David Murphy
1. Find f0(x) if it is known that dxd [f (2x)] = x2.
By the chain rule, dxd [f (2x)] = f0(2x) · 2, so 2f0(2x) = x2. Hence f0(2x) = x2/2, but the left- hand side (f0(2x)) is written in terms of 2x while the right-hand side (x2/2) is written just in terms of x, so we need to write the right-hand side in terms of 2x as this is the argument of f0 on the left. So x2/2 = (2x)2/8 (do some algebra to check this) implies that f0(2x) = (2x)2/8 for all x, so that f0(x) = x2/8.
2. (a) Given g(x) = sin x, find g(42)(x).
We’re looking for a pattern so we can figure out g(42)(x) without actually taking 42 derivatives! So observe that g(1)(x) = cos x, g(2)(x) = − sin x, g(3)(x) = − cos x and g(4)(x) = sin x is equal to g(x) again. So every fourth derivative brings us back to sin x, so g(40)(x) = sin x (10 cycles of 4 derivatives), which implies that g(41)(x) = cos x and g(42)(x) = − sin x.
(b) Given f (x) = 1x, find f(42)(x).
Let’s first rewrite f (x) = x−1 then use the power rule (without simplifying the coef- ficient!) to find the pattern. We have f(1)(x) = (−1)x−2, f(2)(x) = (−1)(−2)x−3, f(3)(x) = (−1)(−2)(−3)x−4, f(4)(x) = (−1)(−2)(−3)(−4)x−5. With this many deriva- tives computed, let’s see if we recognize a pattern yet. Notice that in the FIRST deriva- tive, there is ONE negative sign; in the SECOND derivative, there are TWO negative signs; in the THIRD derivative, there are THREE negative signs; in the FOURTH deriva- tive, there are FOUR negative signs. This is a pattern, which will continue, so in the 42nd derivative there will be 42 negative signs, and (−1)42 = 1. Next, now that we’ve taken care of the negative signs, what’s left of the coefficients? In the FIRST derivative, we have 1; in the SECOND derivative we have 1 · 2; in the THIRD derivative, we have 1 · 2 · 3; in the FOURTH derivative we have 1 · 2 · 3 · 4. What’s the pattern? Well, in the nth derivative, we have 1 · 2 · 3 · · · n = n!, so in the 42nd derivative our coefficient will be (−1)42(42)! = (+1)(42)! = 42! ≈ 1.4 × 1051. (This is a really big number!) Finally, what is the exponent on x in the derivatives of f (x)? In the FIRST derivative, the exponent is −2; in the SECOND, it is −3; in the THIRD, it is −4; in the FOURTH, it is −5.
So in the nth derivative, the exponent is −n − 1, and thus f(n)(x) = (−1)nn!x−n−1. In particular, f(42)= 42!x−43.
(c) If h(x) = 2f (x) + 3g(x), find h(42)(x).
By the sum rule and constant multiple rule, we have h0(x) = 2f0(x) + 3g0(x), h00(x) = 2f00(x) + 3g00(x), and so on. In general, the coefficients 2 and 3 will always be there, so h(42)= 2f(42)(x) + 3g(42)(x) = 2[42!x−43] + 3[− sin x] = 2 · 42!x−43− 3 sin x.
3. For what values of c does the equation ln x = cx2 have exactly one solution? (Don’t forget that c can be either positive or negative.)
Whenever c ≤ 0, the graph of y = cx2 is either the x-axis (when c = 0) or a parabola opening
exactly once. Now, if c > 0, then for c too big the parabola will open and grow faster than ln x so that the two curves will never intersect. For c > 0 but very small, the parabola will be very wide and so will cross the graph of y = ln x twice. The one time that c > 0 and the graph of y = cx2 intersects y = ln x exactly once must occur in such a way, then, that the two curves have the same tangent line at their point of intersection (if not, then the slope of cx2 will be less than that of ln x and so the parabola will intersect ln x twice). So this gives us the second equation 2cx = 1/x which yields x2 = 1/(2c). Plugging this relationship into the first equation, ln x = cx2 = c[1/(2c)] = 1/2, which only happens when x = e1/2 and c = 1/(2x2) implies that c = 1/[2(e1/2)2] = 1/(2e). Thus the parabola y = cx2 intersects the curve y = ln x exactly once when either c ≤ 0 or c = 1/(2e).
