ASSIGNMENT 2 SOLUTION
JAMES MCIVOR
1. Supplementary Problems 2, # 1a,b
2 POINTS Note - they do not have to evaluate the integrals
(a) Let C be the spiral curve which is parametrized by (t cos t, t sin t) as t goes from 0 to 4. Set up
the integral which computes the arc length of C [You don’t need to calculate the integral].
(b) The parametrization (t2 cos t2 , t2 sin t2 ) traces out the same spiral, but at slightly different speeds.
Check that the arc length of C is the same using this parametrization as it was above [Hint: don’t
calculate either of the integrals, but clearly explain why they they would come out to the same
thing if you did calculate them.]
Solution: (a) Using the arc length formula and simplifying gives
Z 4p
t2 + 1dt
L(C) =
0
(b) Plugging in this new parametrization to the arc length formula gives the integral
Z 2 p
L(C) =
2t 1 + t4 dt,
0
where the bounds of the integral are from 0 to 2 because this new parametrization traces out the
same curve in less time: it gets to the point (4 cos 4, 4 sin 4) at time t = 2.
This integral is actually the same as the integral from (a): by making a substitution u = t2 in
R4√
this integral, one obtains L(C) = 0 u2 + 1du, which is the same integral as in (a).
2. Stewart 10.2.52
2 POINTS Note they do not have to evaluate the integrals
Find the distance traveled by a particle moving according to the parametrization (cos2 t, cos t),
0 ≤ t ≤ 4π, and compare this with the arc length of the curve.
Solution: First notice that the x-component of the curve is the square of the y-component.
Therefore the particle stays on the curve cut out by x = y 2 . As t goes from 0 to 4π, the y-coordinate
goes from 1 to -1 and back two full times. This means that the particle moves back and forth along
the same stretch of curve four times - twice down and twice up. Therefore the distance traveled by
the particle is four times the length of the curve. To compute the length of the curve we enter this
parametrization into the arc length formula, which gives
Z 2p
L(C) =
t2 + 1 dt
0
It is not necessary to calculate this integral.
3. Stewart 10.2.73
3 points.
A string is wound around a circle and then unwound while being held taut. The curve traced by
the point P at the end of the string is called the involute of the circle. If the circle has radius r and
center O and the initial position of P is (r, 0), and if the parameter θ is chosen as in the figure,
show that the parametric equations of the involute are
y(θ) = r(sin θ − θ cos θ)
x(θ) = r(cos θ + θ sin θ);
1
2
JAMES MCIVOR
Solution: We need the coordinates of the point P , which will depend on r and θ. The key
observation is that the length of the segment T P is just the amount of string unwound, which is the
length of the arc of the circle corresponding to angle θ. The length of this arc is rθ. Referring to
the figure, we can find the coordinates of P by starting with the coordinates of T and subtracting
the horizontal and vertical components of the segment T P .
First off, the point T has coordinates (r cos θ, r sin θ), since it lies on the circle. Now draw a
horizontal line passing through T . The angle that T P makes with this horizontal line is π/2 − θ.
Therefore the horizontal component of the segment T P is rθ cos(π/2 − θ) (remember that the length
of T P is rθ. This is equal to rθ sin θ, since cos(π/2 − θ) = sin θ. Also the vertical component of T P
is −rθ sin(π/2 − θ) = −rθ(− cos θ) = rθ cos θ. Now if we add and subtract these two from the x and
y coordinates (respectively) of T we get
x(θ) = r cos θ + rθ sin θ = r(cos θ + θ sin θ)
y(θ) = r sin θ − rθ cos θ = r(sin θ − θ cos θ)
4. Stewart 14.1.26
2 points - be merciful with their drawings - I can’t draw very well, either. If they
include correct descriptions along with a bad picture because they cannot draw well,
I think that’s worth full credit, too, as long as the description indicates what type of
surface it is.
Sketch the graph of f (x, y) = 3 − x2 − y 2 .
Solution: Fixing different values of z gives level sets that are circles of radius 3 − z. The surface
is a circular paraboloid opening downward. Wolfram gives
5. Stewart 14.1.34
2 points - be merciful with their drawings, as above
Make of rough sketch of the contour map of the surface
ASSIGNMENT 2 SOLUTION
3
Solution It looks a little bit like the eyes of an owl.
6. Stewart 14.1.38
2 points - be merciful with their drawings, as above
Sketch the graph of the function whose contour map is
Solution: This surface is a pyramid with four walls, like the top half of an octahedron. In fact
it is the graph of a function such as f (x, y) = 3 − |x| − |y|, whose graph is:
7. Stewart 14.1.40
2 points - make sure to take off one point if they do not label each level set with the
appropriate z-value
Draw a contour map of the function f(x, y) = x3 − y showing several level curves.
Solution The top curve is the level set for z = −1, below that, the one for z = 0, etc.
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JAMES MCIVOR