# Confusing Rates of Change

• Apr 9th 2010, 05:24 AM
Sunyata
Confusing Rates of Change
Hi

How do I do this:

A lamp is 40 m above the ground and a stone is dropped from the same height at a point 10 m away. Find the speed of the shadow on the ground given that in t seconds the stone falls $4.9t^2$ m.

Thanks a lot!
• Apr 9th 2010, 05:52 AM
Hello Sunyata
Quote:

Originally Posted by Sunyata
Hi

How do I do this:

A lamp is 40 m above the ground and a stone is dropped from the same height at a point 10 m away. Find the speed of the shadow on the ground given that in t seconds the stone falls $4.9t^2$ m.

Thanks a lot!

Study the diagram I have attached. The distance $x$ m represents the distance of the shadow from $C$, the point directly below $A$, where the stone was dropped.

Using similar triangles:
$\frac{RC}{BC}=\frac{RQ}{PQ}$
What you need to do now:
Substitute into this equation the values I've given you in the diagram.

Solve for $x$ in terms of $t$.

Differentiate to find $\frac{dx}{dt}$. This represents the speed of the shadow.

Can you complete this now?

• Apr 9th 2010, 06:56 AM
Sunyata
Ummm am I meant to get 4.9t^2x + 49t^2 = 400???

if so, when i implicitly differentiate that, I can't seem to get the answer...
• Apr 9th 2010, 07:43 AM
Hello Sunyata
Quote:

Originally Posted by Sunyata
Ummm am I meant to get 4.9t^2x + 49t^2 = 400???

if so, when i implicitly differentiate that, I can't seem to get the answer...

See step 2 above:
Solve for $x$ in terms of $t$
which gives
$x = \frac{400-49t^2}{4.9t^2}$
$=\frac{400}{4.9t^2}-10$
$\Rightarrow \frac{dx}{dt} = \frac{-800t^{-3}}{4.9}$
So the speed of the shadow has magnitude $\frac{8000}{49t^3}$.