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On this page we cover the most common calculus problems using parametric equations. The two main topics are differentiation and integration.

Differentiation

Let's start out with a quick video clip giving us an introduction to finding the derivative \(dy/dx\) if the function \(y=f(x)\) is given in parametric equations \(x(t)\) and \(y(t)\). The theorem is given below.

PatrickJMT - Parametric Differentiation [1min-7secs]

video by PatrickJMT

Theorem: Parametric Derivative

On a smooth curve given by the equations \( x=X(t) \) and \(y = Y(t)\), the slope of the curve at the point \((x,y)\) is \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0 \]

Parametric Derivative Proof

Theorem: Parametric Derivative

On a smooth curve given by the equations \( x=X(t) \) and \(y = Y(t)\), the slope of the curve at the point \((x,y)\) is \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0 \]

We present two proofs here, a short informal version and a longer, more formal version.

Short Informal Proof
Given \( y(t) \) and \( x(t)\), we can write \( y = y(x(t)) \). We want to find \( dy/dx \).
Using the chain rule on \( y(x(t)) \), we have \( dy/dt = dy/dx \cdot dx/dt\).
Solving for \( dy/dx \) we have \( \displaystyle{ dy/dx = \frac{dy/dt}{dx/dt} }\)      [qed]

Longer More Formal Proof
Given the two points \((x_1,y_1) = (X(t), Y(t))\) and \((x_2,y_2) = (X(t+\Delta t), Y(t+\Delta t)) \) and \(\Delta t > 0 \) on a smooth curve, let \( \Delta x = x_2 - x_1 = X(t+\Delta t) - X(t) \) and \( \Delta y = y_2 - y_1 = Y(t+\Delta t) - Y(t) \).

As \( \Delta t \to 0 \), we know that \(\Delta x \to 0 \) and we can write

\(\displaystyle{ \frac{dy}{dx} = \lim_{\Delta x \to 0}{\frac{\Delta y}{\Delta x}} = \lim_{\Delta t \to 0}{\frac{Y(t+\Delta t) - Y(t)}{X(t+\Delta t) - X(t)}} }\)

Now we can multiply the numerator and denominator by \(1/\Delta t\) and use limit laws to give us

\( \begin{array}{rcl} \displaystyle{\frac{dy}{dx}} & = & \displaystyle{\lim_{\Delta t \to 0}{\frac{[Y(t+\Delta t) - Y(t)](1/\Delta t)}{[X(t+\Delta t) - X(t)](1/\Delta t)}} } \\ & = & \displaystyle{ \frac{\lim_{\Delta t \to 0}{[Y(t+\Delta t) - Y(t)]/\Delta t}}{\lim_{\Delta t \to 0}{[X(t+\Delta t) - X(t)]/\Delta t}} } \\ & = & \displaystyle{ \frac{dy/dt}{dx/dt} ~~~~~ \text{ [qed] } } \end{array} \)

Before we go on, let's practice using this theorem.

Find \(dy/dx\) of the parametric curve \(x=t+5\cos(t)\), \(y=3e^t\).

Problem Statement

Find \(dy/dx\) of the parametric curve \(x=t+5\cos(t)\), \(y=3e^t\).

Solution

1373 solution video

video by PatrickJMT

close solution

Find the derivative of the parametric curve \(x=t\sin(t)\), \(y=t^2+t\).

Problem Statement

Find the derivative of the parametric curve \(x=t\sin(t)\), \(y=t^2+t\).

Solution

468 solution video

video by Krista King Math

close solution

Find \(dy/dx\) of the parametric curve \(x=4t+1\), \(y=t^2+2t\).

Problem Statement

Find \(dy/dx\) of the parametric curve \(x=4t+1\), \(y=t^2+2t\).

Solution

1372 solution video

video by PatrickJMT

close solution

Differentiation - Slope and Tangent Lines

To find the equation of a tangent line to a graph given by a set of parametric equations, we need to be able to find the slope by calculating the derivative \( dy/dx \) using the above theorem.