4. A particle moves on a vertical line so that its coordinate at time t is y = t3− 12t + 3, t ≥ 0.
(a) Find the velocity and acceleration functions.
The velocity is the rate of change of the position function, so v(t) = y0(t) = 3t2− 12.
The acceleration is the rate of change of the velocity, so a(t) = v0(t) = 6t.
(b) When is the particle moving upward and when is it moving downward?
The particle is moving upward when its velocity is positive, which happens when v(t) = 3t2− 12 > 0. Thus this occurs while 3t2 > 12, i.e., t2 > 4. This happens when t > 2 or when t < −2, but as we are only observing the particle when t ≥ 0, the particle is moving upward for t > 2 seconds.
The particle is moving downward when v(t) = 3t2− 12 < 0, so whenever t2 < 4, which is true for −2 < t < 2. Again, as we only observe the particle for t ≥ 0, the particle is moving downward when 0 ≤ t < 2 seconds.
(c) Find the total distance that the particle travels in the time interval 0 ≤ t ≤ 3.
The total distance the particle travels is equal to the distance it travels while going down plus the distance it travels while going up again, which is equal to |y(2) − y(0)| + |y(3) − y(2)|, since for 0 ≤ t < 2 the particle was going down and for 2 < t ≤ 3 it is going up.
Thus the total distance is |(8 − 24 + 3) − (0 − 0 + 3)| + |(27 − 36 + 3) − (8 − 24 + 3)| =
|(−13) − (3)| + |(−6) − (−13)| = 16 + 7 = 23.
(d) Sketch graphs of the position, velocity and acceleration functions for 0 ≤ t ≤ 3.
Use your calculator and the formulas from part (a) to do this.
(e) Where is the graph of the position concave up? Concave down? When is the particle speeding up? When is it slowing down?
The graph of the position is concave up when y00(t) > 0, but y00(t) = a(t) = 6t is > 0 whenever t > 0, so the curve is concave up for all t > 0 and is never concave down for t ≥ 0.
The answer to when is the particle speeding up and when is it slowing down depends on our interpretation of “speeding up” and “slowing down” as referring either to the velocity or to the speed of the particle. If we mean the velocity, then “speeding up” will occur whenever the rate of change of v(t) is positive, but the rate of change of v(t) is v0(t) = a(t) = 6t is positive whenever t > 0 and it is never negative so the velocity in never “slowing down” for t ≥ 0.
If, however, by “speeding up” and “slowing down” we are referring to the speed of the particle, which is equal to the absolute value of the velocity, |v(t)| = |3t2− 12|, then the speed is increasing when t > 2 and decreasing for 0 ≤ t < 2.
5. Use the definition of derivative to prove the Reciprocal Rule: If g is differentiable, then
d dx
h 1 g(x)
i
= −[g(x)]g0(x)2.
Using the definition of derivative, dxd h
1 g(x)
i
= limh→0
1
g(x+h)−g(x)1
h = limh→0 g(x)−g(x+h) hg(x)g(x+h) = limh→0hg(x)−g(x+h)
h
i h 1 g(x)g(x+h)
i
=h
limh→0−g(x+h)−g(x) h
i h
limh→0 g(x)g(x+h)1 i
= [−g0(x)]h
1 [g(x)]2
i
=
−g0(x) [g(x)]2.
6. Evaluate the following limits:
(a) limx→πe2 sin xx−π−1:
This has the indeterminate form 0/0, so consider the limit limx→πe2 sin x[2 cos x]
1 = e2(0)1[2(1)] = 2. Thus, by l’Hˆopital’s Rule, limx→π e2 sin xx−π−1 = 2 as well.
