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Calculus 3 : 3-Dimensional Space

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #71 : 3 Dimensional Space

Find the length of the curve $<e^{2t},e^{-2t},2\sqrt{2}t>$ , from t=0 , to t=5

Possible Answers:

$e^{10}$

None of the other answers

$e^{10}-e^{-10}$

$e^{5}-e^{-5}$

Correct answer:

$e^{10}-e^{-10}$

Explanation:

The formula for the length of a parametric curve in 3-dimensional space is $l = \int_{t_0}^{t_1}\sqrt{(\frac{dx_1}{dt})^2+(\frac{dx_1}{dt})^2+(\frac{dx_3}{dt})^2} dt$

Taking dervatives and substituting, we have

$l = \int_{0}^{5}\sqrt{(2e^{2t})^2+(-2e^{-2t})^2+(2\sqrt{2})^2} dt$

$l = \int_{0}^{5}\sqrt{4e^{4t}+4e^{-4t}+8} dt$

$l = 2\int_{0}^{5}\sqrt{e^{4t}+e^{-4t}+2} dt$ . Factor a out of the square root.

$l = 2\int_{0}^{5}\sqrt{e^{4t}+e^{-4t}+2e^{2t}e^{-2t}} dt$ . "Uncancel" an $e^{2t}, e^{-2t}$ next to the . Now there is a perfect square inside the square root.

$l = 2\int_{0}^{5}\sqrt{(e^{2t}+e^{-2t})^2} dt$ . Factor

$l = 2[\frac{1}{2}e^{2t}-\frac{1}{2}e^{-2t}]_{0}^{5}$ . Take the square root, and integrate.

$=2(\frac{e^{10}}{2}-\frac{e^{-10}}{2})-2({0})$

$=e^{10}-e^{-10}$

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Example Question #3 : Arc Length And Curvature

Find the length of the arc drawn out by the vector function $F(t) = <a\cos t, a\sin t, at>$ with a>0 from t = 0 to $t =\pi$ .

Possible Answers:

$\pi$

$\sqrt2 a \pi$

None of the other answers

$2a^2\pi$

$\pi a$

Correct answer:

$\sqrt2 a \pi$

Explanation:

To find the arc length of a function, we use the formula

$\int_{t_{0}}^{t_{1}} \sqrt{(\frac{dx}{dt})^2 +(\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt$ .

Using $t_0 =0, t_1 =\pi$ we have

$=\int_0^\pi \sqrt{(-a\sin t)^2+(a\cos t)^2 +(a)^2} dt$

$=\int_0^\pi \sqrt{a^2\sin^2 t+a^2\cos^2 t +a^2} dt$

$=\int_0^\pi \sqrt{a^2(\sin^2 t+\cos^2 t) +a^2} dt$

$=\int_0^\pi \sqrt{a^2 +a^2} dt$

$=[\sqrt2at]_0^\pi$

$=\sqrt{2} a \pi$

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Example Question #72 : 3 Dimensional Space

Evaluate the curvature of the function y=10x^2 at the point (1,10) .

Possible Answers:

$\frac{20}{401^{3/2}}$

$\frac{20}{400^{2/3}}$

$\frac{20}{400^{3/2}}$

$\frac{20}{401^{2/3}}$

Correct answer:

$\frac{20}{401^{3/2}}$

Explanation:

The formula for curvature of a Cartesian equation is $\kappa(x)= \frac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$ . (It's not the easiest to remember, but it's the most convenient form for Cartesian equations.)

We have f'(x) =20x, f''(x) = 20 , hence

$\kappa(x)= \frac{|20|}{[1+(20x)^2]^{3/2}}$

and $\kappa(1)= \frac{|20|}{[1+(20(1))^2]^{3/2}} = \frac{20}{401^{3/2}}$ .

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Example Question #1 : Arc Length And Curvature

Find the length of the parametric curve

$\vec{r}(t)= \left< \frac{\sqrt{10}}{2}t^2, \frac{1}{3}t^3,5t+14\right>$

for $0\leq t \leq 2$ .

