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Q: A kite $100 \ m$ above the ground moves horizontally at a speed of $8 \ m/s$. At what rate is the angle between the string and the horizontal, i.e. e the ground, decreasing when $200 \ m$ of string has been let out.

The only part I am having trouble with is forming a relationship between the angle $a$ and the horizontal displacement $x$.

I figured that it has something to do with $tan$, but I have difficulty in differentiating an inverse $tan$ function (Which I have not learnt or will learn yet) after finding the relationship between to be $tan \ a=100/x$.

Any help would be appreciated,

Thank you.

mvw
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Chris
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  • About the derivative: $x = f(f^{-1}(x)) \Rightarrow 1 = (f'(f^{-1}(x)) , (f^{-1})'(x)$ and thus $(f^{-1})'(x) = 1 / (f'(f^{-1}(x))$. – mvw Jul 07 '14 at 12:27

3 Answers3

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$$\frac{d}{dt} \arctan(t) = \frac{1}{1+t^2}$$

It is probably worth memorising this in order to be able to integrate $\frac{1}{1+t^2}$ but it can be found if you know the derivatives of $\sin$ and $\cos$:

$$ t = \tan (\theta) = \frac{ \sin (\theta)}{ \cos (\theta)}$$ $$ \frac{dt}{d\theta} = \frac{ \cos(\theta)}{ \cos (\theta)}+ \frac{ \sin^2(\theta)}{ \cos^2 (\theta)}=1+\tan^2(\theta)=1+t^2$$ $$ \frac{d\theta}{dt} = \frac{1}{1+t^2}$$

Henry
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Position vector $$ r(t) = (x(t),y(t))^T = (v_x t, y_0)^T $$

for $v_x = 8 \mbox{m}/\mbox{s}$, $y_0 = 100 \mbox{m}$.

Distance (assuming a straight string): $$ d^2 = x^2 + y^2 $$

For $d = 200 \mbox{m}$ and $y = y_0$ this means $x = 173.21 \mbox{m}$ and with $$ \tan \alpha = \frac{y}{x} = 0.57735 \Rightarrow \alpha = 30^\circ $$

differentiation for $t$ of both sides gives $$ (1 + \tan^2 \alpha) \dot{\alpha} = - \frac{y}{x^2} v_x $$

where we have all numbers to calculate

$$ \dot \alpha = -0.02 \frac{1}{\mbox{s}} = - 1.146 \frac{\circ}{\mbox{s}} $$

mvw
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For the sake of completeness, the numerical approach agrees with the above answer. In python:

from math import *

def angrate(height, length, siderate):
    horiz = sqrt(length**2 - height**2)
    startang = asin(height/length)
    print()
    for i in [2**n for n in range(15)]:
        newang = atan(height/(horiz + siderate/i))
        print(i, i*(newang - startang))
    input("\nEnter to exit. ")  

angrate(100, 200, 8)

Final answer: -0.02