# Homework Help: Related rates waliking away from light towards building

1. Oct 13, 2009

### dethnode

1. The problem statement, all variables and given/known data

a spotlight won the ground shines on a wall 12 m away if am man 2 m tall walks from the spotlight towards the building at a speed of 1.6 m/s how fast is the length of his shadow on the building decreasing when he si 4 m from the building?

2. Relevant equations
using relative triangles

3. The attempt at a solution
trying to learn related rates as well, this is what i got tell me if i am wrong here...

draw triangle ABC with A being the light, B being the base of building, and C being top of shaddow/building.

the second triangle is formend with the man and the light, using ADE, D being the mans feet and E being his head at 2 m height.

using the 2 meter horizontal from the mans height we have two relative triangles.

call the range from the light (line AD) x and call the building/shaddow (line BC) y

using the two triangles we can infer that 2/x=y/12 or xy=24

if we then differentiate relative to time 0=dx/dt * y + dy/dt * x

we then plug in 1.6 for dx/dt and 8 for x(range from the light not the building); and y=3 (using xy=24 @ x=8) and solve for dy/dt= -.6m/s which should be negative.

Is this correct?

2. Oct 13, 2009

### lanedance

wasn't this in the other posting?

3. Oct 13, 2009

### dethnode

yes i had already posted this, i have a new posting out titled related rates kite

4. Oct 13, 2009

### HallsofIvy

Yes, that is correct.

5. Oct 13, 2009

### HallsofIvy

Please don't post the same thing more than once!