(b) limθ→01−cos θθ2
This also has the indeterminate form 0/0, so consider limθ→0 sin θ
2θ . Again, this is inde- terminate of the form 0/0, so we consider limθ→0cos θ
2 = 12. Therefore, by l’Hˆopital’s Rule, we have limθ→0sin θ2θ = 12, so a second application of l’Hˆopital’s Rule implies that limθ→01−cos θθ2 = 12 too.
(c) limθ→0θ+sin θ tan θ
Once more, we have the indeterminate form 0/0, so consider limθ→01+cos θsec2θ = 1+112 = 2.
Hence, applying l’Hˆopital’s Rule, limθ→0 θ+sin θ tan θ = 2.
7. Let f be a function such that f (2) = 1 and whose derivative is known to be f0(x) =√ x5+ 4 (a) Use a linear approximation to estimate the value of f (2.03).
In general, the linear approximation of a function f (x) near x = a is the equation of the tangent line to the curve y = f (x) when x = a, which is
L(x) = f (a) + f0(a)[x − a].
Thus, for our function, the linear approximating function is L(x) = f (2) + f0(2)[x − 2] = (1) + (6)[x − 2] (using f0(x) = √
x5+ 4 evaluated when x = 2 to compute f0(2) = 6).
Therefore, our approximation of f (2.03) ≈ L(2.03) = (1) + (6)[2.03 − 2] = 1.18.
(b) Will the exact value of f (2.03) be less than or greater than your estimate? Clearly explain why.
To determine whether f (2.03) will be less or greater than L(2.03) depends on the con- cavity of the curve y = f (x) when x = 2, for if it is concave up, then the tangent line lies below the curve so that L(x) ≤ f (x) and our approximation L(2.03) < f (2.03) will be too small. Otherwise, if y = f (x) is concave down when x = 2, the tangent line lies above the curve y = f (x) so that L(x) ≥ f (x) and hence L(2.03) ≥ f (2.03) will be too big.
To decide whether y = f (x) is concave up or down when x = 2, we check f00(x) =
d
dx[f0(x)] = dxd[√
x5+ 4] = 12(x5+4)−1/2[5x4], so f00(2) = 20/3 > 0, implies that y = f (x) is concave up when x = 2 so our approximation L(2.03) = 1.18 will be too small.
8. An airplane is traveling in an elliptical holding pattern described by the parametric equations
where x and y have units of miles. The control tower is 5 miles east of the origin. At what point will the airplane be flying directly toward the control tower?
We want to find where the tangent line to the ellipse passes through the point (5, 0), which corresponds to the location of the control tower, and such that the plane is flying toward rather than away from this point. As t increases, the plane goes around the ellipse counter- clockwise, so the time when it is flying toward the control tower will be when it is in the IV-th quadrant rather than when it is in the I-st.
Now the equation for the tangent line to the ellipse at time t is given by y − (3 sin t) =
dy
dx|t[x − (4 cos t)], where dydx|t = dy/dtdx/dt = −4 sin t3 cos t . Hence the equation of the tangent line is y − 3 sin t = −4 sin t3 cos t [x − 4 cos t], and we need this line to pass through the point (5, 0), which is the location of the control tower. Thus, replace y by 0 and x by 5 and let’s see what we get: 0 − 3 sin t = −4 sin t3 cos t [5 − 4 cos t]. Multiplying both sides by −4 sin t and distributing the 3 cos t on the right-hand side we obtain 12 sin2t = 15 cos t − 12 cos2t. Adding 12 cos2t to both sides and recognizing the Pythagorean identity, we have 12 sin2t + 12 cos2t = 12 which will equal 15 cos t, so cos t = 12/15 = 4/5. Thus t is the angle such that the adjacent side has length 4 and the hypotenuse has length 5, but we want t to be an angle in the IV-th quadrant, so the opposite side must be negative, so it is −3. Hence sin t = −3/5. Therefore, the plane will be flying directly toward the control tower when x = 4 cos t = 4(4/5) = 16/5 and y = 3 sin t = 3(−3/5) = −9/5, so when the plane is at the location (16/5, −9/5).
