Length of Curve, Distance traveled, Accumulated Change, Motion of Curve - AP Calculus BC

Card 0 of 60

Question

Give the arclength of the graph of the function $f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, \frac{\pi}{8} \right ]$ .

Answer

The length of the curve of on the interval can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} ; \mathrm{d}x$ .

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , so

$f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x$

The integral becomes

$\int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sec 2x ; \mathrm{d}x$

Use substitution - set . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = \frac{1}{2}\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot \frac{\pi}{8} = \frac{\pi}{4}$ , and the integral becomes

$\frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u ; \mathrm{d}u$

$=\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \ \ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\ \ 0 \end{matrix}$

$=\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0$

$=\frac{1}{2} \ln \left (1 + \sqrt{2} \right )$

Compare your answer with the correct one above

Question

Give the arclength of the graph of the function $f(x) = \frac{2}{3} x \sqrt{x}$ on the interval $\left [ 0,3 \right ]$ .

Answer

The length of the curve of on the interval can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} ; \mathrm{d}x$ .

$f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}$

so

$f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}$ .

The above integral becomes

$\int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} ; \mathrm{d}x$

$\int_{0}^{3} \sqrt{x+1} ; \mathrm{d}x$

Substitute . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , $\mathrm{d} u = \mathrm{d} x$ , and the integral becomes

$\int_{1}^{4} \sqrt{u} ; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} ; \mathrm{d}u$

$= \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \ \ \end{matrix}\right| \begin{matrix} 4\ \ 1 \end{matrix}$

$= \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \ \ \end{matrix}\right| \begin{matrix} 4\ \ 1 \end{matrix}$

$= \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}$

$= \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}$

Compare your answer with the correct one above

Question

Determine the length of the following function between $1\leq t\leq 4$

$v(t)=\frac{2}{3}(t-1)^{\frac{3}{2}}$

Answer

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

$L=\int ds$

where ds is given by the equation below:

$ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx$

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

$\frac{dv}{dt}=(t-1)^{\frac{1}{2}}=\sqrt{t-1}$

Now we can plug this into the given equation to find ds:

$ds=\sqrt{1+(\frac{dv}{dt})^{2}}dt=\sqrt{1+(\sqrt{t-1})^{2}}=\sqrt{1+t-1}=\sqrt{t}$

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

$L=\int ds=\int_{1}^{4}\sqrt{t}dt=[\frac{2}{3}t^{\frac{3}{2}}]_{1}^{4}$

$=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$

Compare your answer with the correct one above

Question

In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent.

This looks like $\small W=\int(F\cdot ds)$ ( $\small W$ is work, $\small F$ is force, and $\small ds$ is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes $\small \small W=\int (Fdx)$ .

If the force on an object as a function of displacement is $\small F(x)=3x^{2}+x$ , what is the work as a function of displacement $\small W(x)$ ? Assume $\small W(0)=0$ and the force is in the direction of the object's motion.

Answer

$\small \small W=\int (Fdx)$ , so $\small W=\int(3x^{2}+x)dx$ .

Both the terms of the force are power terms in the form $\small ax^{n}$ , which have the integral $\small \frac{a}{n+1}x^{n+1}$ , so the integral of the force is $\small x^{3}+x^{2}/2+c$ .

We know

$\small W(0)=0\rightarrow 0^{3}+0^{2}/2+c=0 \therefore c=0$ .

This means

$\small W(x)=x^{3}+x^{2}/2$ .

Compare your answer with the correct one above

Question

What is the length of the curve $g(x)=x^{\frac{3}{2}}$ over the interval $0\le x\le4$ ?

Answer

The general formula for finding the length of a curve over an interval is $\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx$

In this example, the arc length can be found by computing the integral

$\int_{0}^{4}\sqrt{1+(\frac{d}{dx}x^{\frac{3}{2}})^2}\ dx$ .

The derivative of $x^{\frac{3}{2}}$ can be found using the power rule, $\frac{d}{dx} x^n=nx^{n-1}$ , which leads to

$\int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}\ dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx$ .

At this point, a substitution is useful.

Let

$u=1+\frac{9}{4}x,\ du=\frac{9}{4}dx,\frac{4}{9}du=dx$ .

