11.3: Calculus of Parametric Curves (2024)

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

\[\begin{align} x(t) &=2t+3 \label{eq1} \\ y(t) &=3t−4 \label{eq2} \end{align} \]

within \(−2≤t≤3\).

The graph of this curve appears in Figure \(\PageIndex{1}\). It is a line segment starting at \((−1,−10)\) and ending at \((9,5).\)

We can eliminate the parameter by first solving Equation \ref{eq1} for \(t\):

\(x(t)=2t+3\)

\(x−3=2t\)

\(t=\dfrac{x−3}{2}\).

Substituting this into \(y(t)\) (Equation \ref{eq2}), we obtain

\(y(t)=3t−4\)

\(y=3\left(\dfrac{x−3}{2}\right)−4\)

\(y=\dfrac{3x}{2}−\dfrac{9}{2}−4\)

\(y=\dfrac{3x}{2}−\dfrac{17}{2}\).

The slope of this line is given by \(\dfrac{dy}{dx}=\dfrac{3}{2}\). Next we calculate \(x′(t)\) and \(y′(t)\). This gives \(x′(t)=2\) and \(y′(t)=3\). Notice that

\[\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3}{2}. \nonumber \]

This is no coincidence, as outlined in the following theorem.

Equation \ref{paraD} can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function \(y=f(x)\) is any point \(x=x_0\) such that either \(f′(x_0)=0\) or \(f′(x_0)\) does not exist. Equation \ref{paraD} gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function \(y=f(x)\) or not.

Example \(\PageIndex{1}\): Finding the Derivative of a Parametric Curve

Calculate the derivative \(\dfrac{dy}{dx}\) for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.

1. \(x(t)=t^2−3, \quad y(t)=2t−1, \quad\text{for }−3≤t≤4\)
2. \(x(t)=2t+1, \quad y(t)=t^3−3t+4, \quad\text{for }−2≤t≤2\)
3. \(x(t)=5\cos t, \quad y(t)=5\sin t, \quad\text{for }0≤t≤2π\)

Solution

a. To apply Equation \ref{paraD}, first calculate \(x′(t)\) and \(y′(t)\):

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\(x′(t)=2t\)

\(y′(t)=2\).

Next substitute these into the equation:

\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\)

\(\dfrac{dy}{dx}=\dfrac{2}{2t}\)

\(\dfrac{dy}{dx}=\dfrac{1}{t}\).

This derivative is undefined when \(t=0\). Calculating \(x(0)\) and \(y(0)\) gives \(x(0)=(0)^2−3=−3\) and \(y(0)=2(0)−1=−1\), which corresponds to the point \((−3,−1)\) on the graph. The graph of this curve is a parabola opening to the right, and the point \((−3,−1)\) is its vertex as shown.

b. To apply Equation \ref{paraD}, first calculate \(x′(t)\) and \(y′(t)\):

\(x′(t)=2\)

\(y′(t)=3t^2−3\).

Next substitute these into the equation:

\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\)

\(\dfrac{dy}{dx}=\dfrac{3t^2−3}{2}\).

This derivative is zero when \(t=±1\). When \(t=−1\) we have

\(x(−1)=2(−1)+1=−1\) and \(y(−1)=(−1)^3−3(−1)+4=−1+3+4=6\),

which corresponds to the point \((−1,6)\) on the graph. When \(t=1\) we have

\(x(1)=2(1)+1=3\) and \(y(1)=(1)^3−3(1)+4=1−3+4=2,\)

which corresponds to the point \((3,2)\) on the graph. The point \((3,2)\) is a relative minimum and the point \((−1,6)\) is a relative maximum, as seen in the following graph.

c. To apply Equation \ref{paraD}, first calculate \(x′(t)\) and \(y′(t)\):

\(x′(t)=−5\sin t\)

\(y′(t)=5\cos t.\)

Next substitute these into the equation:

\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\)

\(\dfrac{dy}{dx}=\dfrac{5\cos t}{−5\sin t}\)

\(\dfrac{dy}{dx}=−\cot t.\)

This derivative is zero when \(\cos t=0\) and is undefined when \(\sin t=0.\) This gives \(t=0,\dfrac{π}{2},π,\dfrac{3π}{2},\) and \(2π\) as critical points for t. Substituting each of these into \(x(t)\) and \(y(t)\), we obtain

\(t\)	\(x(t)\)	\(y(t)\)
0	5	0
\(\dfrac{π}{2}\)	0	5
\(π\)	−5	0
\(\dfrac{3π}{2}\)	0	−5
\(2π\)	5	0

These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure \(\PageIndex{4}\)). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.