9. Find points P and Q on the parabola y = 1 − x2 so that the triangle ABC formed by the x-axis and the tangent lines at P and Q is an equilateral triangle.
The triangle formed by the three lines will be equilateral when the angles inside at the base vertices are both 60◦. Thus, if P is the point on the left side of the origin and Q is the point on the right side of the origin on the parabola y = 1 − x2, we must have the slope of y = 1 − x2 at P is equal to tan(60◦) = √
3 and the slope of y = 1 − x2 at Q is − tan(60◦) = −√ 3. Yet the slope of y = 1 − x2 for any x is dy/dx = −2x, so the value of x at P satisfies −2x =√
3, so x = −√
3/2 and y = 1 − (−√
3/2)2 = 1/4 at P . At Q, −2x = −√
3, so x = √
3/2 and y = 1/4. Hence P = (−√
3/2, 1/4) and Q = (√
3/2, 1/4).
10. Implicitly differentiate the equation x4+ 4x2y2+ y4= x3y with respect to x and solve for dy dx. Differentiating with respect to x, we have 4x3+ (4x2)[2ydydx] + (y2)[8x] + 4y3 dydx = (x3)[dxdy] + (y)[3x2]. Put everything involving dydx on the left and everything else on right to get 8x2ydxdy+ 4y3 dydx− x3 dydx = [8x2y + 4y3− x3]dydx = 3x2y − 4x3− 8xy2, so dxdy = 3x8x2y−4x2y+4y3−8xy3−x32.
11. Water is flowing at a constant rate into a spherical tank. Let V (t) be the volume of water in the tank and H(t) be the height of the water in the tank at time t.
(a) What are the meanings of V0(t) and H0(t)? Are these derivatives positive, negative or zero?
V0(t) refers to the rate at which the volume is changing, and thus is the constant rate at which water is flowing into the tank.
H0(t) means the rate at which the water level in the tank is changing.
(b) Is V00(t) positive, negative or zero? Explain.
As V0(t) is constant, its derivative V00(t) is zero.
(c) Let t1, t2 and t3 be times when the tank is one-quarter full, half full, and three-quarters full, respectively. Are the values H00(t1), H00(t2), H00(t3) positive, negative or zero?
Why?
When the tank is one-quarter full, as water continues to flow in at a constant rate the width is increasing so the needed increase in depth is decreasing, so H00(t1) < 0.
When the tank is three-quarters full, the widths of the tank are decreasing as we continue to pour water in at a constant rate, so to accommodate the steady increase in volume while the width decreases, the change in height must be increasing, so H00(t3) > 0.
When the tank if half full, we’re changing from the case where widths were increasing so that H00(t) < 0 to where widths are decreasing so that H00(t) > 0. Hence, when the tank is half full H00(t) = 0 must be the transition from H00(t) < 0 to H00(t) > 0.
12. (a) Explain why |x| =
√
x2 for all real numbers x.
By
√
x2, we mean the positive square root of the positive number x2, which will therefore be |x|.
(b) Use part (a) and and the Chain Rule to show that dxd|x| = |x|x . Employing part (a) and the chain rule, dxd[|x|] = dxd[√
x2] = 12(x2)−1/2[2x] = √x
x2 = |x|x . (c) If f (x) = | sin x|, find f0(x) and sketch the graph of f and f0. Where is f not differen-
tiable?
Using the chain rule and the result of part (b), we have f0(x) = | sin x|sin x [cos x], which is de- fined so long as sin x 6= 0. Thus f (x) is not differentiable when x = 0, ±π, ±2π, ±3π, . . . . (d) If g(x) = sin |x|, find g0(x) and sketch the graph of g and g0. Where is g not differentiable?
Using (b) and the chain rule, we have g0(x) = cos |x| · |x|x, which is defined so long as x 6= 0. Thus g(x) = sin |x| is not differentiable only when x = 0.
