12.4E: Exercises for Section 12.4

Determining Arc Length

In questions 1 - 5, find the arc length of the curve on the given interval.

1) $\vecs r(t)=t^2 \,\hat{\mathbf{i}}+(2t^2+1)\,\hat{\mathbf{j}}, \quad 1≤t≤3$. This portion of the graph is shown here:

Answer

$8\sqrt{5}$ units $\approx 17.9$ units

2) $\vecs r(t)=t^2 \,\hat{\mathbf{i}}+14t \,\hat{\mathbf{j}},\quad 0≤t≤7$. This portion of the graph is shown here:

3) $\vecs r(t)=⟨t^2+1,4t^3+3⟩, \quad −1≤t≤0$. This portion of the graph is shown here:

Answer

$\frac{1}{54}(37^{3/2}−1)$ units $\approx 4.15$ units

4) $\vecs r(t)=⟨2 \sin t,5t,2 \cos t⟩,\quad 0≤t≤π$. This portion of the graph is shown here:

5) $\vecs r(t)=⟨e^{−t \cos t},e^{−t \sin t}⟩$ over the interval $[0,\frac{π}{2}]$. Here is the portion of the graph on the indicated interval:

6) Set up an integral to represent the arc length from $t = 0$ to $t = 2$ along the curve traced out by $\vecs r(t) = \langle t, \, t^4\rangle.$ Then use technology to approximate this length to the nearest thousandth of a unit.

7) Find the length of one turn of the helix given by $\vecs r(t)= \frac{1}{2} \cos t \,\hat{\mathbf{i}}+\frac{1}{2} \sin t \,\hat{\mathbf{j}}+\frac{\sqrt{3}}{2}\,t \,\hat{\mathbf{k}}$.

Answer

Length $=2π$ units

8) Find the arc length of the vector-valued function $\vecs r(t)=−t \,\hat{\mathbf{i}}+4t \,\hat{\mathbf{j}}+3t \,\hat{\mathbf{k}}$ over $[0,1]$.

9) A particle travels in a circle with the equation of motion $\vecs r(t)=3 \cos t \,\hat{\mathbf{i}}+3 \sin t \,\hat{\mathbf{j}} +0 \,\hat{\mathbf{k}}$. Find the distance traveled around the circle by the particle.

Answer

$6π$ units

10) Set up an integral to find the circumference of the ellipse with the equation $\vecs r(t)= \cos t \,\hat{\mathbf{i}}+2 \sin t \,\hat{\mathbf{j}}+0\,\hat{\mathbf{k}}$.

11) Find the length of the curve $\vecs r(t)=⟨\sqrt{2}t,\, e^t, \, e^{−t}⟩$ over the interval $0≤t≤1$. The graph is shown here:

Answer

$\left(e−\frac{1}{e}\right)$ units $\approx 2.35$ units

12) Find the length of the curve $\vecs r(t)=⟨2 \sin t, \, 5t, \, 2 \cos t⟩$ for $t∈[−10,10]$.

Unit Tangent Vectors and Unit Normal Vectors

13) The position function for a particle is $\vecs r(t)=a \cos( ωt) \,\hat{\mathbf{i}}+b \sin (ωt) \,\hat{\mathbf{j}}$. Find the unit tangent vector and the unit normal vector at $t=0$.

Solution:

$\begin{align*} \vecs r'(t) &= -aω \sin( ωt) \,\hat{\mathbf{i}}+bω \cos (ωt) \,\hat{\mathbf{j}}\\[5pt] \| \vecs r'(t) \| &= \sqrt{a^2 ω^2 \sin^2(ωt) +b^2ω^2\cos^2(ωt)} \\[5pt] \vecs T(t) &= \dfrac{\vecs r'(t)}{\| \vecs r'(t) \| } = \dfrac{-aω \sin( ωt) \,\hat{\mathbf{i}}+bω \cos (ωt) \,\hat{\mathbf{j}}}{\sqrt{a^2 ω^2 \sin^2(ωt) +b^2ω^2\cos^2(ωt)}}\\[5pt] \vecs T(0) &= \dfrac{bω \,\hat{\mathbf{j}}}{\sqrt{(bω)^2}} = \dfrac{bω \,\hat{\mathbf{j}}}{|bω|}\end{align*}$

