1. Related Rates 1. Ship A is traveling west toward a lighthouse at a speed of 15 kilometers per hour, while ship B is traveling north away from the lighthouse at a speed of 10 kilometers per hour. Write x for the distance between ship A and the lighthouse, and y for the distance between ship B and the lighthouse. a. Find the rate of change of the distance between the two ships when x = 4 km and y = 3 km. Solution: Write d(t) for the distance between the ships. Then by the Pythagorean theorem, d(t)2 = x(t)2 + y(t)2 Differentiating this equation with respect to time gives 2d(t)d0 (t) = 2x(t)x0 (t) + 2y(t)y 0 (t) Now solve for d0 (t): d0 (t) =

x(t)x0 (t) + y(t)y 0 (t) d(t)

Finally, we√need to plug in the values of x, y, etc. At the given time, x = 4 and y = 3, hence d = 9 + 16 = 5. Since ship A is approaching the lighthouse, x(t) is decreasing, so x0 (t) = −15, while y is increasing, so y 0 (t) = 10. Therefore d0 (t) =

4(−15) + 3(10) 30 = − = −6 5 5

in units of km/hr. b. Let θ be the horizontal angle in the right triangle formed by the two ships and the lighthouse. Find the rate of change of θ when x = 4 km and y = 3 km. Solution: tan θ = xy , so differentiating with respect to time gives sec2 (θ)θ0 (t) =

y 0 (t)x(t) − y(t)x0 (t) x(t)2 1

2

JOE HUGHES

Therefore θ0 (t) =

y 0 (t)x(t) − y(t)x0 (t) cos2 (θ) x(t)2

We saw in part (a) that if x = 4 and y = 3 then d = 5, hence cos θ = 54 . Thus 85 17 (10)(4) − 3(−15) 4 2 40 + 45 16 · = · = = θ0 (t) = 16 5 16 25 25 5 in units of radians per hour. 2. A 15 foot long ladder is leaning against a building. Let x(t) denote the distance of the end of the ladder touching the ground from the base of the building, and let y(t) denote the distance of the end of the ladder touching the building from the base. Suppose that x0 (t) = 12 feet per second. a. Find y 0 (t) when x = 9. Solution: The ladder forms a right triangle with the building and the ground. The hypotenuse (i.e. the ladder) has constant length, hence x(t)2 + y(t)2 = 152 Differentiating gives 2x(t)x0 (t) + 2y(t)y 0 (t) = 0 and after solving for y 0 , y 0 (t) = − If x = 9 then y=

√

x(t)x0 (t) y(t)

225 − 81 =

√

144 = 12

so y 0 (t) = −

9 · 21 3 =− 12 8

in units of feet per second. b. Find the rate of change of the area of the triangle when x = 9. Solution: The area of the triangle is A(t) = 12 x(t)y(t). Differentiating using the product rule, we obtain 1 1 A0 (t) = x0 (t)y(t) + x(t)y 0 (t) 2 2 From part (a), we know that if x = 9 then y = 12 and y 0 (t) = − 83 . Therefore 1 1 1 3 27 21 · · 12 + · 9 · (− ) = 3 − = 2 2 2 8 16 16 in units of square feet per second. A0 (t) =

MATH 31A: MIDTERM 2 REVIEW

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2. Critical Points and the first derivative test 2

1. Find the critical points of f (x) = x 3 , and determine whether they are local maxima or minima. 1

1

Solution: f 0 (x) = 23 x− 3 for x 6= 0, and f 0 (0) does not exist. Since 23 x− 3 is non-zero for all x 6= 0, this implies that the only critical point is c = 0. To determine whether 0 is a minimum or a maximum, use the first derivative test: if x < 0 1 1 then x− 3 < 0, while if x > 0 then x− 3 > 0. Therefore c = 0 is a minimum. 2. Let f (x) = x3 + bx2 + cx + d be a cubic polynomial. Show that f has at most 2 local extrema. Solution: f 0 (x) = 3x2 + 2bx + c is a quadratic polynomial, hence has at most two roots. Therefore f has at most two critical points, so at most two local extrema.