For horizontal tangent lines, the slope \(dy/dx\) is zero, so we need \( dy/dt = 0 \) and \( dx/dt \neq 0 \). For vertical tangent lines, the slope is undefined, which means that \( dx/dt = 0 \) when \( dy/dt \neq 0 \). In the case where both \( dx/dt = 0\) and \( dy/dt = 0 \) at the same point, we need to handle that case separately, since nothing can be concluded from \( dy/dx = 0/0 \), which is indeterminate.

Once you have found the slope, you can easily find the equation of a tangent line. Go to the Tangent Lines page for more information.

Find the equation of the tangent line to the parametric curve \(x=2t^2+1\), \(y=3t^3+2\) at \(t=1\).

Problem Statement

Find the equation of the tangent line to the parametric curve \(x=2t^2+1\), \(y=3t^3+2\) at \(t=1\).

Solution

54 solution video

video by Krista King Math

close solution

Find the equation of the tangent line to the parametric curve \(x=3(t-\sin(t))\), \(y=3(1-\cos(t))\) at \(t=\pi/2\).

Problem Statement

Find the equation of the tangent line to the parametric curve \(x=3(t-\sin(t))\), \(y=3(1-\cos(t))\) at \(t=\pi/2\).

Solution

55 solution video

video by Krista King Math

close solution

Find the equation of the tangent line to the parametric curve \(x=t\cos(t)\), \(y=t\sin(t)\) at \(t=\pi\).

Problem Statement

Find the equation of the tangent line to the parametric curve \(x=t\cos(t)\), \(y=t\sin(t)\) at \(t=\pi\).

Solution

467 solution video

video by Krista King Math

close solution

Find the equation of the tangent line to the parametric curve \(x=3t^2-t\), \(y=\sqrt{t}\) at \(t=4\).

Problem Statement

Find the equation of the tangent line to the parametric curve \(x=3t^2-t\), \(y=\sqrt{t}\) at \(t=4\).

Solution

1371 solution video

video by PatrickJMT

close solution

Differentiation - Higher Order Derivatives

In order to determine concavity of a graph and other information, you will need higher order derivatives. We list two of them below from which you can extract a pattern.

Second Derivative

\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d}{dx}\left[ \frac{dy}{dx} \right] = \frac{d\left[ dy/dx \right]/dt}{dx/dt} }\)
You may also find the notation \( \dot{x} = dx/dt \), in which case the above derivative can be written
\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{\dot{x} \ddot{y} - \dot{y} \ddot{x}}{\dot{x}^3} }\)

Third Derivative

\(\displaystyle{ \frac{d^3y}{dx^3} = \frac{d}{dx}\left[ \frac{d^2y}{dx^2} \right] = \frac{d\left[ d^2y/dx^2 \right] /dt}{dx/dt} }\)

Before we go on, let's work some practice problems with what we have learned so far.

Basic Problems

Find the second derivative of the parametric equations \(x=t^2+t\), \(y=2t-1\).

Problem Statement

Find the second derivative of the parametric equations \(x=t^2+t\), \(y=2t-1\).

Solution

56 solution video

video by Krista King Math

close solution

Intermediate Problems

Find the second derivative of the parametric curve \(x=t^3+t\), \(y=t^5+1\).

Problem Statement

Find the second derivative of the parametric curve \(x=t^3+t\), \(y=t^5+1\).

Solution

1374 solution video

video by PatrickJMT

close solution

Find the second derivative of the parametric curve \(x=t-t^3\), \(y= 2t+5\).

Problem Statement

Find the second derivative of the parametric curve \(x=t-t^3\), \(y= 2t+5\).

Solution

1375 solution video

video by PatrickJMT

close solution

Find the horizontal tangent points to the curve \(x=1-t, y=t^2\) and determine the concavity at those points.