We can also express the limits of integration in terms of to simplify computation. When , and when .

Making these substitutions leads to

$\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx=\int_{1}^{10}\frac{4}{9}\sqrt{u}du=\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du$ .

Now use the power rule, which in general is $\int x^n=\frac{x^{n+1}}{n+1}+c$ , to evaluate the integral.

$\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du=\frac{4}{9}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}|_{1}^{10}$

$=\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)$

Compare your answer with the correct one above

Question

Find the total distance traveled by a particle along the curve from to .

Answer

To find the required distance, we can use the arc length expression given by $\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$ .

Taking the derivative of our function, we have . Plugging in our values for our integral bounds, we have

$\int_0^2\sqrt{1+(2x)^2}dx = \int_0^2 \sqrt{1+4x^2}dx$ .

As with most arc length integrals, this integral is too difficult (if not, outright impossible) to evaluate explicitly by hand. So we will just leave it this form, or evaluate it with some computer software.

Compare your answer with the correct one above

Question

Give the arclength of the graph of the function $f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, \frac{\pi}{8} \right ]$ .

Answer

The length of the curve of on the interval can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} ; \mathrm{d}x$ .

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , so

$f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x$

The integral becomes

$\int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sec 2x ; \mathrm{d}x$

Use substitution - set . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = \frac{1}{2}\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot \frac{\pi}{8} = \frac{\pi}{4}$ , and the integral becomes

$\frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u ; \mathrm{d}u$

$=\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \ \ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\ \ 0 \end{matrix}$

$=\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0$

$=\frac{1}{2} \ln \left (1 + \sqrt{2} \right )$

Compare your answer with the correct one above

Question

Give the arclength of the graph of the function $f(x) = \frac{2}{3} x \sqrt{x}$ on the interval $\left [ 0,3 \right ]$ .

Answer

The length of the curve of on the interval can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} ; \mathrm{d}x$ .

$f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}$

so

$f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}$ .

The above integral becomes

$\int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} ; \mathrm{d}x$

$\int_{0}^{3} \sqrt{x+1} ; \mathrm{d}x$

Substitute . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , $\mathrm{d} u = \mathrm{d} x$ , and the integral becomes

$\int_{1}^{4} \sqrt{u} ; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} ; \mathrm{d}u$

$= \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \ \ \end{matrix}\right| \begin{matrix} 4\ \ 1 \end{matrix}$

$= \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \ \ \end{matrix}\right| \begin{matrix} 4\ \ 1 \end{matrix}$

$= \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}$

$= \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}$

Compare your answer with the correct one above

Question

Determine the length of the following function between $1\leq t\leq 4$

$v(t)=\frac{2}{3}(t-1)^{\frac{3}{2}}$

Answer

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

$L=\int ds$

where ds is given by the equation below:

$ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx$

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

$\frac{dv}{dt}=(t-1)^{\frac{1}{2}}=\sqrt{t-1}$

Now we can plug this into the given equation to find ds:

$ds=\sqrt{1+(\frac{dv}{dt})^{2}}dt=\sqrt{1+(\sqrt{t-1})^{2}}=\sqrt{1+t-1}=\sqrt{t}$

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

$L=\int ds=\int_{1}^{4}\sqrt{t}dt=[\frac{2}{3}t^{\frac{3}{2}}]_{1}^{4}$

$=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$

Compare your answer with the correct one above

Question

In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent.

This looks like $\small W=\int(F\cdot ds)$ ( $\small W$ is work, $\small F$ is force, and $\small ds$ is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes $\small \small W=\int (Fdx)$ .

If the force on an object as a function of displacement is $\small F(x)=3x^{2}+x$ , what is the work as a function of displacement $\small W(x)$ ? Assume $\small W(0)=0$ and the force is in the direction of the object's motion.

Answer

$\small \small W=\int (Fdx)$ , so $\small W=\int(3x^{2}+x)dx$ .

Both the terms of the force are power terms in the form $\small ax^{n}$ , which have the integral $\small \frac{a}{n+1}x^{n+1}$ , so the integral of the force is $\small x^{3}+x^{2}/2+c$ .

We know

$\small W(0)=0\rightarrow 0^{3}+0^{2}/2+c=0 \therefore c=0$ .

