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Calculus 2 : Parametric Form

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Example Questions

Example Question #91 : Parametric

Find the arc length of the curve (Round to three significant digits):

$x=\frac{1}{4}t^3,\,y=\frac{1}{10}t^2,\,0\leq t\leq 1$

Possible Answers:

0.460

0.272

0.306

0.422

0.409

Correct answer:

0.272

Explanation:

Finding the length of the curve requires simply applying the formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Where:

a=0, b =1

Since we are also given and , we can easily compute the derivatives of each:

$\frac{dx}{dt}=\frac{d}{dt}[\frac{1}{4}t^3]=\frac{3}{4}t^2$

$\frac{dy}{dt}=\frac{d}{dt}[\frac{1}{10}t^2]=\frac{1}{5}t$

Applying these into the above formula results in:

$L=\int_{0}^{1}\sqrt{(\frac{3}{4}t^2)^2+(\frac{1}{5}t)^2}\,dt=\int_{0}^{1}\sqrt{\frac{9}{16}t^4+\frac{1}{25}t^2}\,dt$

Looking at this integral, it seems pretty intimidating at first, but we can do some algebraic manipulation to make it look like a simple u-substitution problem. First we notice that there is a t^2 term that is common within the two separate functions. If we factor it out then the integral becomes:

$\int_{0}^{1}\sqrt{t^2[\frac{9}{16}t^2+\frac{1}{25}]}\,dt=\int_{0}^{1}t\sqrt{\frac{9}{16}t^2+\frac{1}{25}}\,dt$

Now we can apply the rules of u-substitution:

$u=\frac{9}{16}t^2+\frac{1}{25},\,du=\frac{18}{16}t\,dt=\frac{9}{8}t\,dt$

Therefore:

$\frac{8}{9}du=t\,dt$

Now we must deal with the new bounds of our integral:

$u(0)=\frac{9}{16}(0)^2+\frac{1}{25}=\frac{1}{25}$

$u(1)=\frac{9}{16}(1)^2+\frac{1}{25}=\frac{241}{400}$

Our new integral becomes:

$\frac{8}{9}\int_{\frac{1}{25}}^{\frac{241}{400}}\sqrt{u}\,du=\frac{8}{9}[\frac{2}{3}u^{\frac{3}{2}}]_{\frac{1}{25}}^{\frac{241}{400}}=\frac{16}{27}[(\frac{241}{400})^\frac{3}{2}-(\frac{1}{25})^\frac{3}{2}]$

Plugging this result into a calculator results in:

0.2723945261

Because the problem statement requires us to round this to three significant figures, the final result is:

0.272

This is one of the answer choices.

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Example Question #92 : Parametric Form

Find the arc length of the curve:

$x=8\cos(3t),\,y=8\sin(3t),\,0\leq t\leq \frac{\pi}{2}$

Possible Answers:

$\frac{3\pi}{2}$

$4\pi$

$12\pi$

$8\pi$

$24\pi$

Correct answer:

$12\pi$

Explanation:

Finding the length of the curve requires simply applying the formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Where:

$a=0, b =\frac{\pi}{2}$

Since we are also given and , we can easily compute the derivatives of each:

$\frac{dx}{dt}=\frac{d}{dt}[8\cos(3t)]=-8\sin(3t)\cdot3=-24\sin(3t)$

$\frac{dy}{dt}=\frac{d}{dt}[8\sin(3t)]=8\cos(3t)\cdot3=24\cos(3t)$

Applying these into the above formula results in:

$L=\int_{0}^{\frac{\pi}{2}}\sqrt{(-24\sin(3t))^2+(24\cos(3t))^2}\,dt=\int_{0}^{\frac{\pi}{2}}\sqrt{24^2\sin^2(3t)+24^2\cos^2(3t)}\,dt$ We can factor out the common 24^2 , and pull it outside of the square-root, and we will notice one of the most common trigonometric identities:

$\int_{0}^{\frac{\pi}{2}}\sqrt{24^2[\sin^2(3t)+\cos^2(3t)]}\,dt=\int_{0}^{\frac{\pi}{2}}24\sqrt{\sin^2(3t)+\cos^2(3t)}\,dt$

The term inside the square-root symbol can be simplified to .