Example \(\PageIndex{2}\): Finding a Tangent Line

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

\[x(t)=t^2−3, \quad y(t)=2t−1, \quad\text{for }−3≤t≤4 \nonumber \]

when \(t=2\).

Solution

First find the slope of the tangent line using Equation \ref{paraD}, which means calculating \(x′(t)\) and \(y′(t)\):

\(x′(t)=2t\)

\(y′(t)=2\).

Next substitute these into the equation:

\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\)

\(\dfrac{dy}{dx}=\dfrac{2}{2t}\)

\(\dfrac{dy}{dx}=\dfrac{1}{t}\).

When \(t=2, \dfrac{dy}{dx}=\dfrac{1}{2}\), so this is the slope of the tangent line. Calculating \(x(2)\) and \(y(2)\) gives

\(x(2)=(2)^2−3=1\) and \(y(2)=2(2)−1=3\),

which corresponds to the point \((1,3)\) on the graph (Figure \(\PageIndex{5}\)). Now use the point-slope form of the equation of a line to find the equation of the tangent line:

\(y−y_0=m(x−x_0)\)

\(y−3=\dfrac{1}{2}(x−1)\)

\(y−3=\dfrac{1}{2}x−\dfrac{1}{2}\)

\(y=\dfrac{1}{2}x+\dfrac{5}{2}\).

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 \(y=f(x)\) is defined to be the derivative of the first derivative; that is,

\[\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left[\dfrac{dy}{dx}\right]. \label{eqD2} \]

Since

\[\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}, \nonumber \]

we can replace the \(y\) on both sides of Equation \ref{eqD2} with \(\dfrac{dy}{dx}\). This gives us

\[\dfrac{d^2y}{dx^2}=\dfrac{d}{dx} \left(\dfrac{dy}{dx} \right)=\dfrac{(d/dt)(dy/dx)}{dx/dt}.\label{paraD2} \]

If we know \(dy/dx\) as a function of \(t\), then this formula is straightforward to apply

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 these parametric equations

\[ \begin{align*} x(t) &=t−\sin t \\[4pt] y(t) &=1−\cos t. \end{align*}\]

Suppose we want to find the area of the shaded region in the following graph.

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

\[\begin{align*} x &=x(t) \\[4pt] y &=y(t) \end{align*}\]

where \(a≤t≤b\).

We assume that \(x(t)\) is differentiable and start with an equal partition of the interval \(a≤t≤b\). Suppose \(t_0=a<t_1<t_2<⋯<t_n=b\) and consider the following graph.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is \(y(x(\bar{t_i}))\) for some value \(\bar{t_i}\) in the \(i^\text{th}\) subinterval, and the width can be calculated as \(x(t_i)−x(t_{i−1})\). Thus the area of the \(i^\text{th}\) rectangle is given by

\[A_i=y(x(\bar{t_i}))(x(t_i)−x(t_{i−1})). \nonumber \]

Then a Riemann sum for the area is

\[A_n=\sum_{i=1}^ny(x(\bar{t_i}))(x(t_i)−x(t_{i−1})).\nonumber \]

Multiplying and dividing each area by \(t_i−t_{i−1}\) gives

\[ \begin{align*} A_n &=\sum_{i=1}^ny(x(\bar{t_i})) \left(\dfrac{x(t_i)−x(t_{i−1})}{t_i−t_{i−1}}\right)(t_i−t_{i−1}) \\[4pt] &=\sum_{i=1}^ny(x(\bar{t_i})) \left(\dfrac{x(t_i)−x(t_{i−1})}{Δt}\right)Δt. \end{align*} \nonumber \]

Taking the limit as \(n\) approaches infinity gives

\[A=\lim_{n→∞}A_n=∫^b_ay(t)x′(t)\,dt. \nonumber \]

This leads to the following theorem.

