How Fast Is the Distance Changing Between Mr. John and the Zombie?

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I am stuck on this question and need help asap pleasez.

Mr. John is climbing a tree at a rate of 2m/s. After 1 second of climbing Mr. John is noticed by a 2m tall zombie that is 30m away from the tree. The zombie walks towards Mr. John at a rate of 5m/s. What is the rate of change of the distance between Mr. John and the zombie when the angle of elevation from the zombie to Mr. John is Pi/4.


i kno that
Dy/Dt = 2m/s.
Dx/Dt is 5m/s.
theta is Pi/4.
i just can't find out how to do this pleasez help!
 
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Write down functions j(t) and z(t), then find the derivative of \sqrt{j^2(t) + z^2(t)}. Note that j and z are lines in R^2 space. jx is always 0 and zy is probably 2.
 


Could you please draw out the situation and then post it up? It would be of big assistance.
 


wild said:
I am stuck on this question and need help asap pleasez.

Mr. John is climbing a tree at a rate of 2m/s. After 1 second of climbing Mr. John is noticed by a 2m tall zombie that is 30m away from the tree. The zombie walks towards Mr. John at a rate of 5m/s. What is the rate of change of the distance between Mr. John and the zombie when the angle of elevation from the zombie to Mr. John is Pi/4.i kno that
Dy/Dt = 2m/s.
Dx/Dt is 5m/s.
theta is Pi/4.
i just can't find out how to do this pleasez help!

Some of what you know isn't true! The angle is a function of time, and is ##\pi/4## only at one particular time. Also, since the zombie (who writes these problems?) is approaching the tree, dx/dt = -5 m/s.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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