# Calculus

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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)(x-50)^2

Find the distance traveled by the kite.

y = 150 - (1/40)(x-50)^2
y = 150 - (1/40)(x-50)(x-50)
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.

* Calculus - Count Iblis, Wednesday, June 13, 2007 at 9:50pm

You can calculate the derivative directly as:

y' = -2 (1/40)(x-50)

using the chain rule. Then you find:

1 + y'^2 = 1 + 1/400 (x-50)^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 (x-50)^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)

• Calculus - ,