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Calculus II (UN1102) Section 2
Midterm # 2 Practice Problems with Answers
Semon Rezchikov
Posted: March 27, 2020
Thegoal of this assignment is to give you a few more miscellaneous practice problems for the first midterm.
These will be harder than the problems on the midterm; the problems on the midterm will be of a similar
flavor.
Problem 1.
Take a ball of radius b centered at zero. Remove all points from the sphere that are of distance less than
a from the x axis; the set of such points is a cylinder. A picture is drawn below. Here b is the radius of the
circle that is the crossection of the cylinder, and a is the radius of the large ball.
What is the volume of the resulting solid?
Solution. The figure is the solid of revolution associated to rotating a region in the xy plane along the
x axis. We use the washer method. Intersecting the solid with the xy plane and looking only on the part
with positive y value we get a region described by
2 2 2
x +y b
(i.e. inside the circle of radius a, outside the rectangle of points of positive y value of distance less than b
from the x axis; these two conditions correspond to being inside the ball but outside the cylinder.) We think
√ 2 2
of the outer part (farther from x axis) of this cure as described by the graph of the function f(x) = x −a
while the inner part as described by the function g(x) = b. These curves intersect at two points with x
2 2 2 √ 2 2
values solving the equation b = a −x , i.e. x = ± a −b . Using the formula for the washer method we
get that the volume is
Z √ 2 2 Z √ 2
+ a −b 2 2 + a −b 2 2 2
√ 2 2 π(f(x) −g(x) )dx = √ 2 π(x −a −b )dx.
− a −b − a −b
I leave it to you to take this integral. Note that the integrand is always nonnegative over the region of
integration, which makes sense because volume is always nonnegative. Note also that for the integral to
make sense we need a > b which is sensible since otherwise after cutting out the cylinder we’ve removed the
whole sphere!
4 2
Problem 2. Consider the curve y = (x−1) .
(a) What is the length of the portion of this curve lying in the region y > 0, 1 ≤ x ≤ 2?
1
(b) What is the area of the surface of revolution obtained by revolving the portion of this curve described
in (a) about the y axis?
4 2
(c) Sketch a picture of the curve y = (x − 1) , highlighting the region relevant to part (a). Sketch a
picture of the surface relevant to part (b).
Solution.
a I will write down the integral and indicate how one solves the integral.
2
As 1 ≤ x ≤ 2, we have 0 ≤ (x − 1) ≤ 1 and so (since we just want the positive 4-th root of y)
0 ≤ y ≤ 1. Now taking the square root of the equation and rearranging we have
2 2 2 2
x=y +1, so dx=2ydy and dx =4y dy .
So we will think of the curve as the graph x = g(y) over the y axis of the function g(y) = y2 + 1, in
which case the formula or the length of the curve is
Z 1p 2
1+4y dy.
0
The substitution x = (tanu/2) removes the square root and reduces this integral a trigonometric one.
b Because we are rotating around the y axis, the formula for the surface area is
Z Z 1 2 p 2
2πxds = 2π(y +1) 1+4y dy
0
where we used the fact that we have described our graph as a curve over y. The same substitution
reduces this integral to a sum of trigonometric integrals.
c Here is a sketch of the curve and the surface of revolution. It looks a bit like a funnel; you can see
from the equation that the curve is actually a piece of a parabola y2 = x − 1.
Problem 3 Write the most general solution to the differential equations:
(a) y′′ + 3y′ + 2y = 0.
(b) y′′ − 2y′ − 3y = 0.
2
(c) y′′ + 2y′ + y = 0
For each of the above, compute y(1) when y solves the above differential equations and y(0) = 1, y′(0) = 0.
Solution. The characteristic polynomials are
z2 +3z +2,z2 −2z −3,z2 +2z+1
−2x −x
respectively. The first factors as (z +2)(z +1), and so the solutions to the first one are Ae +Be . The
second factors as (z−3)(z+1), and so the solutions are Ae3x+Be−x. The last is a square, equal to (z+1)2,
−x −x
and so the solutions are Ae +Bxe .
For each of these you need to compute A and B when you have the constraint that y(0) = 1 and y′(0) = 0.
The derivatives of these solutions are
−2x −x 3x −x −x −x
−2Ae −Be ,3Ae −Be ,(−A+B)e −Be .
So for the first problem, plutting x = 0 and setting y and y′ to be 1 and 0 respectively, we get the system
of linear equations:
A+B=1,−2A−B=0
2x x −2 −1
so A = −1, B = 2 and the corresponding solution is −e +2e which has y(1) = −e +2e .
For part 2, the equations you impose are A + B = 1,3A − B = 0, so A = 1/4 and B = 3/4 and the
corresponding y(1) value is e2/4 + 3e/4.
Finally for part 3, the equations you impose are A = 1 and −A+B = 0, so B = −1 and the corresponding
y(1) value is e−1 − e−1 = 0.
Problem 4. Solve the following separable differential equations:
1.
xy3
y′ = √ .
2
1+x
2.
et−yy′ = sec(y)(1+t2).
Solution.
1. Divide by y3 and multiply by dx. We get
Z dy −1 Z √xdx p 2
y3 = 2y2 = 2 = x +1+C.
1+x
Thus the solutions are s
y(x) = ± √ 1 .
2
−2 x +1+C
Note that these solutions go off to infinity in finite time depending on the initial value; at x = 0 you
need C to be greater than 2 for the square root to have a meaning (since otherwise you are taking the
square root of a negative number) and then as x increases the denominator gets closer and closer to
zero and so the y value blows up!
t
2. You divide by e secy and multiply by dt. You get
Z −y Z −t 2
cosye dy = e (1+t )dt
I will save you the integral computation. You can compute both sides by enough integrations by parts!
In the end you get
1 −y −t 2
2e (siny −cosy) = −e (t +2t+3)+C.
You would want to solve for y now but there is no way to do this. So this would be an acceptable final
answer!
3
Problem 5. Solve the following linear differential equation:
3y′ −2y = 4sin3t.
Can you come up with a simple electrical circuit that simulates this differential equation?
Solution. To get into standard linear equation form we divide by 3 to get
y′ − 2y = 4 sin3t.
3 3
The integrating factor is
I(t) = exp Z −2dt =e−(2/3)t.
3
After multiplying by the integrating factor, integrating both sides, and dividing by the integrating factor,
we get that the final answer is R
e−(2/3)t4 sin3tdt
y(t) = 3 .
e−(2/3)t
You can take the integral in the numberator by using the trick where you integrate by parts twice to get
back the same thing up a constant. The answer is
Z e−(2/3)t4 sin3tdt = −4e−(2/3)t(2sin3t+9cos3t)+C.
3 85
Of course the integrating factor cancels the exponential factor after division.
Whatis going on? Well 2sin3t+9cos3t = αcos(3(t+β)) for some α,β. What the equation describes is
the behavior of a circuit with a resistor and an inductor and a voltage source providing an oscillatory voltage
of frequency 3 radians/second. What the solution says is that if you give the circuit an oscillatory voltage, it’s
current will oscillate with the same frequency as the oscillation of the voltage, but with a different amplitude
and phase!
4
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