Possible Answers:

$\frac{38}{3}$

$\frac{25}{3}$

$\frac{10}{3}$

Correct answer:

$\frac{38}{3}$

Explanation:

To find the solution, we need to evaluate

$L = \int_0^2 \left| \vec{r}(t)' \right| dt$ .

First, we find

$\vec{r}(t)' = \left< \sqrt{10}t, t^2, 5\right>$ , which leads to

$\left| \vec{r}(t)' \right| = \sqrt{(\sqrt{10}t)^2+(t^2)^2+(5)^2}$

$\left| \vec{r}(t)' \right| = \sqrt{t^4+10t^2+25}=\sqrt{(t^2+5)^2}=t^2+5$ .

So we have a final expression to integrate for our answer

$L= \int_0^2 \left| \vec{r}(t)' \right| dt = \int_0^2 \left(t^2+5 \right )dt= \frac{(2)^3}{3}+5(2)=\frac{38}{3}$

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Example Question #7 : Arc Length And Curvature

Determine the length of the curve given below on the interval 0<t<2

$\mathbf{r}=[2\sin t,\sqrt{5} t, 2\cos t]$

Possible Answers:

Correct answer:

Explanation:

The length of a curve r is given by:

$L=\int_{t_1}^{t_2}|\mathbf{r'}(t)|dt$

To solve:

$\mathbf{r}'(t)=(2\cos t, \sqrt{5}, 2 \sin t)$

$|\mathbf{r'}(t)|=\sqrt{4\cos^2t+5+4\sin^2t}=\sqrt{4+5}=\sqrt{9}=3$

$L=\int^2_0 3 dt=3t|^2_0=6$

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Example Question #8 : Arc Length And Curvature

Find the arc length of the curve

c(t)=<3t,2sin(2t),2cos(2t)>

on the interval

$0\leq t\leq 10$

Possible Answers:

Correct answer:

Explanation:

To find the arc length of the curve function

c(t)

on the interval $a\leq t\leq b$

we follow the formula

$\int_a^b \sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2}\,dt$

For the curve function in this problem we have

$\frac{\mathrm{dx} }{\mathrm{d} t}=3$

$\frac{\mathrm{dy} }{\mathrm{d} t}=4cos(2t)$

$\frac{\mathrm{dz} }{\mathrm{d} t}=-4sin(2t)$

and following the arc length formula we solve for the integral

$\int_0^{10} \sqrt{(3)^2+(4cos(2t))^2+(-4sin(2t))^2}\,dt$

$=\int_0^{10} \sqrt{9+16sin^2(2t)+16cos^2(2t)}\,dt$

$=\int_0^{10} \sqrt{9+16(sin^2(2t)+cos^2(2t))}\,dt$

$=\int_0^{10} \sqrt{9+16}\,dt$

$=\int_0^{10} \sqrt{25}\,dt$

$=\int_0^{10} 5\,dt$

$=5t|_0^{10}$

=5(10)-5(0)

Hence the arc length is

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Example Question #601 : Calculus 3

Find the arc length of the curve function

$c(t)=<3t,4t,\frac{2}{3}t^{\frac{3}{2}}>$

On the interval $0\leq t\leq5$

Round to the nearest tenth.

