1. ## Related Rates

The shadow casts by a man standing 1 m from a lamppost is 1.2 m long. If the man is 1.8 m tall and walks away from the lamppost at a speed of 120 m/min, at what rate is the shadow lengthening after 5 seconds?

This is my work:

I used Pythagorean theorm
a^2 + b^2 = c^2
differentiating
2a da/dt + 2b db/dt = 2c dc/dt
2(1) da/dt + 2(1.8) db/dt = 2(1/12* -120) dc/dt

2. Originally Posted by skeske1234
The shadow casts by a man standing 1 m from a lamppost is 1.2 m long. If the man is 1.8 m tall and walks away from the lamppost at a speed of 120 m/min, at what rate is the shadow lengthening after 5 seconds?

This is my work:

I used Pythagorean theorm
a^2 + b^2 = c^2
differentiating
2a da/dt + 2b db/dt = 2c dc/dt
Related Rates Example

3. Ok, so I ended up getting 14.4 m/min as my answer.. but the answer on the worksheet says 144 m/min..

Here's how I got my answer: we want to find how the length of the shadow is changing with time.. so s shadow is 1.2 m, and at 5 sec or 1/12 min, then (1.2 ) (1/12) = 14.4 m/min

?

4. Originally Posted by skeske1234
Ok, so I ended up getting 14.4 m/min as my answer.. but the answer on the worksheet says 144 m/min..

Here's how I got my answer: we want to find how the length of the shadow is changing with time.. so s shadow is 1.2 m, and at 5 sec or 1/12 min, then (1.2 ) (1/12) = 14.4 m/min

?
the 5 seconds doesn't matter.

ratio ... (distance from post)/(shadow length)

$\displaystyle \frac{x}{s} = \frac{1}{1.2}$

$\displaystyle s = 1.2 x$

$\displaystyle \frac{ds}{dt} = 1.2 \cdot \frac{dx}{dt}$

$\displaystyle \frac{ds}{dt} = 1.2(120) = 144$ m/min