \ds s=\int\limits_{a}^{b}\sqrt{{\left(\frac{\,dx }{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}\,dt
  • Surface area generated by a parametric curve about a coordinate axis
    \ds S=2\pi \int\limits_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}\,dt\quad\phantom{aaaaa} (revolving about x-axis)
    \ds S=2\pi \int\limits_{a}^{b}x\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}\,dt\quad\phantom{aaaaa} (revolving about y-axis)

Exercises

For the following exercises, find \ds \frac{dy}{dx} as a function of the parameter t.

1. \ds \begin{array}{cc}\ds x=t-\text{cos}\phantom{\rule{0.2em}{0ex}}(t),\hfill &\ds y=\text{sin}\phantom{\rule{0.2em}{0ex}}(3t)\hfill \end{array}

Answer

\ds\frac{3\cos(3t)}{1+\sin(t)}

2. \ds \begin{array}{cc}\ds x=t\sqrt{t},\hfill &\ds y=2t+4\hfill \end{array}

3. \ds \begin{array}{cc}\ds x=\frac{3}{1-t},\hfill &\ds y=\pi+\tan^{-1}(t)\hfill \end{array}

Answer

\ds \frac{(1-t)^2}{3(t^2+1)}

4. \ds \begin{array}{cc}\ds x=e^{t^2},\hfill &\ds y=t^3+et\hfill \end{array}

For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.

5. \ds \begin{array}{cc}\ds x=-5t+7,\hfill &\ds y=3t-1\hfill \end{array}

Answer

\ds \frac{-3}{5}

6. \ds \begin{array}{cc}\ds x=4-3t,\hfill &\ds y=-2+6t\hfill \end{array}

7. \ds \begin{array}{cc}\ds x=8+2t,\hfill &\ds y=1\hfill \end{array}

Answer

0

For the following exercises, determine the slope of the tangent line at the point corresponding to the given value of the parameter.

8. \ds \begin{array}{cc}\ds x=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\hfill &\ds y=3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t);\phantom{\rule{1em}{0ex}}t=\frac{\pi }{4}\hfill \end{array}

9. \ds \begin{array}{cc}\ds x=\text{cos}\phantom{\rule{0.2em}{0ex}}(t),\hfill &\ds y=8\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t);\phantom{\rule{1em}{0ex}}t=\frac{\pi }{2}\hfill \end{array}

Answer

\ds \text{Slope}=0.

10. \ds \begin{array}{cc}\ds x=2t,\hfill &\ds y={t}^{3};\phantom{\rule{1em}{0ex}}t=-1\hfill \end{array}

11. \ds \begin{array}{cc}\ds x=t+\frac{1}{t},\hfill &\ds y=t-\frac{1}{t};\phantom{\rule{1em}{0ex}}t=1\hfill \end{array}

Answer

Slope is undefined.

12. \ds \begin{array}{cc}\ds x=\sqrt{t},\hfill &\ds y=2t;\phantom{\rule{1em}{0ex}}t=4\hfill \end{array}

For the following exercises, find all points on the parametric curve where the tangent line has the given slope.

13. \ds \begin{array}{cc}\ds x=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t),\hfill &\ds y=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t);\phantom{\rule{1em}{0ex}}\text{slope}=0.5\hfill \end{array}

Answer

\ds t=\text{arctan}\left(-2\right)+\pi k, where k is integer, corresponding to the points \ds\left(\frac{4}{\sqrt{5}},-\frac{8}{\sqrt{5}}\right) and \ds\left(-\frac{4}{\sqrt{5}},\frac{8}{\sqrt{5}}\right).

14. \ds \begin{array}{cc}\ds x=1-\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\hfill &\ds y=\sqrt3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t);\phantom{\rule{1em}{0ex}}\text{slope}=3\hfill \end{array}

15. \ds \begin{array}{cc}\ds x=\ln(t),\hfill &\ds y=2+t-t^2;\phantom{\rule{1em}{0ex}}\text{slope}=-1\hfill \end{array}

Answer

t=1 corresponding to the point (0,2) (note that \ds t=-\frac12 is not in the domain of x(t))

16. \ds \begin{array}{cc}\ds x=1+\sqrt{t},\hfill &\ds y=3-4t;\phantom{\rule{1em}{0ex}}\text{slope}=0\hfill \end{array}

For the following exercises, write an equation of the tangent line to the given parametric curve at the point that corresponds to the specified value of the parameter t.

17. \ds \begin{array}{cc}\ds x={e}^{\sqrt{t}},\hfill &\ds y=1-\text{ln}\phantom{\rule{0.2em}{0ex}}({t}^{2});\phantom{\rule{1em}{0ex}}t=1\hfill \end{array}

Answer

\ds y= - \left(\frac{4}{e}\right)x+5

18. \ds \begin{array}{cc}\ds x=t\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}(t),\hfill &\ds y={\text{sin}}^{2}(t);\phantom{\rule{1em}{0ex}}t=\frac{\pi }{4}\hfill \end{array}

19. \ds \begin{array}{cc}\ds x={e}^{t},\hfill &\ds y={\left(t-1\right)}^{2};\phantom{\rule{1em}{0ex}}t=0\hfill \end{array}

Answer

\ds y=3-2x

20. Consider the parametric curve \ds x=5t^2-1,\ y=t^3-12t+2. Find all values of the parameter t that correpsond to the points on the curve where the tangent line is horizontal.

