Exam #2 solutions
Math 220, Section U1
Instructions: do not open the exam until you are told to do so. Be sure to show your work and explain your reasoning for full credit. Calculators, notes, electronic devices, etc. are not allowed. Please use the last sheets of this exam for scrap paper rather than using your own.
NAME
Problem Your Score Possible Score
1 12
2 8
3 8
4 8
5 12
6 8
7 8
Total 64
1.(4 pts. ea.) Compute the derivatives of the following functions:
a. cot(3x) (4x − 2)3
We use the quotient rule:
d dx
cot(3x) (4x − 2)3 =
d
dxcot(3x)(4x − 2)3− dxd((4x − 2)3) cot(3x)
((4x − 2)3)2 = −3 csc2(3x)(4x − 2)3− 3(4x − 2)2cot(3x) (4 − 2)6
b. 1
√1 − e−x2/9
Things are simplified greatly by noticing that 1
√1 − e−x2/9 = (1 − e−x2/9)−1/2. Then, applying the chain rule multiple times, we get:
d
dx(1 − e−x2/9)−1/2= −(1/2)(1 − e−x2/9)−3/2· d
dx(1 − e−x2/9)
= −(1/2)(1 − e−x2/9)−3/2(−e−x2/9) · d
dx(−x2/9) = −(1/2)(1 − e−x2/9)−3/2(−e−x2/9)(−2x/9) c. x
√2x+3
For this one, we need logarithmic differentiation. So, let y = x
√2x+3. Then ln(y) =
√2x + 3 ln(x). Using implicit differentiation, we have:
dy
dx/y = 2 ln(x)/(2√
2x + 3) +√
2x + 3/x So, multiplying both sides by y = x
√2x+3, we get:
dy dx = d
dxx
√2x+3= x
√2x+3((ln(x)/√
2x + 3) +√
2x + 3/x)
2.(4 pts. ea.) Compute the following limits:
a. lim
x→1
log2(x) sin(πx)
log2(1) = sin(π) = 0, so we have a 0/0 indeterminate form; therefore, we can use l’Hospital’s rule:
x→1lim
log2(x)
sin(πx) = lim
x→1 d
dxlog2(x)
d
dxsin(πx) = lim
x→1
1/(ln(2)x) π cos(πx)
Since the limit of the numerator is now (1/ ln(2)) and the limit of the denominator is now −π, this limit is just −1/(π ln(2)).
b. lim
x→0+x2e1/x
This is the product of something going to 0 (x2) with something going to ∞ (e1/x) so we need to do some adjustment before we can use l’Hospital’s; by looking instead at e1/x/(1/x2), we now get ∞/∞ (an indeterminate form). So, using l’Hospital’s, we get:
lim
x→0+
x2e1/x = lim
x→0+
e1/x
1/x2 = lim
x→0+
e1/x(−1/x2)
−2/x3 = lim
x→0+
2e1/xx
This still isn’t solvable (i.e., it still gives us 0 · ∞), but we can tell we’re on the right track as we’ve reduced the exponent of the x piece. So, we set 2e1/xx = 2e1/x1/x and use l’Hospital’s again:
lim
x→0+x2e1/x = lim
x→0+2e1/xx = lim
x→0+2e1/x
1/x = lim
x→0+2e1/x(−1/x2)
−1/x2 = lim
x→0+2e1/x = ∞
3.(8 pts.) Find the maximum area of a triangle with vertices (−1, 0), (1, 0), and a third vertex on the circle x2+ y2 = 1.
Well, this is a triangle with base length 2 (the distance from (−1, 0) to (1, 0)) and height the y-coordinate of the third vertex (assuming the third vertex is in the first or second quadrant; but this is no big deal, as we can just reflect the third vertex of any triangle not satisfying this across the x-axis and get a triangle with the same area). So, let θ be the angle formed by the line from the origin to the third vertex and the positive x-axis. Then the height of our triangle is sin(θ).
So the area of our triangle is given by A(θ) = (1/2)(2)(sin(θ)) = sin(θ). This is maximized when A0(θ) = 0, or when cos(θ) = 0. In the first or second quadrant, this is exactly when θ = π/2.
Quick side note: it is a lot easier to just notice that | sin(θ)| ≤ 1, a fact we’ve been using often (though I didn’t think about this when writing this question!)
4.(8 pts.) At 12 pm, a car starts driving due east at 30 miles per hour from a 4-way intersection. At 1 pm, another car starts driving south from the same intersection at 40 miles per hour. At 3 pm, how fast is the distance between the two cars increasing?
