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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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