Exercises for Section 12.4
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Determining Arc Length
In questions 1 - 6, find the arc length of the curve on the given interval.
1) \(\vecs r(t)=t^2 \,\hat{\mathbf{i}}+(2t^2+1)\,\hat{\mathbf{j}}, \quad 1≤t≤3\)
- Answer:
- \(8\sqrt{5}\) 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) \(\vecs r(t)=⟨t^2+1,4t^3+3⟩, \quad −1≤t≤0\)
- Answer:
- \(\frac{1}{54}(37^{3/2}−1)\) 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)
7) Find the length of one turn of the helix given by \(\vecs r(t)= \frac{1}{2} \cos t \,\hat{\mathbf{i}}+\frac{1}{2} \sin t \,\hat{\mathbf{j}}+\sqrt{\frac{3}{4}}t \,\hat{\mathbf{k}}\).
- 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
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 \(t=0\).
- Answer:
- \(\vecs T(0)= \hat{\mathbf{j}}, \quad \vecs N(0)=−\hat{\mathbf{i}}\)
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 tangent vector \(\vecs T(t)\).
- Answer:
- \(\vecs T(t)=⟨2e^t,e^t \cos t−e^t \sin t,e^t \cos t+e^t \sin t⟩\)
16) Given \(\vecs r(t)=⟨2e^t,e^t \cos t,e^t \sin t⟩\), determine the unit tangent vector \(\vecs T(t)\) evaluated at \(t=0\).
17) 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:
- \(\vecs N(0)=⟨\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2}⟩\)
18) Given \(\vecs r(t)=⟨2e^t,e^t \cos t,e^t \sin t⟩\), find the unit normal vector evaluated at \(t=0\).
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:
- \(\vecs T(t)=\frac{1}{\sqrt{4t^2+2}}<1,2t,1>\)
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:
- \(\vecs T(t)=\frac{1}{\sqrt{100t^2+13}}(3 \,\hat{\mathbf{i}}+10t \,\hat{\mathbf{j}}+2 \,\hat{\mathbf{k}})\)
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:
- \(\vecs T(t)=\frac{1}{\sqrt{9t^4+76t^2+16}}([3t^2−4]\,\hat{\mathbf{i}}+10t \,\hat{\mathbf{j}})\)
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 normal vector \(\vecs N(t)\) for \(\vecs r(t)=⟨2sint,5t,2cost⟩\).
- Answer:
- \(\vecs N(t)=⟨−\sin t,0,−\cos t⟩\)
26) Find the unit tangent vector \(\vecs T(t)\) for \(\vecs r(t)=⟨2 \sin t,5t,2 \cos t⟩\).
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 \(\vecs r(t)\): \(\vecs r(s)=(3−\frac{3s}{5})\mathbf{i}+\frac{4s}{5}\mathbf{j}\)
28) Parameterize the helix \(\mathrm{r(t)= \cos t \mathbf{i}+ \sin t \mathbf{j}+t \mathbf{k}}\) using the arc-length parameter s, from \(\mathrm{t=0}\).
29) Parameterize the curve using the arc-length parameter s, at the point at which \(\mathrm{t=0}\) for \(\mathrm{r(t)=e^t \sin t \mathbf{i} + e^t \cos t \mathbf{j}}\)
- Answer:
- \(\mathrm{(s)=(1+\frac{s}{\sqrt{2}}) \sin ( \ln (1+ \frac{s}{\sqrt{2}})) \mathbf{i} +(1+ \frac{s}{\sqrt{2}}) \cos [ \ln (1+\frac{s}{\sqrt{2}})]\mathbf{j}}\)
Curvature and the Osculating Circle
30) Find the curvature of the curve \(\mathrm{r(t)=5 \cos t \mathbf{i}+4 \sin t \mathbf{j}}\) at \(\mathrm{t=π/3}\). (Note: The graph is an ellipse.)
31) Find the \(x\)-coordinate at which the curvature of the curve \(\mathrm{y=1/x}\) is a maximum value.
- Answer:
- The maximum value of the curvature occurs at \(\mathrm{x=\sqrt[4]{5}}\).