$24\int_{0}^{\frac{\pi}{2}}\sqrt{1}\,dt=24\int_{0}^{\frac{\pi}{2}}dt=24[t]_{0}^{\frac{\pi}{2}}=24[\frac{\pi}{2}-0]=\frac{24\pi}{2}=12\pi$

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Example Question #93 : Parametric Form

Find the arc length of the curve:

$x=e^{3t}\cos(3t),\,y=e^{3t}\sin(3t),\,0 \leq t\leq \frac{\pi}{2}$

Possible Answers:

$\sqrt{2}e^{\frac{3\pi}{2}}$

$3\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

$\frac{1}{3}(e^{\frac{3\pi}{2}}-1)$

$3\sqrt{2}e^{\frac{3\pi}{2}}$

$\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

Correct answer:

$\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

Explanation:

Finding the length of the curve requires simply applying the formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Where:

$a=0, b =\frac{\pi}{2}$

Since we are also given and , we can easily compute the derivatives of each, using the Product Rule:

$\frac{dx}{dt}=\frac{d}{dt}[e^{3t}\cos(3t)]=-3e^{3t}\sin(3t)+3e^{3t}\cos(3t)=3e^{3t}[\cos(3t)-\sin(3t)]$

$\frac{dy}{dt}=\frac{d}{dt}[e^{3t}\sin(3t)]=3e^{3t}\cos(3t)+3e^{3t}\sin(3t)=3e^{3t}[\cos(3t)+\sin(3t)]$

Applying these into the above formula results in:

$L=\int_{0}^{\frac{\pi}{2}}\sqrt{(3e^{3t}[\cos(3t)-\sin(3t)])^2+(3e^{3t}[\cos(3t)+\sin(3t)])^2}\,dt$

Simplifying the above will require these two formulas:

$(a+b)^2=a^2+2ab+b^2 \rightarrow \,(1)$

$(a-b)^2=a^2-2ab+b^2\rightarrow\,(2)$

It may also be useful to know this formula:

$(x^a)^b=x^{ab}\rightarrow\,(3)$

We can factor out the common $(3e^{3t})^2$ to make the above expression easier to look at:

$L=\int_{0}^{\frac{\pi}{2}}\sqrt{(3e^{3t})^2([\cos(3t)-\sin(3t)]^2+[\cos(3t)+\sin(3t)])^2}\,dt$

We can take the $(3e^{3t})^2$ outside of the square-root by cancelling out the representing the "square". Then we can apply formulas (1) & (2) to the trigonometric expressions:

$L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{[\cos^2(3t)-2\cos(3t)\sin(3t)+\sin^2(3t)]+[\cos^2(3t)+2\cos(3t)\sin(3t)+\sin^2(t)]}\,dt$

We can now simplify the terms inside the square-root to get:

$L=3e^{3t}\int_{0}^{\frac{\pi}{2}}\sqrt{[2\cos^2(3t)+2\sin^2(3t)]}\,dt$

If we factor out the common "2" above, we are left with the trigonometric identity, which simplifies to , since:

$\sin^2(t)+\cos^2(t)=1$

Therefore the integral now becomes:

$L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{2[\cos^2(3t)+\sin^2(3t)]}\,dt$

$L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{2(1)}\,dt=\int_{0}^{\frac{\pi}{2}}3\sqrt{2}e^{3t}\,dt$

This is a simple integral which can be solved using u-substitution. But first, we can factor out the constant term $3\sqrt{2}$ , outside of the integral:

$L=3\sqrt{2}\int_{0}^{\frac{\pi}{2}}e^{3t}\,dt$

We will make our substitutions:

$u=3t,\,du=3\,dt\rightarrow\frac{1}{3}du=dt$

We also need to change the bounds of the new integral:

u(0)=3(0)=0

$u(\frac{\pi}{2})=3(\frac{\pi}{2})=\frac{3\pi}{2}$

Our new integral becomes:

$L=3\sqrt{2}\int_{0}^{\frac{3\pi}{2}}\frac{1}{3}e^u\,du=\sqrt{2}[e^u]_{0}^{\frac{3\pi}{2}}=\sqrt{2}(e^{\frac{3\pi}{2}}-e^{0})=\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

This is one of the answer choices.

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Calculus 2 : Parametric Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #91 : Parametric

Find the arc length of the curve (Round to three significant digits):

Example Question #92 : Parametric Form

Find the arc length of the curve:

Example Question #93 : Parametric Form

Find the arc length of the curve:

All Calculus 2 Resources

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