Problem Statement

Find the horizontal tangent points to the curve \(x=1-t, y=t^2\) and determine the concavity at those points.

Final Answer

\((1,0)\), concave up

Problem Statement

Find the horizontal tangent points to the curve \(x=1-t, y=t^2\) and determine the concavity at those points.

Solution

How do you know when you have a horizontal tangent? When the slope is zero.
\(dx/dt=-1, dy/dt=2t\)
\(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{-1} }\)
\(dy/dx = 0\) when \(-2t=0 \to t=0\)
When \(t=0\), \((x,y) = (1,0)\).

To determine concavity at that point, we need to calculate the second derivative.
\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{-2}{-1} = 2 }\)
Since the second derivative is positive everywhere, the curve is concave up.

Final Answer

\((1,0)\), concave up

close solution

For the parametric curve \(x=\sqrt{t}, y=3t-1\), find the derivative and the slope at the point \(t=1\). Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slope-intercept form.

Problem Statement

For the parametric curve \(x=\sqrt{t}, y=3t-1\), find the derivative and the slope at the point \(t=1\). Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slope-intercept form.

Final Answer

\(dy/dx=6\sqrt{t}, m=6, y=6x-4\), concave up

Problem Statement

For the parametric curve \(x=\sqrt{t}, y=3t-1\), find the derivative and the slope at the point \(t=1\). Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slope-intercept form.

Solution

\(dx/dt=(1/2)t^{-1/2}, dy/dt = 3\)
\(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3}{(1/2)t^{-1/2}} = 6\sqrt{t} }\)
slope at \(t=1, m=6\sqrt{1} = 6\)
for the equation of tangent line, we need the \((x,y)\) point at \(t=1\), so \((\sqrt{1},3-1)=(1,2)\)
\(y-y_1=m(x-x_1) \to y-2=6(x-1) \to y=6x-4\)
\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{6(1/2)t^{-1/2}}{(1/2)t^{-1/2}} = 6 }\)
Since the second derivative is positive at \(t=1\) (and everywhere else for that matter), the curve is concave up.

Final Answer

\(dy/dx=6\sqrt{t}, m=6, y=6x-4\), concave up

close solution

Integration - Area

The area under a smooth curve defined parametrically as \( x = X(t) \) and \( y = Y(t) \) from \( t=t_0\) to \( t=t_1\) can be calculated using the integral

\(\displaystyle{s = \int_{t_0}^{t_1}{Y(t)X'(t) ~ dt}}\)     where \( X'(t) = dX/dt \).

You may also find this written in a shorthand form as

\(\displaystyle{s = \int_{t_0}^{t_1}{y~dx}}\)

In this notation, \( dx = (dx/dt) dt \).

Find the area under the curve \(x=1+e^t\), \(y=t-t^2\) and above the x-axis.

Problem Statement

Find the area under the curve \(x=1+e^t\), \(y=t-t^2\) and above the x-axis.

Solution

57 solution video

video by Krista King Math

close solution

Calculate the area enclosed by the parametric curve \(x = t^3 - 3t\), \(y = 4t^2\).

Problem Statement

Calculate the area enclosed by the parametric curve \(x = t^3 - 3t\), \(y = 4t^2\).

Final Answer

Area = \(\displaystyle{ \frac{96}{5}\sqrt{3} }\)

Problem Statement

Calculate the area enclosed by the parametric curve \(x = t^3 - 3t\), \(y = 4t^2\).

Solution

The most difficult and time-consuming part of this problem is setting up the integral. Here is a general overview explaining how we will tackle this problem.
1. Graph the parametric equation.
2. Find the important points that determine the area to be calculated.
3. Determine how to set up the integrals in general to calculate the area.
4. Set up the integrals and evaluate.

1. Graph the parametric equation.
Notice we have included the direction of the graph in the plot. Now we can easily see the area that needs to be calculated. Without this graph, it would have been extremely difficult to set up the integral(s) to determine the area.