Let x represent the distance of the first car from the intersection and y represent the distance of the second car from the intersection. These are both functions of time. By assumption, we have x0(t) = 30mi/h and y0(t) = 40mi/h. At 3 pm, the first car has been driving for three hours, and so x = 30 · 3 = 90 and the second car has been driving for two hours, so y = 40 · 2 = 80. The angle formed by their directions is a right angle, so the distance between them (which we will denote by z) is given by x2 + y2 = z2. At 3, this means z =√
802+ 902.
Finally, taking the derivative of the equation relating x, y, and z, we get:
2xx0+ 2yy0 = 2zz0 or
z0 = (xx0 + yy0)/z = (90 · 30 + 80 · 40)/√
802+ 902
5.(3 pts. ea.) Suppose a function f (x), its derivative f0(x) and its second derivative f00(x) exist and are continuous for all real numbers. The following table gives some values for these functions:
x-value -8 -2 0 1 3 5 7 9 13 f (x) 0 1 2 3 0 -5 -7 -4 0 f0(x) 1 0 1 0 -1 -4 0 2 8 f00(x) 2 0 -3 -2 -1 0 1 0 4
That is, f (−8) = 0, f0(−8) = 1, f00(−8) = 2, and so on. Suppose additionally that this table gives the only zeros of f , f0 and f00. Answer the following questions (being sure to fully justify your responses!):
a. Find the x-coordinate of each local minimum and local maximum of f (x) (be sure to indicate why you know your answer(s) are what you say they are!)
Since f0(x) exists everywhere, local maxs and mins only occur when f0(x) = 0.
This is when x = −2, 1, and 7. For x = −2, f00(−2) = 0, so we cannot use the 2nd derivative test. Using the first derivative test, we see that f (x) is increasing coming into x = −2 and increasing coming out of x = −2 and so x = −2 is the x-coordinate for neither a local min nor a local max of f . Since f00(1) = −2 < 0, we must have that f has a local max at x = 1 and since f00(7) = 1 > 0, we must have that f has a local min at x = 7.
b. Find the absolute minimum and absolute maximum values of f (x) on [0, 3].
As is standard, we compare the value of f at the endpoints of [0, 3] as well as f ’s value at any critical point of f on this interval. The only critical point for f on [0, 3]
is when x = 1. So, we compare f (0) = 2, f (1) = 3, and f (3) = 0 and conclude the absolute maximum value of f on [0, 3] is 3 and the absolute minimum value of f on [0, 3] is 0.
c. Find the intervals on which f is increasing or decreasing and find the intervals on which f is concave up or concave down.
Since f0 and f00 are assumed to be continuous on all real numbers, they can only change sign (and therefore, only switch between increasing/decreasing or concave up/concave down) at these zeros. So, since f0(−8) > 0, we know f is increasing for all x < −2. We can use similar arguments to conclude: f is increasing for x in (−∞, 1) ∪ (7, ∞) and decreasing for x in (1, 7); f is concave up for x in (−∞, −2) ∪ (5, ∞) and concave down for x in (−2, 5).
d. Find the x coordinate of each inflection point of f (x) (and explain why your answer(s) is/are inflection points!)
f has an inflection point where f00(x) changes sign. This occurs only at x = −2 and x = 5.
6.(8 pts.) Find the tangent line to the curve x3+ 2y2 = sin(xy) + 1 at the point (1, 0).
First, we apply implicit differentiation to the above equation. This gives us:
3x2+ 4ydy
dx = cos(xy)(y + xdy dx) When x = 1 and y = 0, this simplifies to:
3 = dy dx
So we want the line with slope 3 through the point (1, 0). This is just y = 3x − 3.
7.(8 pts.) Suppose two differentiable functions f (x) and g(x) satisfy f0(x) = g(x) and g0(x) =
−f (x). Furthermore, suppose that f (0) = 0 and g(0) = 1. Show that [f (x)]2+ [g(x)]2 = 1.
(Hint: use a derivative!)
First, we take the derivative of f (x)2 + g(x)2. This is (using the fact that f0(x) = g(x) and g0(x) = −f (x)):
2f (x)f0(x) + 2g(x)g0(x) = 2f (x)g(x) + 2g(x)(−f (x)) = 0
So, we must conclude that f (x)2 + g(x)2 is constant. Plugging in x = 0, we see it must be the constant f (0)2+ g(0)2 = 02+ 12 = 1.
