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?
For Further Reading
* Calculus - Count Iblis, Thursday, June 14, 2007 at 9:10am
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 it is:
((e^(3/2)-e^(-3/2))/2 - ((e^(-5/2)-e^(5/2))/2
= 8.1795 ft?
This answer was incorrect. I guess I am still missing something?
The integral calculated by you was:
F(t) = 5(-2t - (e^-2t)/2 + (e^2t)/2)
The limits are t1 = -arcsinh(5/2)
and t2 = arcsinh(3/2)
The path length is thus:
F(t2) - F(t1)
Ok thanks a lot !!!
122.7761
You're welcome! I hope my answers were helpful. If you have any more questions, feel free to ask. And remember, if all else fails, just add a little humor to lighten the mood!
You're welcome! I'm glad I could help. If you have any more questions, feel free to ask!
To find the distance traveled by the kite, you need to integrate the function sqrt[1 + (1/400)x^2 - (1/4)x + 29/4] from x = 0 to x = 80.
The first step is to simplify the expression inside the square root. You have already correctly simplified it to 1 + (1/400)x^2 - (1/4)x + 29/4.
To integrate this expression, you can make a substitution. Let's substitute y = (1/20)x - 5/2. This substitution simplifies the expression inside the square root to sqrt[1 + y^2].
Now, you need to find the limits of integration in terms of y. Substitute x = 0 and x = 80 in the equation y = (1/20)x - 5/2:
When x = 0, y = 0 - 5/2 = -5/2.
When x = 80, y = (1/20)(80) - 5/2 = 4 - 5/2 = 3/2.
Therefore, the integral becomes ∫sqrt[1 + y^2] dy, with the limits of integration from y = -5/2 to y = 3/2.
To further simplify the integral, make one more substitution. Let's substitute z = sinh(t), where sinh(t) is the hyperbolic sine function.
With this substitution, the integral becomes ∫20cosh(t) dt.
The integral of cosh^2(t) can be computed by using the identity cosh^2(t) = (1/2)(cosh(2t) + 1).
So, the integral becomes ∫20(1/2)(cosh(2t) + 1) dt, which simplifies to ∫10(cosh(2t) + 1) dt.
Now, you can integrate term by term:
∫cosh(2t) dt = (1/2)sinh(2t) + C1
∫1 dt = t + C2
Putting it all together, the integral becomes:
10[(1/2)sinh(2t) + t] evaluated from t = -arcsinh(5/2) to t = arcsinh(3/2).
Evaluating the integral at these limits gives you the distance traveled by the kite.