If $bω > 0, \; \vecs T(0) = \hat{\mathbf{j}},$ and if $bω < 0, \; T(0)= -\hat{\mathbf{j}}$

Answer

If $bω > 0, \; \vecs T(0)= \hat{\mathbf{j}},$ and if $bω < 0, \; \vecs T(0)= -\hat{\mathbf{j}}$

If $a > 0, \; \vecs N(0)= -\hat{\mathbf{i}},$ and if $a < 0, \; \vecs N(0)= \hat{\mathbf{i}}$

14) Given $\vecs r(t)=a \cos (ωt) \,\hat{\mathbf{i}} +b \sin (ωt) \,\hat{\mathbf{j}}$, find the binormal vector $\vecs B(0)$.

15) Given $\vecs r(t)=⟨2e^t,e^t \cos t,e^t \sin t⟩$, determine the unit tangent vector $\vecs T(t)$.

Answer

$\begin{align*} \vecs T(t) &=\left\langle \frac{2}{\sqrt{6}},\, \frac{\cos t− \sin t}{\sqrt{6}}, \, \frac{\cos t+ \sin t}{\sqrt{6}}\right\rangle \\[4pt] &= \left\langle\frac{\sqrt{6}}{3},\, \frac{\sqrt{6}}{6} (\cos t− \sin t), \, \frac{\sqrt{6}}{6} (\cos t+ \sin t)\right\rangle \end{align*}$

16) Given $\vecs r(t)=⟨2e^t,\, e^t \cos t,\, e^t \sin t⟩$, find the unit tangent vector $\vecs T(t)$ evaluated at $t=0$, $\vecs T(0)$.

17) Given $\vecs r(t)=⟨2e^t,\, e^t \cos t,\, e^t \sin t⟩$, determine the unit normal vector $\vecs N(t)$.

Answer

$\vecs N(t)=⟨0,\, -\frac{\sqrt{2}}{2} (\sin t + \cos t), \, \frac{\sqrt{2}}{2} (\cos t- \sin t)⟩$

18) Given $\vecs r(t)=⟨2e^t,\, e^t \cos t,\, e^t \sin t⟩$, find the unit normal vector $\vecs N(t)$ evaluated at $t=0$, $\vecs N(0)$.

Answer

$\vecs N(0)=⟨0, \;-\frac{\sqrt{2}}{2},\;\frac{\sqrt{2}}{2}⟩$

19) Given $\vecs r(t)=t \,\hat{\mathbf{i}}+t^2 \,\hat{\mathbf{j}}+t \,\hat{\mathbf{k}}$, find the unit tangent vector $\vecs T(t)$. The graph is shown here:

Answer

$\vecs T(t)=\dfrac{1}{\sqrt{4t^2+2}}<1,2t,1>$

20) Find the unit tangent vector $\vecs T(t)$ and unit normal vector $\vecs N(t)$ at $t=0$ for the plane curve $\vecs r(t)=⟨t^3−4t,5t^2−2⟩$. The graph is shown here:

21) Find the unit tangent vector $\vecs T(t)$ for $\vecs r(t)=3t \,\hat{\mathbf{i}}+5t^2 \,\hat{\mathbf{j}}+2t \,\hat{\mathbf{k}}$.

Answer

$\vecs T(t)=\dfrac{1}{\sqrt{100t^2+13}}(3 \,\hat{\mathbf{i}}+10t \,\hat{\mathbf{j}}+2 \,\hat{\mathbf{k}})$

22) Find the principal normal vector to the curve $\vecs r(t)=⟨6 \cos t,6 \sin t⟩$ at the point determined by $t=\frac{π}{3}$.

23) Find $\vecs T(t)$ for the curve $\vecs r(t)=(t^3−4t) \,\hat{\mathbf{i}}+(5t^2−2) \,\hat{\mathbf{j}}$.

