Calculus related rate ball-n-string falling?

Question

Two towers are each 150 feet high and 100 feet apart. Chris is standing atop one tower and is about to drop a ball straight down. A fishing line is attached to the ball and the other end of the line is attached to a fishing reel that Blaire is holding on the top of the other tower so the line can unroll as the ball falls. The ball will fall 16t^2 feet in the first t seconds after it is dropped. How fast is the string unwinding after 3 seconds? (Assume that the bowling ball and the fishing reel are each 150 feet above the ground when the ball is dropped and that the fishing line is taut at all times.) Be descriptive and clear please I want to learn.

Answer 1

First, draw a picture. Let C be Chris, B be Blaire, and A be the ball at some point. Thats a right angled triangle, where side BA is how far the ball has dropped; CB = 100 feet, and CA = the fishing line.
Now, we need to know how fast the ball is dropping. Since its distance is 16t^2, its velocity is 32t (differentiating). So side BA is changing at 32t feet per second at any time.
We want to know the rate at which side CA is changing. But we know that CA^2 = CB^2 + AB^2, so CA = sqrt(CB^2 + AB^2) = sqrt(100^2 + AB^2).
Now, we need to differentiate both sides with respect to time.
Thus, d/dt(CA) = d/dt(sqrt(100^2 + AB^2)).
We need the chain rule for the RHS: sqrt(something) differentiates to 0.5(something)^-0.5, then we multiply by the inside differentiated. AB^2 differentiated (with respect to t) is 2*AB *d/dt (AB) = 2*AB*32t. So the overall equation becomes:
d/dt(CA) = 0.5(100^2 + AB^2)^-0.5 * 2 * AB * 32t.
Now, we know t = 3 and AB = 16*3^2, so substitute that in and you'll get the answer of 78.9 feet/second.

I'm likely to have made a mistake, but hopefully that gives you the idea of what to do.

Answer 2

You know the position of the ball at time t, it's 16t^2 ft below the dropping point. After 3 seconds the ball will have fallen 16*9 = 144 feet so it's still falling.

The ball is at one vertex of an upside-down right triangle, the other two vertices are Chris and Blaire. The vertica side is 16t^2 ft, the horizontal side is 100 ft, so the hypotenuse (the length of the string) is sqrt(16t^2+10000) feet.

Differentiate to find the speed at which the string increases in length. Plug in t=3.

Answer 3

[a] floor area = circumference * height = 2? r * h height = 10cm + 0.1cm * t h = 10 + 0.1t A = 10? * (10 + 0.1t) = 100? + ? t dA/dt = cost of substitute of area with relation to time dA/dt = (100? + ? t)' = ? cm^2/sec [b] volume = area of base * height = ? r^2 * (10 + 0.1t) V = 25? (10 + 0.1t) V = 250? + 2.5? t dV/dt = 2.5? cm^3/sec wish this helps!