12.4E: Exercises for Section 12.4
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- Jan 17, 2020
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Determining Arc Length
In questions 1 - 5, find the arc length of the curve on the given interval.
1) ⇀r(t)=t2ˆi+(2t2+1)ˆj,1≤t≤3. This portion of the graph is shown here:
- Answer
- 8√5 units ≈17.9 units
2) ⇀r(t)=t2ˆi+14tˆj,0≤t≤7. This portion of the graph is shown here:
3) ⇀r(t)=⟨t2+1,4t3+3⟩,−1≤t≤0. This portion of the graph is shown here:
- Answer
- 154(373/2−1) units ≈4.15 units
4) ⇀r(t)=⟨2sint,5t,2cost⟩,0≤t≤π. This portion of the graph is shown here:
5) ⇀r(t)=⟨e−tcost,e−tsint⟩ over the interval [0,π2]. 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 ⇀r(t)=⟨t,t4⟩. 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 ⇀r(t)=12costˆi+12sintˆj+√32tˆk.
- Answer
- Length =2π units
8) Find the arc length of the vector-valued function ⇀r(t)=−tˆi+4tˆj+3tˆk over [0,1].
9) A particle travels in a circle with the equation of motion ⇀r(t)=3costˆi+3sintˆj+0ˆk. 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 ⇀r(t)=costˆi+2sintˆj+0ˆk.
11) Find the length of the curve ⇀r(t)=⟨√2t,et,e−t⟩ over the interval 0≤t≤1. The graph is shown here:
- Answer
- (e−1e) units ≈2.35 units
12) Find the length of the curve ⇀r(t)=⟨2sint,5t,2cost⟩ for t∈[−10,10].
Unit Tangent Vectors and Unit Normal Vectors
13) The position function for a particle is ⇀r(t)=acos(ωt)ˆi+bsin(ωt)ˆj. Find the unit tangent vector and the unit normal vector at t=0.
- Solution:
- ⇀r′(t)=−aωsin(ωt)ˆi+bωcos(ωt)ˆj‖⇀r′(t)‖=√a2ω2sin2(ωt)+b2ω2cos2(ωt)⇀T(t)=⇀r′(t)‖⇀r′(t)‖=−aωsin(ωt)ˆi+bωcos(ωt)ˆj√a2ω2sin2(ωt)+b2ω2cos2(ωt)⇀T(0)=bωˆj√(bω)2=bωˆj|bω|
If bω>0,⇀T(0)=ˆj, and if bω<0,T(0)=−ˆj
- Answer
- If bω>0,⇀T(0)=ˆj, and if bω<0,⇀T(0)=−ˆj
If a>0,⇀N(0)=−ˆi, and if a<0,⇀N(0)=ˆi
14) Given ⇀r(t)=acos(ωt)ˆi+bsin(ωt)ˆj, find the binormal vector ⇀B(0).
15) Given ⇀r(t)=⟨2et,etcost,etsint⟩, determine the unit tangent vector ⇀T(t).
- Answer
- ⇀T(t)=⟨2√6,cost−sint√6,cost+sint√6⟩=⟨√63,√66(cost−sint),√66(cost+sint)⟩
16) Given ⇀r(t)=⟨2et,etcost,etsint⟩, find the unit tangent vector ⇀T(t) evaluated at t=0, ⇀T(0).
17) Given ⇀r(t)=⟨2et,etcost,etsint⟩, determine the unit normal vector ⇀N(t).
- Answer
- ⇀N(t)=⟨0,−√22(sint+cost),√22(cost−sint)⟩
18) Given ⇀r(t)=⟨2et,etcost,etsint⟩, find the unit normal vector ⇀N(t) evaluated at t=0, ⇀N(0).
- Answer
- ⇀N(0)=⟨0,−√22,√22⟩
19) Given ⇀r(t)=tˆi+t2ˆj+tˆk, find the unit tangent vector ⇀T(t). The graph is shown here:
- Answer
- ⇀T(t)=1√4t2+2<1,2t,1>
20) Find the unit tangent vector ⇀T(t) and unit normal vector ⇀N(t) at t=0 for the plane curve ⇀r(t)=⟨t3−4t,5t2−2⟩. The graph is shown here:
21) Find the unit tangent vector ⇀T(t) for ⇀r(t)=3tˆi+5t2ˆj+2tˆk.