21. Consider the parametric curve \ds x=\text{sin}\left(2t\right),\ y=2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t), \ds 0\le t<2\pi. Find all values of the parameter t that correpsond to the points on the curve where the tangent line is vertical.

Answer

\ds t=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}.

For the following exercises, find all points on the given parametric curve where the tangent line is horizontal or vertical.

22. \ds x=\ln(t^2+1),\ y=12t+3t^2-2t^3

23. \ds x=\text{sec}\phantom{\rule{0.2em}{0ex}}(\theta) ,\ y=\text{tan}\phantom{\rule{0.2em}{0ex}}(\theta)

Answer

No horizontal tangents. Vertical tangents at \ds \left(1,0\right),\left(-1,0\right).

24. \ds \begin{array}{cc}\ds x=\frac{t}{1+{t}^{2}},\hfill &\ds y=\frac{{t}^{2}}{1+{t}^{2}}\hfill \end{array}

25. \ds \begin{array}{cc}\ds x=t\left({t}^{2}-3\right),\hfill &\ds y=3\left({t}^{2}-3\right)\hfill \end{array}

Answer

Horizontal tangent at \ds \left(0,-9\right); vertical tangents at \ds \left(\pm2,-6\right).

For the following exercises, find \ds \frac{{d}^{2}y}{d{x}^{2}}.

26. \ds \begin{array}{cc}\ds x={t}^{4}-1,\hfill &\ds y=t-{t}^{2}\hfill \end{array}

27. \ds \begin{array}{cc}\ds x=\text{sin}\left(\pi t\right),\hfill &\ds y=\text{cos}\left(\pi t\right)\hfill \end{array}

Answer

\ds - {\text{sec}}^{3}\left(\pi t\right)

28. \ds \begin{array}{cc}\ds x={e}^{-t},\hfill &\ds y=t{e}^{2t}\hfill \end{array}

29. \ds \begin{array}{cc}\ds x=\frac1t+2,\hfill &\ds y=t-\ln(t)\hfill \end{array}

Answer

\ds 2t^3-t^2

For the following exercises, find \ds \frac{{d}^{2}y}{d{x}^{2}} at the specified value of the parameter.

30. \ds \begin{array}{ccc}x=\ds\frac{1}{2}{t}^{2}-t,\hfill &\ds y=\frac{1}{3}{t}^{3}+1;\hfill &\ds t=2\hfill \end{array}

31. \ds x=\sqrt{t}+\sqrt2,\phantom{\rule{1em}{0ex}}y=2t-1;\phantom{\rule{1em}{0ex}}t=1

Answer

4

For the following exercises, find t intervals on which the given parametric curve is concave up and t intervals on which it is concave down.

32. \ds x=3{t}^{2},\ y={t}^{3}-t

33. \ds x=2t+\text{ln}\phantom{\rule{0.2em}{0ex}}(t),\ y=2t-\text{ln}\phantom{\rule{0.2em}{0ex}}(t).

Answer

Concave up on \ds (0,\infty).

34. \ds x=1-\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(2t),\ y=\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(2t)

35. \ds x=2+\text{sec}\phantom{\rule{0.2em}{0ex}}(t),\ y=1+2\phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}(t),\ -\frac{\pi}2<t<\frac{\pi}2

Answer

Concave up on \ds\left(-\frac{\pi}2,0\right) and concave down on \ds\left(0,\frac{\pi}2\right).

36. Sketch and find the area under one arch of the cycloid \ds x=r\left(\theta -\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right),\ y=r\left(1-\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right). Here r\symbol{"3E}0 is a fixed real number and \theta is a parameter.

37. Find the area below the curve \ds x=t^2-1,\ y={e}^{t},\ 0\le t\le 1 and above the x-axis.

Answer

2

38. Find the area enclosed by the ellipse \ds x=a\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta),\ y=b\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta),\ 0\le \theta <2\pi .

39. Find the area of the region below the curve \ds x=2\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}(\theta),\ y=2\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}(\theta) \phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) and above the x-axis over the interval \ds \left[0,\frac{\pi }{2}\right].

Answer

\ds \frac{3\pi }{2}

For the following exercises, find the total area of the regions between the parametric curves and the x-axis. In exercises 41-43 a\symbol{"3E}0 is a fixed real number.

40. \ds x=2\phantom{\rule{0.2em}{0ex}}\text{cot}\phantom{\rule{0.2em}{0ex}}(\theta),\ y=2\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}(\theta),\ 0<\theta< \pi

41.* \ds x=2a\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)-a\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2t\right),\ y=2a\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t)-a\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2t\right),\ 0\le t<2\pi.

