1
$\begingroup$

A light on the ground is 30 feet away from a building. A 4 foot tall man is walking from the light to the building at a rate of 3 feet per second. He is casting a shadow on the side of the building. At what rate is his shadow shrinking when he is 5 feet from the building?

This is my figure: enter image description here Is this correct?

How do I solve this problem? I tried using similar triangles but couldn't succeed. Could someone explain this with a figure?

Modified figure: light is on the ground. enter image description here

$\endgroup$
0

2 Answers 2

Reset to default
2
$\begingroup$

Perhaps you can do the drawing. Let $L$ be the location of the light. (It is on the ground!) Let $F$ be the location of the man's feet, and let $H$ be the top of his head. Let $B$ be the bottom of the building. Draw the horizontal line $LB$. Draw the vertical line $FH$ to represent the short thin man.

Draw the line through $L$ and $H$. Let it meet the wall of the building at $S$. Then $BS$ is the length of the shadow.

It is convenient to call the distance $LF$ by the name $x$. We are told the man is walking at $3$ feet per second, so $\frac{dx}{dt}=3$.

Let $y=BS$. Note that triangles $LBS$ and $LFH$ are similar, so $$\frac{y}{30}=\frac{4}{x}.$$ We may want to rewrite this as $$y=\frac{120}{x}.$$ Differentiate. We get $$\frac{dy}{dt}=-\frac{120}{x^2}\frac{dx}{dt}.$$ Now it is almost over.

$\endgroup$
4
  • $\begingroup$ Why did you write the derivative of 120/x to be -120/x^2 *dx/dt and not just -120/x^2 $\endgroup$
    – Minu
    Jul 3, 2014 at 21:43
  • $\begingroup$ Because $x$ is a function of time, and I was differentiating with respect to $t$. The answer I got uses the Chain Rule. The derivative of $120/x^2$ with respect to $t$ is the derivative with respect to $x$, times $\frac{dx}{dt}$. Chain Rule comes up lots in related rates problems. $\endgroup$
    – André Nicolas
    Jul 3, 2014 at 21:45
  • $\begingroup$ Ok. X is a function of t. Thank you $\endgroup$
    – Minu
    Jul 3, 2014 at 21:46
  • $\begingroup$ You are welcome. Just saw your altered picture. It is right. $\endgroup$
    – André Nicolas
    Jul 3, 2014 at 21:48
0
$\begingroup$

My approach would be to define a function which gives us the shadow height (S) in dependence of his walked distance (x):

x/4 = 30/S -> S(x) = 120/x

Now we know that x(t) = 3*t -> S(t)= 40/t.

All you have to do now is to find S'(25)!

but don't take my word for it and ask if something is unclear.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .