Calculus challenge help?

Question

i've been trying to do problems out of a calculus challenge textbook and I'm having some difficulty with various problems...this one seems particularly interesting to me, unfortunately I can't seem to solve it.

The path of a basketball being shot into a hoop follows y=2.15+2.09x - .41x^2, where 0<=x<=4.6

x is the horizontal distance in meters from the point of release and y is the vertical distance above the floor. what is the distance the ball travels from release to the time it enters the hoop?

Any help solving this would be greatly appreciated. Is this an arclength problem?

Answer 1

Yes, it is an arclength problem:

dy/dx = 2.09 -.82x

C =∫ √(1 + (2.09 -.82x)²)dx between 4.6 and 0

Let u = 2.09 -.82x  ==&gt; du = -.82dx ==&gt; -1/.82 du = dx

u(4.6) = -1.682; u(0) = 2.09

C = (1/.82)∫ √(1 + u²)du between 2.09 and -1.682

Now let u = tanθ  ==&gt; du = sec²θ dθ

u(1.1245) = 2.09; u(-1.0344) = -1.68

C = (1/.82)∫sec³θ dθ between 1.1245 and -1.0344

= (1/.82)(1/2)[secθtanθ + ln|secθ + tanθ| between 1.1245 and -1.0344

(I didn&#039;t check, but you&#039;ll get the idea.)

All the best.