Arc Length of a Parametric Curve

In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point \(A\) to point \(B\) along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

Given a plane curve defined by the functions \(x=x(t),\quad y=y(t),\quad \text{for }a≤t≤b\), we start by partitioning the interval \([a,b]\) into \(n\) equal subintervals: \(t_0=a<t_1<t_2<⋯<t_n=b\). The width of each subinterval is given by \(Δt=(b−a)/n\). We can calculate the length of each line segment:

\[d_1=\sqrt{(x(t_1)−x(t_0))^2+(y(t_1)−y(t_0))^2} \nonumber \]

\[d_2=\sqrt{(x(t_2)−x(t_1))^2+(y(t_2)−y(t_1))^2} \nonumber \]

etc.

Then add these up. We let \(s\) denote the exact arc length and \(s_n\) denote the approximation by \(n\) line segments:

\[s≈\sum_{k=1}^ns_k=\sum_{k=1}^n\sqrt{(x(t_k)−x(t_{k−1}))^2+(y(t_k)−y(t_{k−1}))^2}. \label{arc5} \]

If we assume that \(x(t)\) and \(y(t)\) are differentiable functions of \(t\), then the Mean Value Theorem applies, so in each subinterval \([t_{k−1},t_k]\) there exist \(\hat{t_k}\) and \(\tilde{t_k}\) such that

\[x(t_k)−x(t_{k−1})=x′(\hat{t_k})(t_k−t_{k−1})=x′(\hat{t_k})\,Δt \nonumber \]

\[y(t_k)−y(t_{k−1})=y′(\tilde{t_k})(t_k−t_{k−1})=y′(\tilde{t_k})\,Δt. \nonumber \]

Therefore Equation \ref{arc5} becomes

\[\begin {align*} s ≈\sum_{k=1}^ns_k &=\sum_{k=1}^n\sqrt{(x′(\hat{t_k})Δt)^2+(y′(\tilde{t_k})Δt)^2} \\[4pt]
&=\sum_{k=1}^n\sqrt{(x′(\hat{t_k}))^2(Δt)^2+(y′(\tilde{t_k}))^2(Δt)^2} \\[4pt]
&=\sum_{k=1}^n\sqrt{(x′(\hat{t_k}))^2+(y′(\tilde{t_k}))^2}Δt. \end{align*}\]

This is a Riemann sum that approximates the arc length over a partition of the interval \([a,b]\). 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

\[ \begin{align*} s &=\lim_{n→∞}\sum_{k=1}^ns_k \\[4pt]
&=\lim_{n→∞}\sum_{k=1}^n\sqrt{(x′(\hat{t_k}))^2+(y′(\tilde{t_k}))^2}Δt \\[4pt]
&=∫^b_a\sqrt{(x′(t))^2+(y′(t))^2}\,dt . \end{align*}\]

When taking the limit, the values of \(\hat{t_k}\) and \(\tilde{t_k}\) are both contained within the same ever-shrinking interval of width \(Δt\), so they must converge to the same value.

We can summarize this method in the following theorem.

At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function \(y=F(x)\). Then \(y(t)=F(x(t))\) and the Chain Rule gives

\[y′(t)=F′\big(x(t)\big)x′(t). \nonumber \]

Substituting this into Equation \ref{arcP} gives

\[\begin {align*} s &=∫^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(F′(x)\dfrac{dx}{dt}\right)^2}\,dt \\[4pt]
&=∫^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2(1+\left(F′(x)\right)^2)}\,dt \\[4pt]
&=∫^{t_2}_{t_1}x′(t)\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dt. \end{align*}\]

Here we have assumed that \(x′(t)>0\), which is a reasonable assumption. The Chain Rule gives \(dx=x′(t)\,dt,\) and letting \(a=x(t_1)\) and \(b=x(t_2)\) we obtain the formula

\[s=∫^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx, \nonumber \]

which is the formula for arc length obtained in the Introduction to the Applications of Integration.

