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 $\displaystyle 4.9t^2$ m.

Thanks a lot!

Printable View

- Apr 9th 2010, 05:24 AMSunyataConfusing 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 $\displaystyle 4.9t^2$ m.

Thanks a lot! - Apr 9th 2010, 05:52 AMGrandad
Hello SunyataStudy the diagram I have attached. The distance $\displaystyle x$ m represents the distance of the shadow from $\displaystyle C$, the point directly below $\displaystyle A$, where the stone was dropped.

Using similar triangles:$\displaystyle \frac{RC}{BC}=\frac{RQ}{PQ}$What you need to do now:

Substitute into this equation the values I've given you in the diagram.Can you complete this now?

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

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

Grandad - Apr 9th 2010, 06:56 AMSunyata
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 AMGrandad
Hello SunyataSee step 2 above:

Solve for $\displaystyle x$ in terms of $\displaystyle t$which gives

$\displaystyle x = \frac{400-49t^2}{4.9t^2}$So the speed of the shadow has magnitude $\displaystyle \frac{8000}{49t^3}$.$\displaystyle =\frac{400}{4.9t^2}-10$$\displaystyle \Rightarrow \frac{dx}{dt} = \frac{-800t^{-3}}{4.9}$

Grandad