# Homework Help: Trigonometric Calculus Problem Solving Question

1. May 7, 2014

### sonic25

1. The problem statement, all variables and given/known data
A spotlight on the ground shines on a wall 14m away. A person of height 2 m walks toward the wall on a direct path between the spotlight and the wall at a rate of 5/3m/s. How fast is the height of the shadow on the wall changing when the person is 4m from the wall?

2. Relevant equations
Not sure? I think implicit differentiation needs to be employed though.

3. The attempt at a solution
I tried to construct a right-angled triangle to use as a diagram to represent the situation, however this didn't really help as I cannot identify which variables I need to use to answer the problem.

Thanks in advance for any help:)

2. May 7, 2014

### Sunil Simha

Hi sonic,

You are supposed to find how the height of the shadow varies with time right? You know that the height depends on the distance of the person from the source. Just write down the relationship between the two and see how it varies with time.

3. May 7, 2014

### sonic25

Yeah I'm just not sure as to how the 2m height of the person affects this relationship (if it does at all)?

Last edited: May 7, 2014
4. May 7, 2014

### xiavatar

Think of a straight line that goes from the spotlight to the top of the guys head and onto the wall. You will have similar triangles, and then you can construct the ratio.

Hint*: What is the length the man has traveled after time t.

5. May 7, 2014

### sonic25

Ok. Here is my latest attempt at a solution:
I devised a diagram like you said and tried to use a ratio.
Let x be the distance from the wall and h be the height of the shadow above the wall.
tan(theta)=h/14
So, in terms of h: h=14tan(theta)
But tan(theta)=2/(14-x)
Substitute this into h=14tan(theta),
h=14(2/(14-x))
I then implicitly differentiated this with respect to time.
dh/dt=-28((14-x)^(-2))(-dx/dt)
I then substituted the values given in the question:
dh/dt=35/12

Is this correct?

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Last edited: May 7, 2014
6. May 7, 2014

### xiavatar

Your answer is incorrect. First note that you want the height to be a function of time,$t$. Your diagram is correct, but forget $x$ and $\theta$. After a time t, how far is the man from the spotlight at time $t$. Now this will be the length the man is from the spotlight. Now you have two similar triangles. One of these triangles is the man and his distance from the spotlight. And the other triangle composed of the height of the wall and the distance the wall is from the spotlight.

Last edited: May 7, 2014
7. May 7, 2014

### sonic25

I used the similar triangles and gathered that the ratio of (5/3)t:14 was equal to 2:h.
By rearrangement, h=84/(5t)
Therefore, dh/dt=-84/(5t^2)
Since it takes 6 seconds for the man to reach the point 4m from the wall, I substituted t=6.
So, my end result for dh/dt was -7/15.