Example \(\PageIndex{5}\): Finding the Arc Length of a Parametric Curve

Find the arc length of the semicircle defined by the equations

\[x(t)=3\cos t, \quad y(t)=3\sin t, \quad \text{for }0≤t≤π. \nonumber \]

Solution

The values \(t=0\) to \(t=π\) trace out the blue curve in Figure \(\PageIndex{8}\). To determine its length, use Equation \ref{arcP}:

\[\begin{align*} s &=\int^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\\[4pt]
&=∫^π_0\sqrt{(−3\sin t)^2+(3\cos t)^2}\,dt\\[4pt]
&=∫^π_0\sqrt{9\sin^2t+9\cos^2t}\,dt\\[4pt]
&=∫^π_0\sqrt{9(\sin^2t+\cos^2t)}\,dt\\[4pt]
&=∫^π_03\,dt=3t\Big|^π_0\\[4pt]
&=3π \text{ units}. \end{align*}\]

Note that the formula for the arc length of a semicircle is \(πr\) and the radius of this circle is \(3\). This is a great example of using calculus to derive a known formula of a geometric quantity.

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 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

\[x(t)=140t, \quad y(t)=−16t^2+2t \nonumber \]

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 \(x(t)\) and \(y(t)\) using v as an independent variable, so as to eliminate any confusion with the parameter \(t\):

\[x(v)=140v, \quad y(v)=−16v^2+2v. \nonumber \]

Then we write the arc length formula as follows:

\[\begin{align*} s(t) &=∫^t_0\sqrt{(\dfrac{dx}{dv})^2+(\dfrac{dy}{dv})^2}\,dv\\[4pt]
&=∫^t_0\sqrt{140^2+(−32v+2)^2}\,dv \end{align*}\]

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 we can use a formula from Appendix A,

\[∫\sqrt{a^2+u^2}\,du=\dfrac{u}{2}\sqrt{a^2+u^2}+\dfrac{a^2}{2}\ln ∣u+\sqrt{a^2+u^2}∣+C. \nonumber \]

We set \(a=140\) and \(u=−32v+2.\) This gives \(du=−32\,dv,\) so \(dv=−\dfrac{1}{32}\,du.\) Therefore

\[\begin{align*} ∫\sqrt{140^2+(−32v+2)^2}\,dv &=−\dfrac{1}{32}∫\sqrt{a^2+u^2}\,du\\[4pt]
&=−\dfrac{1}{32}\left[\dfrac{(−32v+2)}{2}\sqrt{140^2+(−32v+2)^2}+\dfrac{140^2}{2}\ln ∣(−32v+2)+\sqrt{140^2+(−32v+2)^2}|+C\right] \end{align*}\]

and

\[\begin{align*} s(t) &=-\dfrac{1}{32}\left[\dfrac{(-32t+2)}{2}\sqrt{140^2+(-32t+2)^2}+\dfrac{140^2}{2}\ln \left |(-32t+2)+\sqrt{140^2+(-32t+2)^2}\right |\right] +\dfrac{1}{32}\left[\sqrt{140^2+2^2}+\dfrac{140^2}{2}\ln \left |2+\sqrt{140^2+2^2}\right |\right] \\[4pt]
&=\left(\dfrac{t}{2}-\dfrac{1}{32}\right)\sqrt{1024t^2-128t+19604}-\dfrac{1225}{4}\ln \left |(-32t+2)+\sqrt{1024t^2-128t+19604}\right |+\dfrac{\sqrt{19604}}{32}+\dfrac{1225}{4}\ln (2+\sqrt{19604}) \end{align*}. \nonumber \]

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:

\[\dfrac{d}{dx}∫^x_af(u)\,du=f(x). \nonumber \]

Therefore

\[ \begin{align*} s′(t) &= \dfrac{d}{dt}\Big[s(t)\Big]\\[4pt]
&=\dfrac{d}{dt}\left[∫^t_0\sqrt{140^2+(−32v+2)^2}\,dv\right]\\[4pt]
&=\sqrt{140^2+(−32t+2)^2}\\[4pt]
&=\sqrt{1024t^2−128t+19604}\\[4pt]
&=2\sqrt{256t^2−32t+4901}. \end{align*} \nonumber \]