13. A hockey team plays in an arena with a seating capacity of 15,000 spectators. With ticket prices at $12, average attendance at a game has been 11,000. A market survey indicates that for each dollar that ticket prices are lowered, the average attendance will increase by 1000.
How should the owners of the team set ticket prices to maximize their revenue from ticket sales?
Currently demand satisfies p = 12, q = 11000 but our market survey tells us that when p = 11, we’ll have q = 12000. Based on this, assuming demand is linear, using the point-slope equation for the line we have q − 11000 = 12000−11000
11−12 [p − 12] = −1000[p − 12]. Hence q = 11000 − 1000[p − 12] = 11000 − 1000p + 12000 = 23000 − 1000p. Therefore the owner’s revenue is R(p) = pq = p[23000 − 1000p] = 23000p − 1000p2. To find when this is maximized, consider R0(p) = 23000 − 2000p, which is always defined but is zero when p = 23000/2000 = 11.50.
The revenue is indeed maximized (in fact we have an absolute maximum) when p = 11.50 as R00(p) = −2000 is always negative so the revenue function is always concave down. Therefore, the owner’s should lower ticket prices to $11.50 in order to maximize their revenue.
14. Let f (x) = −x3+ 3x2+ 9x − 18.
(a) Find the relative extrema point(s) of f .
First, consider f0(x) = −3x2+ 6x + 9 = −3(x2− 2x − 3) = −3(x − 3)(x + 1). This is defined for all x, so our only critical numbers are when it is zero, which is when x = 3 or x = −1. To determine if we have relative max or min at these points, consider the Second Derivative Test: f00(x) = −6x + 6, so f00(3) = −12 < 0 implies f has a local max when x = 3 and f00(−1) = +12 > 0 implies that f has a local min when x = −1. Thus, f has
(b) Determine the inflection point(s) of f .
The inflection point(s) of f occur where the concavity changes, which it determined by the second derivative changing sign. Now f00(x) = −6x + 6 = 0 only when x = 1 and for x < 1, f00(x) > 0 (e.g., f00(0) = 6 > 0) so f is concave up while for x > 1, f00(x) < 0 (e.g., f00(2) = −6 < 0) so f is concave down. Thus f changes from being concave up to concave down at x = 1, so f has an inflection point at (1, f (1)) = (1, −7).
(c) When is the graph of y = f (x) concave up? When is it concave down?
We just answered this question above: f is concave down when f00 < 0, on the interval (1, ∞) and f is concave up when f00> 0 on the interval (−∞, 1).
(d) When is the graph of y = f (x) increasing? When is it decreasing?
The function is increasing when f0 > 0 and decreasing when f0 < 0. Our critical numbers were x = −1 and x = 3, so checking f0(−2) = −15 < 0 implies f is decreasing on the interval (−∞, −1); f0(0) = 9 > 0 implies f is increasing on the interval (−1, 3); and f0(4) = −15 < 0 implies that f is decreasing on the interval (3, ∞).
15. A metal storage tank with volume 1000 m3is to be constructed in the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal?
Let r be the radius of the cylinder (and hence also of the hemisphere) and h be the height of the cylinder. Then the volume of the tank is V = πr2h + 12(43πr3) = πr2h +23πr3. Hence 1000 = πr2h + 23πr3, so πr2h = 1000 − 23πr3 which means h = 1000−
2 3πr3 πr2 .
In our first interpretation of this problem, let’s assume the tank has no bottom but only the sides and top. Then the amount of metal used to make the tank is equal to the lateral surface area of the cylinder plus the surface area of the hemisphere, which is
A = 2πrh +1
2(4πr2) = 2πr1000 −23πr3
πr2 + 2πr2= 2000r−1−4
3πr2+ 2πr2= 2000r−1+2 3πr2. To minimize this A, consider
A0(r) = −2000r−2+4 3πr =
4
3πr3− 2000
r2 ,
which is not defined when r = 0 (but r = 0 is not in the domain of A(r) since the volume is 1000 and if r = 0 then the volume would be 0 also) and is zero when 43πr3 = 2000, so r = (1500/π)1/3 and h = 0 (using the formula relating h to r above). This means that our tank is really just the hemisphere without the cylinder base, which is fine. To confirm that this is a minimum, look at A00(r) = 4000r−3 + 43π, which is > 0 for all r > 0, so A(r) is concave up on the entire domain {r > 0} and hence we have an absolute minimum surface area when r = (1500/π)1/3, h = 0.
