Calculus
posted by COFFEE on .
A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by the following.
y = 150  (1/40)(x50)^2
Find the distance traveled by the kite.
y = 150  (1/40)(x50)^2
y = 150  (1/40)(x50)(x50)
y = 150  (1/40)x^2 + (5/2)x + 125/2
y = (1/40)x^2 + (5/2)x + 425/2
y' = (1/20)x + 5/2
(y')^2 = ((1/20)x + 5/2)^2
(y')^2 = (1/400)x^2  (1/4)x + 25/4
Length = Integral from 0 to 80 of:
Sqrt[1+(1/400)x^2  (1/4)x + 25/4]
Sqrt[(1/400)x^2  (1/4)x + 29/4]
How would I integrate this? Is this the correct procedure? Thanks.
For Further Reading
* Calculus  Count Iblis, Wednesday, June 13, 2007 at 9:50pm
You can calculate the derivative directly as:
y' = 2 (1/40)(x50)
using the chain rule. Then you find:
1 + y'^2 = 1 + 1/400 (x50)^2
Which is the same as what you got. However, to compute the integral, you need to write it in this form anyway:
Integral sqrt[1 + 1/400 (x50)^2] dx
put x = 20 y + 50:
Integral 20 sqrt[1 + y^2] dy
Substitute y = Sinh(t) in here. The square root beomes a hyperbolic cosine, you get another hyperbolic cosine from the integration measure dy.
The integral of cosh^2 can be computed by using that it is a sum of exp(t) and
exp(t) The square of this is just a sum of exponentials which you can integrate term by term.

So I have: 20*Integral[Cosh(t)]^2 dt
=5(2t  (e^2t)/2 + (e^2t)/2) evaluated from 0 to 80. If you evaluate at 80 and 0 you get 7.6746e69 which seems unlikely. Is this correct?
We've put:
x = 20 y + 50:
So, you must first evaluate the integration range for y:
x = 0 > y = 5/2
x = 80 > y = 3/2
Then we've put:
y = Sinh(t)
So:
y = 5/2 > t = arcsinh(5/2)
y = 3/2 > t = arcsinh(3/2)

So what is the answer?