# Math Help - Shadow Problem

Didn't know if this was the right category to put this into but anyway.

I got the following problem. There is a lantern that has a height of 11 meters
a man with a height of 2.20 meters is standing at the lantern's base. The man starts moving away from the lantern with a velocity of v=2 m/s. I need to find with what velocity his shadow grows. Please this is urgent. Thx from now.

2. Originally Posted by pokopikos
Didn't know if this was the right category to put this into but anyway.

I got the following problem. There is a lantern that has a height of 11 meters
a man with a height of 2.20 meters is standing at the lantern's base. The man starts moving away from the lantern with a velocity of v=2 m/s. I need to find with what velocity his shadow grows. Please this is urgent. Thx from now.
Search the forum for "shadow length". You'll get 57 examples completely done for your problem. Pick the one which is most convenient for you.

3. Originally Posted by pokopikos
Didn't know if this was the right category to put this into but anyway.

I got the following problem. There is a lantern that has a height of 11 meters
a man with a height of 2.20 meters is standing at the lantern's base. The man starts moving away from the lantern with a velocity of v=2 m/s. I need to find with what velocity his shadow grows. Please this is urgent. Thx from now.
Firstly you need to find the general equation relating the variables. You can draw a diagram to represent the general situation. At this time man's height is 2.20 meters and the lantern that has a height of 11 m. What varies in the general case? Well, man is moving away from the lantern so the distance from him to the base of latern is changing. We'll label that distance "x". Of course we are interested in the shadow, so let's label the length of shadow "y".

Now if you see the diagram you made, notice that there are two right triangles in the picture, one with base "y" and height 2.20, the other with base "x+y" and and height 11. These two triangles are similar triangles and their angles are the same so that their ratios of their corresponding edges should be equal:

$\frac{2.20}{y} = \frac{11}{x+y}$

$2.20(x+y) = 11y$

$2.20x = 8.8y$

$x=4y$

The rate of change of the length of the shadow is the derivative of y. So we want to find $\frac{dy}{dt}$. Since he is moving away from the lantern at 2 m/s, we know that $\frac{dx}{dt}=2$ (if he was moving towards the lantern it'd be -2)

Now you must differentiate the general equation implicitly

$\frac{dx}{dt} = 4 \frac{dy}{dt}$

Do you see how it's done?