2. Find the important points that determine the area to be calculated.
First, we need to find the point where the graph crosses itself. It looks like this occurs at the point \((0,12)\) but we can't guarantee that just by looking at the graph. So we need to find \(t_1\) and \(t_2\) where the \(x\) and \(y\) values are equal.
At time \(t_1\), we have \( x_1 = {t_1}^3 - 3t_1 \) and \( y_1 = 4{t_1}^2\).
Similarly, at time \(t_2\), we have \( x_2 = {t_2}^3 - 3t_2 \) and \( y_2 = 4{t_2}^2\).
When the graph crosses, we have \( x_1 = x_2\) and \( y_1=y_2\). So we can write
\(\begin{array}{rcl} {t_1}^3 - 3t_1 & = & {t_2}^3 - 3t_2 \\ 4{t_1}^2 & = & 4{t_2}^2 \end{array}\)
From the second equation, we may be tempted to write \( t_1 = t_2 \) only, but since \(t\) is squared, we also have the case where \( t_1 = -t_2 \). The time \( t_1 = t_2 \) does not help us since it is true by default. We need the case \( t_1 = -t_2 \). So we substitute \( t_1 = -t_2 \) into the first equation.
\( (-t_2)^3 - 3(-t_2) = {t_2}^3 - 3t_2 \)
For simplicity, we will drop the subscript temporarily in this equation and just use \(t\), i.e. \( t=t_2\).
\(\begin{array}{rcl} (-t)^3 - 3(-t) = t^3 - 3t \\ -t^3 + 3t = t^3 - 3t \\ 2t^3 - 6t & = & 0 \\ 2t(t^2-3) & = & 0 \end{array}\)
Using the zero-product rule we have \( 2t = 0 \) and \(t^2-3=0 ~~ \to ~~ t^2 = 3 ~~ \to ~~ t = \pm \sqrt{3} \)
Now we will go back to the subscript to be clear. \( t_2 = \pm \sqrt{3} \)
So we have \(t_1 = -t_2 \) and \( t_2 = \pm \sqrt{3} \).
When \( t_2 = \sqrt{3}\), \( t_1 = -\sqrt{3} \) and when \( t_2 = -\sqrt{3}\), \( t_1 = \sqrt{3} \). So the graph crosses itself at \( t_1 = -\sqrt{3} \) and \(t_2 = \sqrt{3} \).
(Since both \(\sqrt{3}\) and \(-\sqrt{3}\) are covered here, it doesn't matter which set we choose. This may not always be the case, so be careful. We chose \( t_1 < t_2 \) just to be consistent with the idea that time is usually increasing.)
Let's find the \((x,y)\) point where the graph crosses itself. Let \( t = t_1 = -\sqrt{3}\) first.
\( x = {t_1}^3 - 3t_1 = (-\sqrt{3})^3 - 3(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0 ~~ \to ~~ x = 0 \)
\( y = 4{t_1}^2 = 4(-\sqrt{3})^2 = 4(3) = 12 ~~ \to ~~ y = 12 \)
So the point where the graph crosses itself is \((0,12)\). Just to confirm our work, let's find the point when \( t = t_2 = \sqrt{3} \).
\( x = (\sqrt{3})^3 - 3(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0 \)
\( y = 4(\sqrt{3})^2 = 4(3) = 12 \)
So the t-values give us the same \( (x,y)=(0,12)\) point.

3. Determine how to set up the integrals in general to calculate the area.
Let's discuss how we are going to approach this. First, we may be tempted to find the area of half and then multiply by \(2\). The graph certainly looks symmetric about the y-axis but we would need to prove that it is and that's not easy to do. So we need to find both areas.

If we do the obvious and just integrate from \(t_1\) to \(t_2\), we will be calculating part of the area multiple times, since the integral calculates the area under the curve and above the x-axis. So we need to integrate from the left edge (the furthest x-value on the left) to the right edge to get the area shown in the left figure below. Then we need to subtract the areas beneath the curve (outside the region) shown in the right figure below. This will give us the area we are asked to calculate.