This means

$\small W(x)=x^{3}+x^{2}/2$ .

Compare your answer with the correct one above

Question

What is the length of the curve $g(x)=x^{\frac{3}{2}}$ over the interval $0\le x\le4$ ?

Answer

The general formula for finding the length of a curve over an interval is $\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx$

In this example, the arc length can be found by computing the integral

$\int_{0}^{4}\sqrt{1+(\frac{d}{dx}x^{\frac{3}{2}})^2}\ dx$ .

The derivative of $x^{\frac{3}{2}}$ can be found using the power rule, $\frac{d}{dx} x^n=nx^{n-1}$ , which leads to

$\int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}\ dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx$ .

At this point, a substitution is useful.

Let

$u=1+\frac{9}{4}x,\ du=\frac{9}{4}dx,\frac{4}{9}du=dx$ .

We can also express the limits of integration in terms of to simplify computation. When , and when .

Making these substitutions leads to

$\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx=\int_{1}^{10}\frac{4}{9}\sqrt{u}du=\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du$ .

Now use the power rule, which in general is $\int x^n=\frac{x^{n+1}}{n+1}+c$ , to evaluate the integral.

$\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du=\frac{4}{9}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}|_{1}^{10}$

$=\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)$

Compare your answer with the correct one above

Question

Find the total distance traveled by a particle along the curve from to .

Answer

To find the required distance, we can use the arc length expression given by $\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$ .

Taking the derivative of our function, we have . Plugging in our values for our integral bounds, we have

$\int_0^2\sqrt{1+(2x)^2}dx = \int_0^2 \sqrt{1+4x^2}dx$ .

As with most arc length integrals, this integral is too difficult (if not, outright impossible) to evaluate explicitly by hand. So we will just leave it this form, or evaluate it with some computer software.

Compare your answer with the correct one above

Question

Give the arclength of the graph of the function $f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, \frac{\pi}{8} \right ]$ .

Answer

The length of the curve of on the interval can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} ; \mathrm{d}x$ .

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , so

$f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x$

The integral becomes

$\int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sec 2x ; \mathrm{d}x$

Use substitution - set . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = \frac{1}{2}\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot \frac{\pi}{8} = \frac{\pi}{4}$ , and the integral becomes

$\frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u ; \mathrm{d}u$

$=\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \ \ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\ \ 0 \end{matrix}$

$=\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0$

$=\frac{1}{2} \ln \left (1 + \sqrt{2} \right )$

Compare your answer with the correct one above

Question

Give the arclength of the graph of the function $f(x) = \frac{2}{3} x \sqrt{x}$ on the interval $\left [ 0,3 \right ]$ .

Answer

The length of the curve of on the interval can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} ; \mathrm{d}x$ .

$f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}$

so

$f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}$ .

The above integral becomes

$\int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} ; \mathrm{d}x$

$\int_{0}^{3} \sqrt{x+1} ; \mathrm{d}x$

Substitute . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , $\mathrm{d} u = \mathrm{d} x$ , and the integral becomes

$\int_{1}^{4} \sqrt{u} ; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} ; \mathrm{d}u$

$= \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \ \ \end{matrix}\right| \begin{matrix} 4\ \ 1 \end{matrix}$

$= \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \ \ \end{matrix}\right| \begin{matrix} 4\ \ 1 \end{matrix}$

$= \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}$

$= \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}$

Compare your answer with the correct one above

Question

Determine the length of the following function between $1\leq t\leq 4$

$v(t)=\frac{2}{3}(t-1)^{\frac{3}{2}}$

Answer

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

$L=\int ds$

where ds is given by the equation below:

$ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx$

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

$\frac{dv}{dt}=(t-1)^{\frac{1}{2}}=\sqrt{t-1}$

Now we can plug this into the given equation to find ds:

$ds=\sqrt{1+(\frac{dv}{dt})^{2}}dt=\sqrt{1+(\sqrt{t-1})^{2}}=\sqrt{1+t-1}=\sqrt{t}$

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

$L=\int ds=\int_{1}^{4}\sqrt{t}dt=[\frac{2}{3}t^{\frac{3}{2}}]_{1}^{4}$

$=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$

Compare your answer with the correct one above

Question

In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent.