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

\[\begin{align*} s\left(\frac{1}{3}\right) &=\left(\frac{1/3}{2}−\frac{1}{32}\right)\sqrt{1024\left(\frac{1}{3}\right)^2−128\left(\frac{1}{3}\right)+19604}\\
&−\frac{1225}{4}\ln \Bigg|\left(−32\left(\frac{1}{3}\right)+2\right)+\sqrt{1024\left(\frac{1}{3}\right)^2−128\left(\frac{1}{3}\right)+19604}\Bigg|\\
&+\frac{\sqrt{19604}}{32}+\frac{1225}{4}\ln(2+\sqrt{19604})\\[4pt]
&≈46.69 \text{ feet}.\end{align*}\]

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

\(s′\left(\frac{1}{3}\right)=2\sqrt{256\left(\frac{1}{3}\right)^2−32\left(\frac{1}{3}\right)+4901}≈140.27\) 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 volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function \(y=f(x)\) from \(x=a\) to \(x=b,\) revolved around the \(x\)-axis:

\[S=2π∫^b_af(x)\sqrt{1+(f′(x))^2}\,dx. \nonumber \]

We now consider a volume of revolution generated by revolving a parametrically defined curve \(x=x(t), \quad y=y(t), \quad \text{for }a≤t≤b\) around the \(x\)-axis as shown in Figure \(\PageIndex{9}\).

The analogous formula for a parametrically defined curve is

\[S=2π∫^b_ay(t)\sqrt{(x′(t))^2+(y′(t))^2}\,dt \label{ParSurface} \]

provided that \(y(t)\) is not negative on \([a,b]\).

Example \(\PageIndex{6}\): Finding Surface Area

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

Solution

We start with the curve defined by the equations

\[x(t)=r\cos t, \quad y(t)=r\sin t, \quad \text{for }0≤t≤π. \nonumber \]

This generates an upper semicircle of radius \(r\) centered at the origin as shown in the following graph.

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 Equation \ref{ParSurface}:

\[\begin{align*} S &= 2π∫^b_ay(t)\sqrt{(x′(t))^2+(y′(t))^2}\,dt\\[4pt]
&=2π∫^π_0r\sin t\sqrt{(−r\sin t)^2+(r\cos t)^2}\,dt\\[4pt]
&=2π∫^π_0r\sin t\sqrt{r^2\sin^2t+r^2\cos^2t}\,dt\\[4pt]
&=2π∫^π_0r\sin t\sqrt{r^2(\sin^2t+\cos^2t)}\,dt\\[4pt]
&=2π∫^π_0r^2\sin t\,dt\\[4pt]
&=2πr^2\left(−\cos t\Big|^π_0\right)\\[4pt]
&=2πr^2(−\cos π+\cos 0)\\[4pt]
&=4πr^2 \text{ units}^2.\end{align*} \nonumber \]

This is, in fact, the formula for the surface area of a sphere.

Key Concepts

• The derivative of the parametrically defined curve \(x=x(t)\) and \(y=y(t)\) can be calculated using the formula \(\dfrac{dy}{dx}=\dfrac{y′(t)}{x′(t)}\). Using the derivative, we can find the equation of a tangent line to a parametric curve.
• The area between a parametric curve and the \(x\)-axis can be determined by using the formula \(\displaystyle A=∫^{t_2}_{t_1}y(t)x′(t)\,dt.\)
• The arc length of a parametric curve can be calculated by using the formula \[s=∫^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt. \nonumber \]
• The surface area of a volume of revolution revolved around the \(x\)-axis is given by \[S=2π∫^b_ay(t)\sqrt{(x′(t))^2+(y′(t))^2}\,dt. \nonumber \]
• If the curve is revolved around the \(y\)-axis, then the formula is \[S=2π∫^b_a x(t)\sqrt{(x′(t))^2+(y′(t))^2}\,dt. \nonumber \]

Key Equations

• Derivative of parametric equations

\[\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{y′(t)}{x′(t)} \nonumber \]

• Second-order derivative of parametric equations

\[\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=\dfrac{(d/dt)(dy/dx)}{dx/dt} \nonumber \]

• Area under a parametric curve

\[A=∫^b_ay(t)x′(t)\,dt \nonumber \]

• Arc length of a parametric curve

\[s=∫^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \nonumber \]

• Surface area generated by a parametric curve

\[S=2π∫^b_ay(t)\sqrt{(x′(t))^2+(y′(t))^2}\,dt \nonumber \]