Our next step is to calculate the far left and far right points on the graph. To do that we notice that, at those points, the graph has vertical asymptotes, meaning that the slope is undefined. In the slope equation
\(\displaystyle{\frac{dy}{dx} = \frac{dy/dt}{dx/dt}}\)
\( dx/dt = 0 \) is where the slope is undefined (as long as \(dy/dt \neq 0 \)).
So we have \( x = t^3 - 3t ~~ \to ~~ dx/dt = 3t^2-3 \)
Setting \(dx/dt\) equal to zero and solving for \(t\), we find \( t = \pm 1 \).
We need to check that \(dy/dt \neq 0\) at these points.
\( y = 4t^2 ~~ \to ~~ dy/dt = 8t \) and at \( t=1 ~~ dy/dt = 8 \neq 0 \) and \( t=-1 ~~ dy/dt = -8 \neq 0 \). So, \( t = \pm 1 \) are the \(t\)-values when the graph has vertical asymptotes, which are the left and right boundaries of the area we are calculating.

The \((x,y)\) points are
\( t = -1: ~~ x = (-1)^3 - 3(-1) = -1+3 = 2 ~~ y = 4(-1)^2 = 4 ~~ (2,4) \)
\( t = 1: ~~ x = (1)^3 - 3(1) = 1-3 = -2 ~~ y = 4(1)^2 = 4 ~~ (-2,4) \)

Side Note: Isn't it interesting that as \(t\) sweeps from \( t = -\sqrt{3}\) to \( t = \sqrt{3}\), the graph is swept out clockwise? You need to know this to know how to correctly set up the integrals.

4. Set up the integrals and evaluate.
So now we have all the information we need to set up the integrals. Generally, the integral is
\(\displaystyle{\int_{t_1}^{t_2}{y(t)x'(t)dt}}\)

Area = \(\displaystyle{ \int_{-\sqrt{3}}^{-1}{4t^2(3t^2-3)dt} + \int_{1}^{\sqrt{3}}{4t^2(3t^2-3)dt} - \int_{1}^{-1}{4t^2(3t^2-3)dt} }\)

The first integral calculates the area under the top part of the curve on the right side of the y-axis down to the x-axis.
The second integral calculates the same area on the left side of the y-axis.
The last integral calculates the area below the bottom part of the curve on both sides of the y-axis and subtracts it from the other values, giving us the area enclosed by the curve. Notice that we need to sweep left to right for all the integrals. This is where the direction the curve is being swept is important.

All the integrals are the same expect for the limits of integration. So we will integrate once and then substitute the limits in. We may get some cancellation which will simplify our algebra. So we will watch for that.
\( \displaystyle{ \int{4t^2(3t^2-3)dt} = \int{12t^4-12t^2 dt} = 12 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right] + C }\)
We can drop the \(+C\) when we substitute the limits of integration since it will cancel in the subsequent calculations.

Area = \(\displaystyle{ 12 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right]_{-\sqrt{3}}^{-1} + 12 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right]_{1}^{\sqrt{3}} - 12 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right]_{1}^{-1} }\)