Answer

$\vecs T(t)=\dfrac{1}{\sqrt{9t^4+76t^2+16}}\big((3t^2−4)\,\hat{\mathbf{i}}+10t \,\hat{\mathbf{j}}\big)$

24) Find $\vecs N(t)$ for the curve $\vecs r(t)=(t^3−4t)\,\hat{\mathbf{i}}+(5t^2−2)\,\hat{\mathbf{j}}$.

25) Find the unit tangent vector $\vecs T(t)$ for $\vecs r(t)=⟨2 \sin t,\, 5t,\, 2 \cos t⟩$.

Answer

$\vecs T(t)=⟨\frac{2\sqrt{29}}{29}\cos t,\, \frac{5\sqrt{29}}{29},\,−\frac{2\sqrt{29}}{29}\sin t⟩$

26) Find the unit normal vector $\vecs N(t)$ for $\vecs r(t)=⟨2\sin t,\,5t,\,2\cos t⟩$.

Answer

$\vecs N(t)=⟨−\sin t,\, 0,\, −\cos t⟩$

Arc Length Parameterizations

27) Find the arc-length function $\vecs s(t)$ for the line segment given by $\vecs r(t)=⟨3−3t,\, 4t⟩$. Then write the arc-length parameterization of $r$ with $s$ as the parameter.

Answer

Arc-length function: $s(t)=5t$; The arc-length parameterization of $\vecs r(t)$: $\vecs r(s)=\left(3−\dfrac{3s}{5}\right)\,\hat{\mathbf{i}}+\dfrac{4s}{5}\,\hat{\mathbf{j}}$

28) Parameterize the helix $\vecs r(t)= \cos t \,\hat{\mathbf{i}}+ \sin t \,\hat{\mathbf{j}}+t \,\hat{\mathbf{k}}$ using the arc-length parameter $s$, from $t=0$.

29) Parameterize the curve using the arc-length parameter $s$, at the point at which $t=0$ for $\vecs r(t)=e^t \sin t \,\hat{\mathbf{i}} + e^t \cos t \,\hat{\mathbf{j}}$

Answer

$\vecs r(s)=\left(1+\dfrac{s}{\sqrt{2}}\right) \sin \left( \ln \left(1+ \dfrac{s}{\sqrt{2}}\right)\right)\,\hat{\mathbf{i}} +\left(1+ \dfrac{s}{\sqrt{2}}\right) \cos \left( \ln \left(1+\dfrac{s}{\sqrt{2}}\right)\right)\,\hat{\mathbf{j}}$

Curvature and the Osculating Circle

30) Find the curvature of the curve $\vecs r(t)=5 \cos t \,\hat{\mathbf{i}}+4 \sin t \,\hat{\mathbf{j}}$ at $t=π/3$. (Note: The graph is an ellipse.)

31) Find the $x$-coordinate at which the curvature of the curve $y=1/x$ is a maximum value.

Answer

The maximum value of the curvature occurs at $x=1$.

32) Find the curvature of the curve $\vecs r(t)=5 \cos t \,\hat{\mathbf{i}}+5 \sin t \,\hat{\mathbf{j}}$. Does the curvature depend upon the parameter $t$?

33) Find the curvature $κ$ for the curve $y=x−\frac{1}{4}x^2$ at the point $x=2$.

Answer

$\frac{1}{2}$

34) Find the curvature $κ$ for the curve $y=\frac{1}{3}x^3$ at the point $x=1$.

35) Find the curvature $κ$ of the curve $\vecs r(t)=t \,\hat{\mathbf{i}}+6t^2 \,\hat{\mathbf{j}}+4t \,\hat{\mathbf{k}}$. The graph is shown here:

Answer

$κ≈\dfrac{49.477}{(17+144t^2)^{3/2}}$

36) Find the curvature of $\vecs r(t)=⟨2 \sin t,5t,2 \cos t⟩$.

37) Find the curvature of $\vecs r(t)=\sqrt{2}t \,\hat{\mathbf{i}}+e^t \,\hat{\mathbf{j}}+e^{−t} \,\hat{\mathbf{k}}$ at point $P(0,1,1)$.