Answer

\ds 6\pi {a}^{2}

42. \ds x=a\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2t\right),\ y=b\phantom{\rule{0.2em}{0ex}}\text{sin}\left(t\right),\ 0\le t<2\pi (the “hourglass”)

43.[T] \ds x=2a\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(t)-a\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2t\right),\ y=b\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\ 0\le t<2\pi (the “teardrop”)

Answer

\ds 2\pi ab

For the following exercises, find the arc length of the given parametric curve.

44. \ds x=4t+3,\phantom{\rule{1em}{0ex}}y=3t-2,\phantom{\rule{1em}{0ex}}0\le t\le 2

45. \ds \begin{array}{ccc}\ds x=\frac{1}{3}{t}^{3},\hfill &\ds y=\frac{1}{2}{t}^{2},\hfill &\ds 0\le t\le 1\hfill \end{array}

Answer

\ds \frac{1}{3}\left(2\sqrt{2}-1\right)

46. \ds \begin{array}{ccc}x=\text{cos}\left(2t\right),\hfill &\ds y=\text{sin}\left(2t\right),\hfill &\ds 0\le t\le \frac{\pi }{2}\hfill \end{array}

47. \ds \begin{array}{ccc}x=3+3{t}^{2},\hfill &\ds y=3t-t^3,\hfill &\ds 0\le t\le 1\hfill \end{array}

Answer

\ds 4

48. \ds \begin{array}{ccc}x={e}^{t}\text{cos}\phantom{\rule{0.2em}{0ex}}(t),\hfill &\ds y={e}^{t}\text{sin}\phantom{\rule{0.2em}{0ex}}(t),\hfill &\ds 0\le t\le \frac{\pi }{2}\hfill \end{array}

49. \ds x=\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{3}(\theta),\ y=\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{3}(\theta),\ 0\le\theta<2\pi (the hypocycloid)

Answer

\ds 6

50. Find the length of one arch of the cycloid \ds x=4\left(t-\text{sin}\phantom{\rule{0.2em}{0ex}}(t)\right),\ y=4\left(1-\text{cos}\phantom{\rule{0.2em}{0ex}}(t)\right).

51. Find the distance traveled by a particle with position \ds \left(x,y\right) as t varies in the given time interval: \ds \begin{array}{ccc}x={\text{sin}}^{2}(t),\hfill &\ds y={\text{cos}}^{2}(t),\hfill &\ds 0\le t\le 3\pi \hfill \end{array}.

Answer

\ds 6\sqrt{2}

52. Find the length of the curve \ds x={e}^{t}-t,\ y=4{e}^{t\text{/}2},\ -8\le t\le 3.

For the following exercises, set up but do not evaluate the integral that represents the area of the surface obtained by rotating the given parametric curve about the x-axis.

53. \ds x=t^5+4t^2+3,\ y=2t+t^3,\ 0\le t\le 2

Answer

\ds 2\pi\int\limits_0^2 (2t+t^3)\sqrt{(5t^4+8t)^2+(2+3t^2)^2}\,dt

54. \ds x=t\sin(t),\ y=\cos(2t),\ t\in\left[-\frac{\pi}4,\frac{\pi}4\right].

55. \ds x=\sqrt{t^2+1},\ y=\frac{t+1}t,\ t\in\left[1,2\right].

Answer

\ds 2\pi \int\limits_1^2 \frac{t+1}t\sqrt{\frac{t^2}{t^2+1}+\frac1{t^4}}\,dt

56. \ds x=e^t\cos(t),\ y=e^{-t}+1,\ t\in\left[-1,0\right].

For the following exercises, find the area of the surface obtained by rotating the given parametric curve about the x-axis.

57. \ds \begin{array}{ccc}x={t}^{3},\hfill &\ds y={t}^{2},\hfill &\ds 0\le t\le 1\hfill \end{array}

Answer

\ds \frac{2\pi \left(247\sqrt{13}+64\right)}{1215}

58. \ds \begin{array}{ccc}x=a\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{3}(\theta) ,\hfill &\ds y=a\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{3}(\theta) ,\hfill &\ds 0\le \theta \le \hfill \end{array}\frac{\pi }{2}

For the following exercises, set up but do not evaluate the integral that represents the area of the surface obtained by rotating the given parametric curve about the y-axis.

59. \ds x=\ln(3t-1),\ y=t^2+t^4,\ 1\le t\le 10

Answer

\ds 2\pi\int\limits_1^{10} \ln(3t-1)\sqrt{\left(\frac3{3t-1}\right)^2+(2t+4t^3)^2}\,dt

60. \ds x=t^2\sqrt t+t,\ y=\frac1{\sqrt t}-2,\ 2\le t\le 3

61. Find the area of the surface generated by revolving \ds x=\sqrt 3 t^2,\ y=t^3-t,\ 0\le t\le 1 about the y-axis.

Answer

\ds \frac{28\pi}{5\sqrt3}

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Calculus: Volume 2 (Second University of Manitoba Edition) Copyright © 2021 by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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