Area =\(\displaystyle{12 \left[ \frac{(-1)^5}{5} - \frac{(-1)^3}{3} \right] - 12 \left[ \frac{(-\sqrt{3})^5}{5} - \frac{(-\sqrt{3})^3}{3} \right]}\)
\(\displaystyle{ + 12 \left[ \frac{(\sqrt{3})^5}{5} - \frac{(\sqrt{3})^3}{3} \right] - 12 \left[ \frac{1}{5} - \frac{1}{3} \right] - 12 \left[ \frac{(-1)^5}{5} - \frac{(-1)^3}{3} \right] + 12 \left[ \frac{(1)^5}{5} - \frac{(1)^3}{3} \right] }\)
Let's factor out 12 and simplify the negative signs to see if we get any cancellation.
Area = \(\displaystyle{ 12 \left[ \frac{-1}{5} + \frac{1}{3} + \frac{(\sqrt{3})^5}{5} - \frac{(\sqrt{3})^3}{3} + \frac{(\sqrt{3})^5}{5} - \frac{(\sqrt{3})^3}{3} -\frac{1}{5} + \frac{1}{3} + \frac{1}{5} - \frac{1}{3} + \frac{1}{5} - \frac{1}{3} \right] }\)
So all the terms without roots cancel giving us
Area = \(\displaystyle{ 12 \left[ \frac{2}{5}(\sqrt{3})^5 - \frac{2}{3}(\sqrt{3})^3 \right] }\)
Simplifying and combining terms, we get Area = \(\displaystyle{ \frac{96}{5}\sqrt{3} }\)
Note:
Although we mentioned above that integrating from \(-\sqrt{3}\) to \( \sqrt{3}\) would not work, that is only partially true. In general cases, it would not work since integration is sensitive to direction, i.e. left to right gives you the positive area under the curve; right to left gives you the negative area under the curve. However, IN THIS PROBLEM ONLY, because of the direction the curve is swept, using one integral, sweeping from \(-\sqrt{3}\) to \( \sqrt{3}\) would give the correct answer. However, you would need to explain why. Do you see why?

Whew! That was a lot of work. I hope you didn't get lost in all the calculations. If you lost sight of the big picture, go back and scan the overview at the top of the solution to make sure you understand the main structure of this solution.

Final Answer

Area = \(\displaystyle{ \frac{96}{5}\sqrt{3} }\)

close solution

Calculate the area enclosed by the line \(y=2.5\) and the parametric curve \(x=t-1/t\), \(y=t+1/t\).

Problem Statement

Calculate the area enclosed by the line \(y=2.5\) and the parametric curve \(x=t-1/t\), \(y=t+1/t\).

Final Answer

\(\displaystyle{6-\frac{15}{8}-2\ln(2)}\)

Problem Statement

Calculate the area enclosed by the line \(y=2.5\) and the parametric curve \(x=t-1/t\), \(y=t+1/t\).

Solution

In the video, he sets up the integral but does not evaluate it. Here are some of those steps.

\(\displaystyle{ \int_{1/2}^{2}{ \left[ 2.5-(t+1/2) \right]( 1+1/t^2)~dt } }\)

\(\displaystyle{ \int_{0.5}^{2}{ 2+2t^{-2} - t - t^{-1}~dt } }\)

\(\displaystyle{ \left[ 2t - \frac{2}{t} - \frac{t^2}{2} - \ln(t) \right]_{0.5}^{2} }\)

\(\displaystyle{ 6 - \frac{15}{8} - 2\ln(2) }\)

1376 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{6-\frac{15}{8}-2\ln(2)}\)

close solution

Calculate the area under one arc of the parametric curve \( x = r \theta - d \sin\theta \), \( y = r - d \cos\theta \).

Problem Statement

Calculate the area under one arc of the parametric curve \( x = r \theta - d \sin\theta \), \( y = r - d \cos\theta \).

Solution

1281 solution video

video by Krista King Math

close solution

Integration - Arc Length

When a smooth curve is defined parametrically as \( x=X(t) \) and \( y = Y(t) \), the arc length between the points \( t=t_0 \) and \( t = t_1 \) can be calculated using the integral

\(\displaystyle{ s = \int_{t_0}^{t_1}{\sqrt{[X'(t)]^2 + [Y'(t)]^2} dt}}\)
Notice we use a small s here to represent the arc length. This is the standard symbol you will see in many textbooks. We reserve a capital S to represent surface area.

Here is a quick video clip going over this equation in more detail.

