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Math 105 A,B Winter 2008 Midterm II NAME:
- 16 pts For each of the following functions f(x), find f′(x) (its deriva- tive) and F (x) (its most general anti-derivative):
a. f(x) = 2x^ + x^2 + 2 f′(x) = 2x^ ln 2 + 2x F (x) = (^) ln 2^2 x + x 33 + 2x + C b. f(x) = (^5) x − 2 ln x f′(x) = − (^) x^52 − (^2) x F (x) = 5 ln x − 2(x ln x − x) + C
- 4 pts Let f(x) = 3x^ and g(x) = log 3 x. Find the derivative of h(x) = f(g(x)). Notice that h(x) = 3log^3 x^ = x (in other words, f and g are inverses). So h′(x) = 1.
- 4 pts Find a solution to the Initial Value Problem (IVP) F ′^ = x+cos x; F (π/2) = 1. F (x) = x
2 2 + sin^ x^ +^ C F (π/2) = (^
π 2 ) 2 2 + sin^
π 2 +^ C^ =^
π^2 8 + 1 +^ C^ = 1 So C = −π
2 8 Therefore F (x) = x
2 2 + sin^ x^ −^
π^2 8
- 4 pts Does the function f(x) = x + cos x have a local minimum on the interval [0,2]? If so, find it. If not, explain why not. f′(x) = 1 − sin x f′(x) ≥ 0, so f is non-decreasing everywhere. Therefore, there is no local minimum (or maximum).
- 8 pts A mold grows at a rate proportional to the amount present. Initially (at time t = 0), its weight is 3 grams. After 2 days, it weighs 5 grams. How much does it weigh after 8 days?
The situation described is modeled by the DE: y′^ = ky. The solution to this DE is y = y(0)ekt. Using the initial condition given (y(0) = 3) we find y = 3ekt. Using the fact that y(2) = 5, we solve for k: y(2) = 3e^2 k^ = 5 e^2 k^ = (^53) Taking the natural log of both sides: 2k = ln (^53)
k = ln^
(^53) 2 y = 3e
ln^53 2 t^ = 3(^53 ) t 2
Now, find y(8) = 3(^53 ) 28 = 3(^53 )^4 ≈ 23 .1 grams.
- 16 pts Find the derivatives of the following functions (HINT: Think before you derive.)
a. h(x) = tan x csc x h(x) = sin cos^ xx sin^1 x = (^) cos^1 x = sec x So, h′(x) = sec x tan x or h′(x) = (^) cossin 2 x (^) x b. g(t) = (t + 4)^3 g′(t) = 3(t + 4)^2 c. f(w) = sin cos^ w w Noticing that f(w) = tan w, we see that f′(w) = sec^2 w d. P (t) = tan sec tt P (t) = sin cos^ tt cos t = sin t P ′(t) = cos t
- 12 pts The equation esin^ y^ = x implicitly defines y as one or more functions of x.
a. Find dy dx. Taking the derivative of both sides w.r.t. y: esin^ y^ cos y dy dx = 1
a. f(x) = g(xn) f′(x) = g′(xn)nxn−^1 b. f(x) = arcsin(g(x)) f′(x) = √ 1 −^1 (g(x)) 2 g′(x)
c. f(x) = g(arcsin x) f′(x) = g′(arcsin x) √ 11 −x 2
- 8 pts A farmer wants to fence a rectangular area, but she must stay within a given budget. The fences run east-west and north-south. As- sume fencing material costs $ 5 per foot in the east-west direction, and $ 7 per foot in the north-south direction. The total budget she has available is $ 1000. What are the best dimensions for the fence that enclose the largest area? (HINT: Maximize the area subject to the budget constraint.) Let x represent the vertical sides, and y represent the horizontal sides. Objective Function: A(x, y) = xy The cost will by represented by the expression 5y + 5y + 7x + 7x = 10 y + 14x. Constraint Equation: 10y + 14x = 1000 We want to maxmize A(x, y). To do this, we use the constraint equation to solve for y: y = 100010 −^14 x We plug this into A(x, y) to get the area in terms of one variable: A(x) = x(1000 10 − 14 x)= 100x − 1. 4 x^2 We take the derivative w.r.t. x and set it equal to zero, to find the stationary points of A: A′(x) = 100 − 2. 8 x = 0 x = (^1002). 8 = 2507 ≈ 35 .7 feet. Knowing that the graph of A(x) is an “upside down” parabola, we know its stationary point is a global maximum. Or use the second derivative
test to check this. A′′(x) = − 2 .8, so A(x) is concave down everywhere, again confirming we have found a global maximum.
Therefore the dimensions that enclose the largest area are the x we
found above, and y = 1000 −14(^
(^2507) ) 10 =^
1000 − 500 10 = 50 feet.