32) Find the curvature of the curve \(\mathrm{r(t)=5 \cos t \mathbf{i}+5 \sin t \mathbf{j}}\). Does the curvature depend upon the parameter t?
33) Find the curvature \(κ\) for the curve \(\mathrm{y=x−\frac{1}{4}x^2}\) at the point \(\mathrm{x=2}\).
- Answer:
- \(\mathrm{\frac{1}{2}}\)
34) Find the curvature \(κ\) for the curve \(\mathrm{y=\frac{1}{3}x^3}\) at the point \(\mathrm{x=1}\).
35) Find the curvature \(κ\) of the curve \(\mathrm{r(t)=t \mathbf{i}+6t^2 \mathbf{j}+4t \mathbf{k}}\). The graph is shown here:
- Answer:
- \(\mathrm{κ≈\frac{49.477}{(17+144t^2)^{3/2}}}\)
36) Find the curvature of \(\mathrm{r(t)=⟨2 \sin t,5t,2 \cos t⟩}\).
37) Find the curvature of \(\mathrm{r(t)=\sqrt{2}t \mathbf{i}+e^t \mathbf{j}+e^{−t} \mathbf{k}}\) at point \(\mathrm{P(0,1,1)}\).
- Answer:
- \(\mathrm{\frac{1}{2\sqrt{2}}}\)
38) At what point does the curve \(\mathrm{y=e^x}\) have maximum curvature?
39) What happens to the curvature as \(\mathrm{x→∞}\) for the curve \(\mathrm{y=e^x}\)?
- Answer:
- The curvature approaches zero.
40) Find the point of maximum curvature on the curve \(\mathrm{y=\ln x}\).
41) Find the equations of the normal plane and the osculating plane of the curve \(\mathrm{r(t)=⟨2 \sin (3t),t,2 \cos (3t)⟩}\) at point \(\mathrm{(0,π,−2)}\).
- Answer:
- \(\mathrm{y=6x+π}\) and \(\mathrm{x+6=6π}\)
42) Find equations of the osculating circles of the ellipse \(\mathrm{4y^2+9x^2=36}\) at the points \(\mathrm{(2,0)}\) and \(\mathrm{(0,3)}\).
43) Find the equation for the osculating plane at point \(\mathrm{t=π/4}\) on the curve \(\mathrm{r(t)=\cos (2t) \mathbf{i}+ \sin (2t) \mathbf{j}+t}\).
- Answer:
- \(\mathrm{x+2z=\frac{π}{2}}\)
44) Find the radius of curvature of \(\mathrm{6y=x^3}\) at the point \(\mathrm{(2,\frac{4}{3}).}\)
45) Find the curvature at each point \(\mathrm{(x,y)}\) on the hyperbola \(\mathrm{r(t)=⟨a \cosh( t),b \sinh (t)⟩}\).
- Answer:
- \(\mathrm{\frac{a^4b^4}{(b^4x^2+a^4y^2)^{3/2}}}\)
46) Calculate the curvature of the circular helix \(\mathrm{r(t)=r \sin (t) \mathbf{i}+r \cos (t) \mathbf{j}+t \mathbf{k}}\).
47) Find the radius of curvature of \(\mathrm{y= \ln (x+1)}\) at point \(\mathrm{(2,\ln 3)}\).
- Answer:
- \(\mathrm{\frac{10\sqrt{10}}{3}}\)
48) Find the radius of curvature of the hyperbola \(\mathrm{xy=1}\) at point \(\mathrm{(1,1)}\).
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 \(\mathrm{[0,2]}\).
- Answer:
- \(\mathrm{\frac{38}{3}}\)
50) Find the curvature of the plane curve at \(\mathrm{t=0,1,2}\).
51) Describe the curvature as t increases from \(\mathrm{t=0}\) to \(\mathrm{t=2}\).
- Answer:
- The curvature is decreasing over this interval.
The surface of a large cup is formed by revolving the graph of the function \(\mathrm{y=0.25x^{1.6}}\) from \(\mathrm{x=0}\) to \(\mathrm{x=5}\) 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:
- \(\mathrm{κ=\frac{6}{x^{2/5}(25+4x^{6/5})}}\)
54) [T] Use technology to graph the curvature function.