A second interpretation of the problem says the tank should include a circular base, so that the area is now
A(r) = 2000r−1+ 2
3πr2+ πr2 = 2000r−1+5 3πr2
(the same as the area formula above except that we now add another πr2 for the circular bottom). Then this area is minimized when
A0(r) = −2000r−2+10 3 πr =
10
3πr3− 2000 r2
is not defined (when r = 0, which is still not in the domain so this isn’t a critical number) or when it is zero. But A0(r) = 0 when 103πr3= 2000, so r = (600/π)1/3= 5.7588 and h = 5.7588 as well (again using the formula for h above). To ensure that the area is minimized for this value of r, check that A00(r) = 4000r−3+ 103π is > 0 for all r > 0, which is the domain of A(r), so that the function is always concave up and hence our critical point is at an absolute minumum.
16. Suppose that the price p and the number of copies q (measured in 100s) of an old video game satisfy the demand equation 6p + q + qp = 94. Determine the rate at which the quantity is changing when q = 4, p = 9 and the price is falling $1.50 per week.
This is a related rates problem, so differentiate with respect to time, t: 6dpdt + dqdt + (q)[dpdt] + (p)[dqdt] = 0. Now evaluate when q = 4, p = 9,a dn dpdt = −1.5 to find dqdt: 6(−1.5) + dqdt + (4)[−1.5] + (9)dqdt = −15 + 10dqdt = 0 implies that dqdt = 15/10 = 1.5.
17. Suppose that the cost function of producing R radios is given by the formula C(R) = 100+√ R.
Find the marginal cost at production level R = 100.
The marginal cost is the derivative of the cost function, so is C0(R) = 12R−1/2. Hence the marginal cost when producing R = 100 radios is C0(100) = 1
2√
100 = 1/20 = 0.05.
18. A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom and 80 cm wide at the top, and has a height 50 cm. If the trough is being filled with water at the rate of 0.2 ft3/min, how fast is the water level rising when the water is 30 cm deep?
The volume of the trough is going to be the area of the trapezoid times 10, since the trough is 10 m long. Let h be the depth of water in the trough and w be the width of the water’s surface in the trough when the depth is h. We need to convert all of the cm measurements to m, since the length and the change in volume, dV /dt = 0.2 are expressed in terms of meters rather than centimeters, so the base of the trapezoid is 30 cm = 0.3 m, the top is 80 cm = 0.8 m and the height is 50 cm = 0.5 m. This isoceles trapezoid can be broken up into a rectangle of width 0.3 and height 0.5 and two congruent triangles (because the trapezoid is isoceles) of height 0.5 m.
As the top of the trapezoid has length 0.8 and 0.3 m of that is used up by the rectangle, the remaining 0.5 m must be evenly divided between the two triangles on either end, so the tops of the triangles have length 0.25 m. This is half of their height, so when the water is h meters deep in the trough, the width of the water’s surface within one of these triangles (using a similar triangle arguement) is half the depth. Thus w = 0.5h + 0.3 + 0.5h = h + 0.3. Therefore the area of the trapezoid when the water is h meters deep is A = 12[(0.3) + (h + 0.3)]h and the volume is thus V = 10A = 5[0.6 + h]h = 3h + 5h2. Therefore dVdt = 3dhdt+ 10hdhdt = (3 + 10h)dhdt. Thus 0.2 = [3 + 10(0.3)]dhdt = 6dhdt, so dhdt = 0.2/6 = 0.0333 m/min, which is 3.33 cm/min.