Answer

$\frac{1}{2\sqrt{2}}$

38) At what point does the curve $y=e^x$ have maximum curvature?

39) What happens to the curvature as $x→∞$ for the curve $y=e^x$?

Answer

The curvature approaches zero.

40) Find the point of maximum curvature on the curve $y=\ln x$.

41) Find the equations of the normal plane and the osculating plane of the curve $\vecs r(t)=⟨2 \sin (3t),t,2 \cos (3t)⟩$ at point $(0,π,−2)$.

Answer

$y=6x+π$ and $x+6y=6π$

42) Find equations of the osculating circles of the ellipse $4y^2+9x^2=36$ at the points $(2,0)$ and $(0,3)$.

43) Find the equation for the osculating plane at point $t=π/4$ on the curve $\vecs r(t)=\cos (2t) \,\hat{\mathbf{i}}+ \sin (2t) \,\hat{\mathbf{j}}+t\,\hat{\mathbf{k}}$.

Answer

$x+2z=\frac{π}{2}$

44) Find the radius of curvature of $6y=x^3$ at the point $(2,\frac{4}{3}).$

45) Find the curvature at each point $(x,y)$ on the hyperbola $\vecs r(t)=⟨a \cosh( t),b \sinh (t)⟩$.

Answer

$\dfrac{a^4b^4}{(b^4x^2+a^4y^2)^{3/2}}$

46) Calculate the curvature of the circular helix $\vecs r(t)=r \sin (t) \,\hat{\mathbf{i}}+r \cos (t) \,\hat{\mathbf{j}}+t \,\hat{\mathbf{k}}$.

47) Find the radius of curvature of $y= \ln (x+1)$ at point $(2,\ln 3)$.

Answer

$\frac{10\sqrt{10}}{3}$

48) Find the radius of curvature of the hyperbola $xy=1$ at point $(1,1)$.

A particle moves along the plane curve $C$ described by $\vecs r(t)=t \,\hat{\mathbf{i}}+t^2 \,\hat{\mathbf{j}}$. Use this parameterization to answer questions 49 - 51.

49) Find the length of the curve over the interval $[0,2]$.

Answer

$\frac{1}{4}\big[ 4\sqrt{17} + \ln\left(4+\sqrt{17}\right)\big]\text{ units }\approx 4.64678 \text{ units}$

50) Find the curvature of the plane curve at $t=0,1,2$.

51) Describe the curvature as t increases from $t=0$ to $t=2$.

Answer

The curvature is decreasing over this interval.

The surface of a large cup is formed by revolving the graph of the function $y=0.25x^{1.6}$ from $x=0$ to $x=5$ about the $y$-axis (measured in centimeters).

52) [T] Use technology to graph the surface.

53) Find the curvature $κ$ of the generating curve as a function of $x$.

Answer

$κ=\dfrac{30}{x^{2/5}\left(25+4x^{6/5}\right)^{3/2}}$

Note that initially your answer may be:
$\dfrac{6}{25x^{2/5}\left(1+\frac{4}{25}x^{6/5}\right)^{3/2}}$

We can simplify it as follows:
$\begin{align*} \dfrac{6}{25x^{2/5}\left(1+\frac{4}{25}x^{6/5}\right)^{3/2}} &= \dfrac{6}{25x^{2/5}\big[\frac{1}{25}\left(25+4x^{6/5}\right)\big]^{3/2}}\\[4pt] &= \dfrac{6}{25x^{2/5}\left(\frac{1}{25}\right)^{3/2}\big[25+4x^{6/5}\big]^{3/2}} \\[4pt] &= \dfrac{6}{\frac{25}{125}x^{2/5}\big[25+4x^{6/5}\big]^{3/2}} \\[4pt] &= \dfrac{30}{x^{2/5}\left(25+4x^{6/5}\right)^{3/2}}\end{align*}$

54) [T] Use technology to graph the curvature function.

Contributors:

• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

• Paul Seeburger (Monroe Community College) created question 6.