This looks like $\small W=\int(F\cdot ds)$ ( $\small W$ is work, $\small F$ is force, and $\small ds$ is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes $\small \small W=\int (Fdx)$ .

If the force on an object as a function of displacement is $\small F(x)=3x^{2}+x$ , what is the work as a function of displacement $\small W(x)$ ? Assume $\small W(0)=0$ and the force is in the direction of the object's motion.

Answer

$\small \small W=\int (Fdx)$ , so $\small W=\int(3x^{2}+x)dx$ .

Both the terms of the force are power terms in the form $\small ax^{n}$ , which have the integral $\small \frac{a}{n+1}x^{n+1}$ , so the integral of the force is $\small x^{3}+x^{2}/2+c$ .

We know

$\small W(0)=0\rightarrow 0^{3}+0^{2}/2+c=0 \therefore c=0$ .

This means

$\small W(x)=x^{3}+x^{2}/2$ .

Compare your answer with the correct one above

Question

What is the length of the curve $g(x)=x^{\frac{3}{2}}$ over the interval $0\le x\le4$ ?

Answer

The general formula for finding the length of a curve over an interval is $\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx$

In this example, the arc length can be found by computing the integral

$\int_{0}^{4}\sqrt{1+(\frac{d}{dx}x^{\frac{3}{2}})^2}\ dx$ .

The derivative of $x^{\frac{3}{2}}$ can be found using the power rule, $\frac{d}{dx} x^n=nx^{n-1}$ , which leads to

$\int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}\ dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx$ .

At this point, a substitution is useful.

Let

$u=1+\frac{9}{4}x,\ du=\frac{9}{4}dx,\frac{4}{9}du=dx$ .

We can also express the limits of integration in terms of to simplify computation. When , and when .

Making these substitutions leads to

$\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx=\int_{1}^{10}\frac{4}{9}\sqrt{u}du=\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du$ .

Now use the power rule, which in general is $\int x^n=\frac{x^{n+1}}{n+1}+c$ , to evaluate the integral.

$\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du=\frac{4}{9}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}|_{1}^{10}$

$=\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)$

Compare your answer with the correct one above

Question

Find the total distance traveled by a particle along the curve from to .

Answer

To find the required distance, we can use the arc length expression given by $\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$ .

Taking the derivative of our function, we have . Plugging in our values for our integral bounds, we have

$\int_0^2\sqrt{1+(2x)^2}dx = \int_0^2 \sqrt{1+4x^2}dx$ .

As with most arc length integrals, this integral is too difficult (if not, outright impossible) to evaluate explicitly by hand. So we will just leave it this form, or evaluate it with some computer software.

Compare your answer with the correct one above

Question

Give the arclength of the graph of the function $f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, \frac{\pi}{8} \right ]$ .

Answer

The length of the curve of on the interval can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} ; \mathrm{d}x$ .

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , so

$f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x$

The integral becomes

$\int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } ; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sec 2x ; \mathrm{d}x$

Use substitution - set . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = \frac{1}{2}\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot \frac{\pi}{8} = \frac{\pi}{4}$ , and the integral becomes

$\frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u ; \mathrm{d}u$

$=\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \ \ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\ \ 0 \end{matrix}$

$=\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0$

$=\frac{1}{2} \ln \left (1 + \sqrt{2} \right )$

Compare your answer with the correct one above

Question

Give the arclength of the graph of the function $f(x) = \frac{2}{3} x \sqrt{x}$ on the interval $\left [ 0,3 \right ]$ .

Answer

The length of the curve of on the interval can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} ; \mathrm{d}x$ .

$f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}$

so

$f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}$ .

The above integral becomes

$\int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} ; \mathrm{d}x$

$\int_{0}^{3} \sqrt{x+1} ; \mathrm{d}x$

Substitute . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , $\mathrm{d} u = \mathrm{d} x$ , and the integral becomes

$\int_{1}^{4} \sqrt{u} ; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} ; \mathrm{d}u$

$= \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \ \ \end{matrix}\right| \begin{matrix} 4\ \ 1 \end{matrix}$

$= \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \ \ \end{matrix}\right| \begin{matrix} 4\ \ 1 \end{matrix}$

$= \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}$

$= \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}$

Compare your answer with the correct one above

Tap the card to reveal the answer