PatrickJMT - Arc Length [47secs]

video by PatrickJMT

Basic Problems

Compute the arc length of the curve with parametric equations \(x(t)=3t^2-9\), \(y(t)=t^3-3t\), \(0\leq t\leq 3\).

Problem Statement

Compute the arc length of the curve with parametric equations \(x(t)=3t^2-9\), \(y(t)=t^3-3t\), \(0\leq t\leq 3\).

Final Answer

36

Problem Statement

Compute the arc length of the curve with parametric equations \(x(t)=3t^2-9\), \(y(t)=t^3-3t\), \(0\leq t\leq 3\).

Solution

2001 solution video

video by Dr Chris Tisdell

Final Answer

36

close solution

Calculate the arc length of the parametric curve \(x(t) = t \sin(t)\) \(y(t) = t \cos(t)\) on the interval \(0 \leq t \leq1\).

Problem Statement

Calculate the arc length of the parametric curve \(x(t) = t \sin(t)\) \(y(t) = t \cos(t)\) on the interval \(0 \leq t \leq1\).

Final Answer

arc length = \(\displaystyle{ \frac{1}{2}\left[ \sqrt{2}+\ln(\sqrt{2}+1) \right] }\)

Problem Statement

Calculate the arc length of the parametric curve \(x(t) = t \sin(t)\) \(y(t) = t \cos(t)\) on the interval \(0 \leq t \leq1\).

Solution

483 solution video

video by Krista King Math

Final Answer

arc length = \(\displaystyle{ \frac{1}{2}\left[ \sqrt{2}+\ln(\sqrt{2}+1) \right] }\)

close solution

Calculate the arc length of the parametric curve \(x = 1+3t^2\), \(y = 4+2t^3\), \(0 \leq t \leq 1\).

Problem Statement

Calculate the arc length of the parametric curve \(x = 1+3t^2\), \(y = 4+2t^3\), \(0 \leq t \leq 1\).

Solution

1377 solution video

video by PatrickJMT

close solution

Calculate the arc length of the parametric curve \(x = |6-t|\), \(y = t\), \(0 \leq t \leq 3\).

Problem Statement

Calculate the arc length of the parametric curve \(x = |6-t|\), \(y = t\), \(0 \leq t \leq 3\).

Solution

1379 solution video

video by PatrickJMT

close solution

Intermediate Problems

Find the arc length of the parametric curve \(x = 2t\), \(y = (2/3)t^{3/2}\) for \(5 \leq t \leq 12\).

Problem Statement

Find the arc length of the parametric curve \(x = 2t\), \(y = (2/3)t^{3/2}\) for \(5 \leq t \leq 12\).

Solution

58 solution video

video by Krista King Math

close solution

Calculate the arc length of the parametric curve \(x = t^3\), \(y = t^2\) from \((0,0)\) to \((8,4)\).

Problem Statement

Calculate the arc length of the parametric curve \(x = t^3\), \(y = t^2\) from \((0,0)\) to \((8,4)\).

Solution

469 solution video

video by Dr Chris Tisdell

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Calculate the arc length of the curve \(x(\theta)=a\cos^3(\theta)\), \(y(\theta)=a\sin^3(\theta)\), \(0\leq\theta\leq2\pi\).

Problem Statement

Calculate the arc length of the curve \(x(\theta)=a\cos^3(\theta)\), \(y(\theta)=a\sin^3(\theta)\), \(0\leq\theta\leq2\pi\).

Solution

Upon initial inspection, this video does not seem to go with the problem statement. However, if you eliminate the parameter θ from the equation, you will get the equation that he starts with in the video.

1380 solution video

video by Dr Chris Tisdell

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Calculate the arc length of the parametric curve \(x = e^t+e^{-t}\), \(y = 5-2t\), \(0 \leq t \leq 3\).

Problem Statement

Calculate the arc length of the parametric curve \(x = e^t+e^{-t}\), \(y = 5-2t\), \(0 \leq t \leq 3\).

