Implicit Differentiation & Related Rates

[Gloamglozer]

1.  We have the equation .  Find the points on the curve which the tangent line is parallel to .

I've differentiated it and got:











If it is to be parallel to , then that means that .

So the next thing I should do would be:

  Right?

2.  Just to note, this is an assignment question.  Feel free to help me as much or as less as you wish. 

I think the main problem is identifying the rates.

A street light is mounted at the top of a 5.4 metre pole.  A man 1.8 m tall walks away from the light in a straight path at a speed of 1 m/s.  How fast is the top of the his shadow moving when he is 10 m from the pole?

I've drawn a picture:


From that, I've calculated the length of the shadow, which I can't find how it would relate to the question:

Let x be the length of the shadow.








I think what I need to do next is to get the m/s but I can't figure out what to do next with setting up the chain rule.

[Martoman]

For the first question, I see no problems in what you have done.

[m@tty]

For 1. yeah, let then and .

Sub that into the original relation, and solve.

[m@tty]

2.

You want to find the rate of change of length of shadow with respect to time.

Let distance from man to pole = d. As he is walking at 1 m/s,  d=t, where t is time in seconds.

Let length of shadow = l.

So you want which by the chain rule is equivalent to .

You should be able to figure it out from there, keeping in mind your earlier workings...


I did the working then remembered it is an assignment, and you may not want me to do it for you... But it's here anyway, because I didn't want to waste typing it all out...

\frac{dd}{dt}=1 So \frac{dl}{dt}=\frac{dl}{dd}

From the similar triangles you set up \frac{1.8}{l}=\frac{5.4}{d+l}

1.8d+1.8l=5.4l

l=\frac{1.8}{3.6}d

\frac{dl}{dd}=\frac{1}{2}.

Therefore, from earlier working, \frac{dl}{dt}=\frac{1}{2} \cdot 1 ms^{-1}

[Chavi]

Question 2 involves a few steps:
a) let the distance between the dude and the pole = d (not 10, as this figure is subject to change)
b) express x in terms of d: (similar triangles) solving for x gives
c) find the speed at which the length of the shadow is increasing:



from before so

as

d) Express, in terms of x, the distance fo the end point, P, of the shadow



and thus the change in shadow size/movement with respect to time:



hope that helps
m@tty beat me to it

[m@tty]

Ah, yes. You have to add the speed of movement of the guy relative to the pole as well. So my final answer is not quite correct.

[Ilovemathsmeth]

Yep I got x = 3/2^0.25 and y = -3/2^0.25

I also got x = -3/2^0.25 and y = 3/2^0.25

I got 3/2 m/sec for the second part.

[Gloamglozer]

Yep I got x = 3/2^0.25 and y = -3/2^0.25

I also got x = -3/2^0.25 and y = 3/2^0.25

I got 3/2 m/sec for the second part.

Awesomeness.  Good work ILLM.  H1 on the way for you. 

[Ilovemathsmeth]

I really hope so because my other subjects are going downhill... :S Good luck to you too!