3. The mean value theorem *1. Let f (x) be continuous on [0, 1] and differentiable on (0, 1), with f (0) = 0 and f (1) = 1. Show that there is a point c in (0, 1) such that f 0 (c) = 2c. Solution: Let g(x) = f (x) − x2 . Then g(0) = 0 and g(1) = f (1) − 1 = 1 − 1 = 0. By the mean value theorem applied to g, there is a point c in (0, 1) such that g 0 (c) =

g(1) − g(0) =0 1−0

Therefore 0 = g 0 (c) = f 0 (c) − 2c which shows that f 0 (c) = 2c.

4. Concavity and the second derivative test 1. Let f (x) = x3 − 3x. Find the inflection points of f , the intervals on which f (x) is concave up and concave down, and use the second derivative test to find the local maxima and minima of f . Solution: f 0 (x) = 3x2 − 3 = 3(x − 1)(x + 1), so the critical points of f are at ±1. f 00 (x) = 6x, so the only inflection point is x = 0. f is concave down for x < 0, and concave up for x > 0. The second derivative test therefore implies that x = −1 is a local maximum, and x = 1 is a local minimum.

4

JOE HUGHES

5. Optimization 1. A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs 10 dollars per square meter for the base and 5 dollars per square meter for the sides, what are the dimensions of the least expensive tank, and what is its cost? Solution: Let l, w, h denote the length, width, and height of the tank. Then the cost of the tank is C = 10lw + 2 · 5lh + 2 · 5wh = 10(lw + lh + wh) This our objective function, i.e. the function we want to minimize. Of course, to apply the techniques we’ve learned, we need this function to depend on only one variable. The first step is to recall that w = 4, so the objective function becomes C = 10(4l + lh + 4h) Next, the volume of the tank is 36 cubic meters, so 36 = lwh = 4lh Therefore l = h9 , so 36 + 9 + 4h) h Now take the derivative and set it equal to zero: 360 0 = C 0 (h) = − 2 + 40 h Solving for h gives 360 =9 h2 = 40 so h = 3 (the negative square root can be ignored, since the tank can’t have negative height). C(h) = 10(

Now take the second derivative and evaluate it at h = 3: 720 >0 C 00 (3) = 27 so the second derivative test guarantees that this is a minimum. Thus l=

9 9 = =3 h 3

and the cost of the tank is 10( (in dollars).

36 + 9 + 4h) = 10(12 + 9 + 12) = 330 h

MATH 31A: MIDTERM 2 REVIEW

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6. Antiderivatives 1. Evaluate

R

2 sin θ cos θ dθ.

Solution: To find the antiderivative, recall the following trig identity: sin(2θ) = 2 sin θ cos θ Therefore

Z

Z 2 sin θ cos θ dθ =

sin(2θ) dθ = −

cos(2θ) +C 2

2. Let F (x) be a function such that F 00 (x) = f (x). If G(x) also satisfies G00 (x) = f (x), show that G(x) = F (x) + C1 x + C2 for constants C1 , C2 . Solution: Let A(x) = G0 (x) − F 0 (x). Then by the linearity of the derivative, A0 (x) = G00 (x) − F 00 (x) = f (x) − f (x) = 0 This implies that A(x) is constant, say A(x) = C1 . Now let B(x) = G(x) − F (x) − C1 x. Then B 0 (x) = G0 (x) − F 0 (x) − C1 = A(x) − C1 = C1 − C1 = 0 so B is also constant, say B(x) = C2 . Thus G(x) = F (x) + C1 x + B(x) = F (x) + C1 x + C2 This problem shows that in order to specify a particular function F such that F 00 = f , we need to know the value of f at a point and the value of f 0 at a point.