Solution

1378 solution video

video by PatrickJMT

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Calculate the arc length of the parametric curve \(x = e^t \cos(t)\), \(y = e^t \sin(t)\), \(0 \leq t \leq \pi\).

Problem Statement

Calculate the arc length of the parametric curve \(x = e^t \cos(t)\), \(y = e^t \sin(t)\), \(0 \leq t \leq \pi\).

Solution

1381 solution video

video by PatrickJMT

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Integration - Surface Area

To calculate the surface area defined by revolving a parametric curve defined as \( x=X(t)\) and \(y=Y(t)\) from \(t=t_0\) to \(t=t_1\) about the x-axis, we use this integral.

\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{ Y(t) \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~ dt}}\)

If you compare this integral to the equation for arc length (in the prevous section) you will see the common factor \( \sqrt{[X'(t)]^2 + [Y'(t)]^2} \)

Some textbooks write the surface integral differently, taking this into account. You may see it written as

\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{Y(t) ~ ds}}\)
where \( ds = \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~dt \)

When rotating about the y-axis, the integral we use is

\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{X(t) ~ ds}}\)

Notice we use a capital S to represent surface area. This is the standard symbol you will see in many textbooks. We reserve a lowercase s to represent arc length.

Find the surface area of revolution of the parametric curve \(x = 3t^2\), \(y = 2t^3\), \(0 \leq t \leq 5\) rotated about the y-axis.

Problem Statement

Find the surface area of revolution of the parametric curve \(x = 3t^2\), \(y = 2t^3\), \(0 \leq t \leq 5\) rotated about the y-axis.

Final Answer

\(\displaystyle{ S_y = \frac{24\pi}{5} \left[ 949\sqrt{26}+1 \right] }\)

Problem Statement

Find the surface area of revolution of the parametric curve \(x = 3t^2\), \(y = 2t^3\), \(0 \leq t \leq 5\) rotated about the y-axis.

Solution

594 solution video

video by Krista King Math

Final Answer

\(\displaystyle{ S_y = \frac{24\pi}{5} \left[ 949\sqrt{26}+1 \right] }\)

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Find the surface area of revolution of the parametric curve \(x = 1-t\), \(y = 2\sqrt{t}\), \(1 \leq t \leq 4\) rotated about the x-axis.

Problem Statement

Find the surface area of revolution of the parametric curve \(x = 1-t\), \(y = 2\sqrt{t}\), \(1 \leq t \leq 4\) rotated about the x-axis.

Solution

59 solution video

video by Krista King Math

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Integration - Volume

If we have a parametric curve defined as \( x = X(t) \) and \( y = Y(t) \), we can determine the volume of the solid object defined by revolving this curve about an axis. We will limit our curve from \( t=t_0 \) to \( t=t_1 \).

Revolution About the x-axis

When revolved about the x-axis, the integral is
\(\displaystyle{V_x = \pi \int_{t_0}^{t_1}{y^2 [dx/dt] dt}}\)
This is sometimes written as \(\displaystyle{V_x = \pi \int_{t_0}^{t_1}{y^2 ~ dx}}\) where \( dx = [dx/dt] dt \)

Revolution About the y-axis

Similar to the x-axis integral, we have
\(\displaystyle{V_y = \pi \int_{t_0}^{t_1}{x^2 [dy/dt] dt} }\) also written as \(\displaystyle{V_y = \pi \int_{t_0}^{t_1}{x^2 ~ dy}}\) where \( dy = [dy/dt] dt \)

Calculate the volume of revolution when the curve \(x = t^3\), \(y = 2t^2+1\), \(-1 \leq t \leq 1\) is rotated about the x-axis.

Problem Statement

Calculate the volume of revolution when the curve \(x = t^3\), \(y = 2t^2+1\), \(-1 \leq t \leq 1\) is rotated about the x-axis.

Solution

466 solution video

video by Krista King Math

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