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.
For Further Reading
* 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)
So what is the answer?
To find the distance traveled by the kite, we first need to evaluate the integral of sqrt[1 + 1/400 (x-50)^2] with respect to x.
1. Substitute x = 20y + 50 to change the variable from x to y.
2. The integral becomes 20 times the integral of sqrt[1 + y^2] with respect to y.
3. Substitute y = sinh(t), where sinh is the hyperbolic sine function.
4. The integral then becomes 20 times the integral of cosh^2(t) with respect to t.
5. The integral of cosh^2(t) can be computed as a sum of exponentials using the identity cosh^2(t) = (e^t + e^(-t))/2.
6. Integrate term by term and evaluate the resulting expression at the upper and lower limits of t.
7. Finally, multiply the result by 20 to get the total distance traveled by the kite.
It seems like you may have made a mistake when evaluating the limits for y and t. Make sure to evaluate those correctly and then compute the integral using the steps outlined above.
To integrate the expression you have, you need to perform a change of variables. Let's go through the steps again:
1. Start with the expression: ∫√[(1/400)x^2 - (1/4)x + 29/4] dx
2. Rewrite the expression using a new variable y: x = 20y + 50
3. Find the new limits of integration for y by replacing x in the original limits:
- When x = 0, y = (-5/2)
- When x = 80, y = (3/2)
4. Substitute the new expression for x into the original:
∫√[(1/400)(20y + 50)^2 - (1/4)(20y + 50) + 29/4] dx
5. Simplify:
∫√[(1/400)(400y^2 + 2000y + 2500) - (1/4)(20y + 50) + 29/4] dx
= ∫√[(y^2 + 5y + 25/4)] dx
6. Substitute y = sinh(t) to further simplify:
∫√[(sinh^2(t) + 5*sinh(t) + 25/4)] dx
7. The square root becomes a hyperbolic cosine, so the integral becomes:
20 ∫cosh(t) dt
8. Integrate cosh^2(t) by using the identity cosh^2(t) = sinh^2(t) + 1:
20 ∫[(sinh^2(t) + 1)] dt = 20[t + sinh(t)cosh(t)] + C
9. Evaluate the definite integral by substituting the limits:
20[(t + sinh(t)cosh(t))] from -arcsinh(5/2) to arcsinh(3/2)
10. Calculate the values at the upper and lower limits:
20[(arcsinh(3/2) + sinh(arcsinh(3/2))*cosh(arcsinh(3/2)))] - 20[(-arcsinh(5/2) + sinh(-arcsinh(5/2))*cosh(-arcsinh(5/2)))]
11. Simplify the expression and evaluate the result.
The result you calculated, 7.6746e69, seems unlikely. Please double-check your calculations and make sure you entered the limits correctly.