Calculus Study Guide Help?

Question

Hello there, my Cal 1 professor gave this:
http://www.flickr.com/photo_zoom.gne?id=296714838&size=l
and said that if we can finish this we should be ready for our final coming at the end of the month. The problem is these are super-hard.

Can you guys out there shed some light on them.

ps: Don't worry about Problem A. I solved that It is an interesting one

ooops I had forgotten to make the photo public. now it is...! :)

Answer 1

It is difficult to answer when one cannot even see the question. Try making the photo public so we can see it, or type the question into Y!A directly.

Edit: You say you have solved problem A, and you already recieved help on the yam problem, so I shall focus on the remaining three.

Problem C:
First, we need an expression for f(x). It consists of two line segments, the first from (-3, 1) to (0, 5), and the second from (0, 5) to (2, 0). The slope of the first segment is (5-1)/(0-(-3)) or 4/3, and the slope of the second is (0-5)/(2-0) or -5/2. The intercept of both lines is the same. Thus we can define the function piecewise as:

f(x) = {4/3x+5 for -3≤x≤0, -5/2x+5 for 0≤x≤2}

Now we find the derivatives of the given expressions. Since it asks us to evaluate them at only one point, we need only find the derivative of the expression for the piece of the function containing that point, (e.g. there is no need to find the derivative of √(x+f(x)) over the interval (0, 2), since the point we are asked to evaluate isn't on that interval). So:

d/dx √(x+f(x)) at -1
d/dx √(x+4/3x+5) at -1
d/dx √(7/3x+5) at -1
1/(2√(7/3x+5)) * 7/3 at -1 (using the chain rule)
1/(2√(7/3(-1)+5)) * 7/3
7/(6√(8/3))
7/(12√(2/3))
7√3/(12√2)
7√6/24

Similarly for the second expression:

d/dx f(2 sin x) at π/6
d/dx -5/2(2 sin x)+5 at π/6
d/dx -5 sin x +5 at π/6
-5 cos x at π/6
-5 cos π/6
-5√3/2

Problem D:

The area of the rectangle is (9-x)√x (can you see why?). Thus we need to find the maximum of this function. Finding the derivative:
d/dx 9√x - x^(3/2)
9/(2√x) - 3√x/2
Setting this equal to zero:
9/(2√x) - 3√x/2 = 0
9/2 - 3x/2 = 0
9-3x=0
9=3x
x=3

So the maximum area occurs at x=3.

Problem E:

Your instructor probably wants you to set up a function giving θ in terms of t, and then find the maximum value for |dθ/dt| (which will occur where d²θ/dt² = 0), then find the height of the elevator at that point. The reason for finding the absolute value is that we only want the angular speed of the elevator, not its angular velocity. However, we shall take a slightly faster route. First note that the h-100 ft from the elevator to your eye level, and the 150 feet from you to the hotel form two legs of a right triangle. So:

tan θ = (h-100)/150

Since the specifics of the problem don't really matter, I shall simplify this as:

tan θ ∝ h-100

The symbol ∝ means "is proportional to". Formally, a ∝ b means that there exists a real number k>0 such that a=kb. I use this notation to emphasize that we really can ignore most of the specifics of this type of problem (i.e. the solution will be the same no matter how far you are from the building, or how fast the elevator is moving). Now we differentiate both sides to get:

sec² θ dθ/dt ∝ dh/dt

But since dh/dt is a negative constant:

sec² θ dθ/dt ∝ -1
dθ/dt ∝ -cos² θ
|dθ/dt| ∝ cos² θ

Since the maximum of cos² θ is 1, and this occurs at θ=0 (and at θ=π, but the elevator is in front of you, not behind you, so θ=π will never occur), it follows that dθ/dt is at a maximum where θ=0. And at θ=0, h=100.

So the elevator appears to be moving fastest at 100 ft above the ground.

Edit: corrected answer to problem E. I had forgotten to mention the velocity-speed distinction, with the result that I had implicitly assumed dh/dt was positive, when it is clearly not (the elevator is moving down). Although this error did not affect the answer, I felt it was best to correct it.

Edit: corrected typo in last edit.