Possible Answers:

35.5

45.4

20.5

26.2

Correct answer:

26.2

Explanation:

To find the arc length of the curve function

c(t)

on the interval $a\leq t\leq b$

we follow the formula

$\int_a^b \sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2}\,dt$

For the curve function in this problem we have

$\frac{\mathrm{dx} }{\mathrm{d} t}=3$

$\frac{\mathrm{dy} }{\mathrm{d} t}=4$

$\frac{\mathrm{dz} }{\mathrm{d} t}=t^{\frac{1}{2}}$

and following the arc length formula we solve for the integral

$\int_0^{5} \sqrt{(3)^2+(4)^2+(t^{\frac{1}{2}})^2}\,dt$

$=\int_0^{10} \sqrt{t+25}\,dt$

Using u-substitution, we have

u=t+25 and du=dt

The integral then becomes

$=\int_{25}^{30} \sqrt{u}\,du$

$=\frac{2}{3}u^{\frac{3}{2}}|_{25}^{30}$

$=\frac{2}{3}([30^{\frac{3}{2}}]-[25^{\frac{3}{2}}])$

Hence the arc length is 26.2

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Example Question #1 : Arc Length And Curvature

Given that a curve is defined by $\vec{r}=<2\sin 2t,2\cos 2t,4t>$ , find the arc length in the interval $0\leq t \leq \frac{\pi}{2}$

Possible Answers:

$4\sqrt{\pi}$

$\sqrt{2}\pi$

$4\sqrt{2}\pi$

$2\sqrt{2}\pi$

Correct answer:

$2\sqrt{2}\pi$

Explanation:

Untitled

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Example Question #603 : Calculus 3

Find the arc length of the parametric curve

$c(t)=<\frac{2}{3}t^{3/2}, 5t, 3>$

on the interval $0\leq t \leq 11$ .

Round to the nearest tenth.

Possible Answers:

81.3

57.1

96.6

60.7

Correct answer:

60.7

Explanation:

To find the arc length of the curve function

c(t)

on the interval

$a\leq t \leq b$

we follow the formula

$\int_{a}^{b}\sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2}\,dt$

For the curve function in this problem we have

$\frac{\mathrm{d} x}{\mathrm{d} t}=t^{1/2}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=5$

$\frac{\mathrm{d} z}{\mathrm{d} t}=0$

and following the arc length formula we solve for the integral

$\int_{0}^{11}\sqrt{(t^{1/2})^2+(5)^2+(0)^2}\,dt$

$\int_{0}^{11}\sqrt{t+25}\,dt$

And using u-substitution, we set u=t+25 and then solve the integral

$\int_{25}^{36}\sqrt{u}\,du$

$=\frac{2}{3}u^{3/2}|_{25}^{36}$

$=\frac{2}{3}([36]^{3/2}-[25]^{3/2})$

$=\frac{2}{3}(216-125)$

$=\frac{2}{3}(91)$

Which is approximately

60.7 units

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Example Question #601 : Calculus 3

Determine the curvature of the vector $\left \langle 3t^2,5t,0\right \rangle$ .

Possible Answers:

$k=\frac{30}{(36t^2+25)^{\frac{1}{2}}}$

$k=\frac{30}{(36t^2+25)}$

$k=\frac{30}{(36t^2+25)^{\frac{3}{2}}}$

$k=\frac{25}{(36t^2+25)^{\frac{3}{2}}}$

Correct answer:

$k=\frac{30}{(36t^2+25)^{\frac{3}{2}}}$

Explanation:

Using the formula for curvature $k=\frac{\left \| r'(t)\times r''(t) \right \|}{\left \| r'(t)\right \|^3}$ . r'(t)=6ti+5j, r''(t)=6i , $r'(t)\times r''(t) =-30k, \left \| r'(t)\times r''(t)\right \|=30$ , and $\left \| r'(t)\right \|\sqrt{36t^2+25}$ . Plugging into the formula, we get $k=\frac{30}{(36t^2+25)^{\frac{3}{2}}}$

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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #71 : 3 Dimensional Space

Example Question #3 : Arc Length And Curvature

Example Question #72 : 3 Dimensional Space

Example Question #1 : Arc Length And Curvature

Example Question #7 : Arc Length And Curvature

Example Question #8 : Arc Length And Curvature

Example Question #601 : Calculus 3

Example Question #1 : Arc Length And Curvature

Example Question #603 : Calculus 3

Example Question #601 : Calculus 3

All Calculus 3 Resources

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