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Answer Key for Exam 2
1(a) We have lim x→ 3 x^3 − x^2 − 21 x + 45 x^3 − 5 x^2 + 3x + 9
, so we need to do something. x − 3 must be
a factor of both top and bottom, and once we know this it is not too hard to see that
x^3 − x^2 − 21 x + 45 x^3 − 5 x^2 + 3x + 9
(x − 3)(x^2 + 2x − 15) (x − 3)(x^2 − 2 x − 3)
(x − 3)((x − 3)(x + 5) (x − 3)(x − 3)(x + 1)
Therefore
xlim→ 3
x^3 − x^2 − 21 x + 45 x^3 − 5 x^2 + 3x + 9 = lim x→ 3 x + 5 x + 1
More likely you used L’Hˆopital’s rule, which is a good idea since the limit is 00. We have
xlim→ 3
x^3 − x^2 − 21 x + 45 x^3 − 5 x^2 + 3x + 9
LH = lim x→ 3
3 x^2 − 2 x − 21 3 x^2 − 10 x + 3
so we’re not done yet, and we can use L’Hˆopital’s rule again. This time we get
xlim→ 33 x
(^2) − 2 x − 21 3 x^2 − 10 x + 3
LH = lim x→ 3
6 x − 2 6 x − 10
= 6 ·^3 −^2
=^16
Therefore the original limit equals 2. One can also give a “mixed” solution:
xlim→ 3 x
(^3) − x (^2) − 21 x + 45 x^3 − 5 x^2 + 3x + 9
LH = lim x→ 3
3 x^2 − 2 x − 21 3 x^2 − 10 x + 3 = lim x→ 3 (3x^ + 7)(x^ −^ 3) (3x + 1)(x − 3) = lim x→ 33 x^ + 7 3 x + 1
=^16
1(b) We have sin 0 = 0, cos 0 = 1 and ln 1 = 0, so ln (cos 0) = ln 1 = 0. Therefore lim x→ 0 ln (cos x) sin x
, and L’Hˆopital’s rule is our only good option:
xlim→ 0
ln (cos x) sin x
LH = lim x→ 0
1 cos x (−^ sin^ x) cos x = lim x→ 0 − sin x cos^2 x
- If x^4 + y^4 = 8xy, then taking the derivative of both sides with respect to x we get
4 x^3 + 4y^3 dy dx
d dx xy = 8
x dy dx
, or x^3 + y^3 dy dx = 2x dy dx
after dividing both sides by 4. Rearranging this we get
x^3 − 2 y = 2x dy dx − y^3 dy dx
2 x − y^3
) (^) dy dx
so dy dx =^
x^3 − 2 y 2 x − y^3 ,^ or^
dy dx =
2 y − x^3 y^3 − 2 x is also correct.
(2, 2) is on x^4 + y^4 = 8xy since both sides equal 32 at that point, and the slope there is
2 · 2 − 23 23 − 2 · 2
For the extra credit, note that the left side of x^4 + y^4 = 8xy is always nonnegative, so the right side must be too. Therefore x and y must have the same sign (in other words, y equals x times something nonnegative),
so it makes sense to use a parametric form that reflects this. Taking y = xt^2 has another advantage which we’ll see presently. We get
x^4 + x^4 t^8 = 8x(xt^2 ) =⇒ x^4
1 + t^8
= 8x^2 t^2.
Dividing both sides by x^2 (1 + t^8 ) we get x^2 = 8 t
2 1 + t^8 , and taking square roots we have
x =
√^2 t 1 + t^8
, and y =
√^2 t^3 1 + t^8
since y = xt^2.
We don’t have to use a ± on the square root since we can take t to be positive or negative. With y = xt we would have had a ± on the square roots and could only have used nonnegative t’s.
3(i) dx dt = cos t and dy dt = sin t (− sin t) + cos t (cos t) = cos^2 t − sin^2 t, so dy dx
dy dt dx dt
= cos
(^2) t − sin^2 t cos t
This is as good a form of the answer as any, though there are several ways to rewrite it.
3(ii) and (iii): If sin t = 0 then x = 0, and y equals sin t times something finite so it also becomes zero. If sin t = 0 then cos t can only be ±1 since cos^2 t + sin^2 t = 1. If t is an even number times π then sin t = 0 and cos t = 1, and if t is an odd number times π then sin t = 0 and cos t = −1, but it isn’t necessary to know what the angles are.
3(iv) The curve goes through the origin when x and y are both 0. In (ii) we saw that this happens when sin t = 0, and if sin t 6 = 0 then x 6 = 0 so this is the only way to go through the origin. When sin t = 0 either
cos t = 1, in which case the slope is^1
= 1; or cos t = −1, in which case the slope is (−1)
Thus the curve has two slopes at (0, 0), 1 and −1.
For the extra credit note that y = x cos t. Therefore y^2 x^2 = cos
(^2) t = 1 − sin (^2) t = 1 − x (^2) , and a non-parametric
form of the curve is y^2 = x^2 − x^4.
- Let x be the length of the row of evergreen trees, and let y be the length of the rows of rubber flowers perpendicular to the row of trees. Then x is the length of the third row of flowers, and the total cost of trees and flowers is C = 30x + 12y + 12y + 12x = 42x + 24y, which we want to make as small as possible. There is a constraint on x and y: xy = 252 since this is the area of the classroom. Therefore we can reduce C to a function of one variable:
C = 42x + 24y = 42x + 24
x
x + 24
x
x + 144x−^1
The only restriction on x is 0 < x < ∞, but a very small or very large x would lead to a very thin, very expensive classroom. So there has to be some medium-sized x which makes C a minimum, and it has to come from a place where dCdx = 0. So we calculate:
dC dx
1 − 144 x−^2
x^2
This equals zero when 1 =
x^2 , which implies^ x
(^2) = 144, which implies x = 12 since x is positive. If x = 12
then y = 25212 = 21. This must be the minimum since it is the only candidate, but we can check the sign of dC dx : it is positive if^ x >^ 12 and negative if 0^ < x <^ 12, so^ x^ = 12 has the sign pattern of a local minimum. Scores: Half were 88 or higher, and half 87 or lower. The mean was about 84.1.