- Answer
- ⇀T(t)=1√100t2+13(3ˆi+10tˆj+2ˆk)
22) Find the principal normal vector to the curve ⇀r(t)=⟨6cost,6sint⟩ at the point determined by t=π3.
23) Find ⇀T(t) for the curve ⇀r(t)=(t3−4t)ˆi+(5t2−2)ˆj.
- Answer
- ⇀T(t)=1√9t4+76t2+16((3t2−4)ˆi+10tˆj)
24) Find ⇀N(t) for the curve ⇀r(t)=(t3−4t)ˆi+(5t2−2)ˆj.
25) Find the unit tangent vector ⇀T(t) for ⇀r(t)=⟨2sint,5t,2cost⟩.
- Answer
- ⇀T(t)=⟨2√2929cost,5√2929,−2√2929sint⟩
26) Find the unit normal vector ⇀N(t) for ⇀r(t)=⟨2sint,5t,2cost⟩.
- Answer
- ⇀N(t)=⟨−sint,0,−cost⟩
Arc Length Parameterizations
27) Find the arc-length function ⇀s(t) for the line segment given by ⇀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 ⇀r(t): ⇀r(s)=(3−3s5)ˆi+4s5ˆj
28) Parameterize the helix ⇀r(t)=costˆi+sintˆj+tˆk 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 ⇀r(t)=etsintˆi+etcostˆj
- Answer
- ⇀r(s)=(1+s√2)sin(ln(1+s√2))ˆi+(1+s√2)cos(ln(1+s√2))ˆj
Curvature and the Osculating Circle
30) Find the curvature of the curve ⇀r(t)=5costˆi+4sintˆj 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 ⇀r(t)=5costˆi+5sintˆj. Does the curvature depend upon the parameter t?
33) Find the curvature κ for the curve y=x−14x2 at the point x=2.
- Answer
- 12
34) Find the curvature κ for the curve y=13x3 at the point x=1.
35) Find the curvature κ of the curve ⇀r(t)=tˆi+6t2ˆj+4tˆk. The graph is shown here:
- Answer
- κ≈49.477(17+144t2)3/2
36) Find the curvature of ⇀r(t)=⟨2sint,5t,2cost⟩.
37) Find the curvature of ⇀r(t)=√2tˆi+etˆj+e−tˆk at point P(0,1,1).
- Answer
- 12√2
38) At what point does the curve y=ex have maximum curvature?
39) What happens to the curvature as x→∞ for the curve y=ex?
- Answer
- The curvature approaches zero.
40) Find the point of maximum curvature on the curve y=lnx.
41) Find the equations of the normal plane and the osculating plane of the curve ⇀r(t)=⟨2sin(3t),t,2cos(3t)⟩ at point (0,π,−2).
- Answer
- y=6x+π and x+6y=6π
42) Find equations of the osculating circles of the ellipse 4y2+9x2=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 ⇀r(t)=cos(2t)ˆi+sin(2t)ˆj+tˆk.
- Answer
- x+2z=π2
44) Find the radius of curvature of 6y=x3 at the point (2,43).
45) Find the curvature at each point (x,y) on the hyperbola ⇀r(t)=⟨acosh(t),bsinh(t)⟩.
- Answer
- a4b4(b4x2+a4y2)3/2
46) Calculate the curvature of the circular helix ⇀r(t)=rsin(t)ˆi+rcos(t)ˆj+tˆk.
47) Find the radius of curvature of y=ln(x+1) at point (2,ln3).
- Answer
- 10√103
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 ⇀r(t)=tˆi+t2ˆj. Use this parameterization to answer questions 49 - 51.
49) Find the length of the curve over the interval [0,2].
- Answer
- 14[4√17+ln(4+√17)] units ≈4.64678 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=0.25x1.6 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=30x2/5(25+4x6/5)3/2
Note that initially your answer may be:
625x2/5(1+425x6/5)3/2
We can simplify it as follows:
625x2/5(1+425x6/5)3/2=625x2/5[125(25+4x6/5)]3/2=625x2/5(125)3/2[25+4x6/5]3/2=625125x2/5[25+4x6/5]3/2=30x2/5(25+4x6/5)3/2
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.