1. ## Related Rates Question /Shadow of a Man

A man 6ft tall is walking towards a streetlight 18ft high at a rate of 3ft/second.

a). At what rate is his shadow length changing?

=>
s is the length of his shadow. p is the length from him to the lamp.

Using proportional triangles:
18s = 6s + 6p
18ds/dt= 6ds/dt + 6dp/dt

3ds/dt = ds/dt + dp/dt
2ds/dt = dp/dt

2ds/dt = 3

ds/dt = 3/2 ft/s

b). How fast is the tip of his shadow moving ?

I don't understand what they mean by the tip of his shadow. If the man is walking towards the right, is the very far left of his shadow ? And wouldn't it be moving at the same speed that the length is changing ?

2. the tip of the shadow moves along the ground with a speed = ds/dt + dp/dt

btw ... both ds/dt and dp/dt are negative values because the respective distances are decreasing.

3. Hello, StarlitxSunshine!

. . but you got the right answer.

A man 6 ft tall is walking towards a streetlight 18 ft high at a rate of 3 ft/sec.

a) At what rate is his shadow length changing?
Code:
. . C *
. . - |  *
. . - |     *
. . - |        *  A
. .18 |           *
. .   |           |  *
. .   |          6|     *
. .   |           |        *
. .   * - - - - - * - - - - - *
. .   D     p     B     s     E

The man is $AB = 6.$
The streelight is $CD = 18.$
His distance from the streetlight is: . $p = BD.$
. . $\frac{dp}{dt} \:=\:3$ (decreasing)

The length of his shadow is: . $s = BE.$
. . We want: $\frac{dx}{dt}$

From the similar triangles: . $\frac{s}{6} \:=\:\frac{p+s}{18} \quad\Rightarrow\quad s \:=\:\tfrac{1}{2}p$

Differentiate with respect to time: . $\frac{ds}{dt} \:=\:\tfrac{1}{2}\,\frac{dp}{dt}$

Therefore: . $\frac{ds}{dt} \:=\:\tfrac{1}{2}(3) \:=\:\frac{3}{2}$ ft/sec.

b) How fast is the tip of his shadow moving?

I don't understand what they mean by the tip of his shadow.
Wouldn't it be moving at the same speed that the length is changing?
No, this is a different situation.

Code:
. . C *
. .   |  *
. .   |     *
. . - |        *  A
. .18 |           *
. .   |           |  *
. .   |          6|     *
. .   |           |        *
. .   * - - - - - * - - - - - *
. .   D     p     B    x-p    E
. .   : - - - - - x - - - - - :

The man is $AB = 6.$
The streelight is $CD = 18.$
His distance from the streetlight is: . $p = BD.$
. . $\frac{dp}{dt} \:=\:3$ (decreasing)

The distance of the tip of his shadow to the streetlight is: $x \:=\:DE$

From the similar triangles: . $\frac{x-p}{6} \:=\:\frac{x}{18} \quad\Rightarrow\quad x \:=\:\tfrac{3}{2}p$

Differentiate with respect to time: . $\frac{dx}{dt} \:=\:\tfrac{3}{2}\frac{dp}{dt}$

Therefore: . $\frac{dx}{dt} \:=\:\tfrac{3}{2}(3) \:=\:\frac{9}{2}$ ft/sec.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Some explanation . . .

In part (a), they ask for the rate of change of the length of the shadow.
That is, how fast is the tip of his shadow $(E)$ moving towards his feet $(B)$?
. . We found this to be: . $1\tfrac{1}{2}$ ft/sec.

In part (b), they ask for the rate of change of point $E$ relative to the world.
Relative to the streetlight, the distance is $x = DE.$
. . We found that: . $\frac{dx}{dt} \:=\:4\tfrac{1}{2}$ ft/sec.

To put it yet another way:

The tip of his shadow is moving towards his feet at 1½ ft/sec.

But his feet are moving at 3 ft/sec.

So, relative to world, the tip of his shadow is moving at 4½ ft/sec.

4. Originally Posted by Soroban
Hello, StarlitxSunshine!

. . but you got the right answer.

Code:
. . C *
. . - |  *
. . - |     *
. . - |        *  A
. .18 |           *
. .   |           |  *
. .   |          6|     *
. .   |           |        *
. .   * - - - - - * - - - - - *
. .   D     p     B     s     E
The man is $AB = 6.$
The streelight is $CD = 18.$
His distance from the streetlight is: . $p = BD.$
. . $\frac{dp}{dt} \:=\:3$ (decreasing)

The length of his shadow is: . $s = BE.$
. . We want: $\frac{dx}{dt}$

From the similar triangles: . $\frac{s}{6} \:=\:\frac{p+s}{18} \quad\Rightarrow\quad s \:=\:\tfrac{1}{2}p$

Differentiate with respect to time: . $\frac{ds}{dt} \:=\:\tfrac{1}{2}\,\frac{dp}{dt}$

Therefore: . $\frac{ds}{dt} \:=\:\tfrac{1}{2}(3) \:=\:\frac{3}{2}$ ft/sec.

No, this is a different situation.

Code:
. . C *
. .   |  *
. .   |     *
. . - |        *  A
. .18 |           *
. .   |           |  *
. .   |          6|     *
. .   |           |        *
. .   * - - - - - * - - - - - *
. .   D     p     B    x-p    E
. .   : - - - - - x - - - - - :
The man is $AB = 6.$
The streelight is $CD = 18.$
His distance from the streetlight is: . $p = BD.$
. . $\frac{dp}{dt} \:=\:3$ (decreasing)

The distance of the tip of his shadow to the streetlight is: $x \:=\E" alt="x \:=\E" />

From the similar triangles: . $\frac{x-p}{6} \:=\:\frac{x}{18} \quad\Rightarrow\quad x \:=\:\tfrac{3}{2}p$

Differentiate with respect to time: . $\frac{dx}{dt} \:=\:\tfrac{3}{2}\frac{dp}{dt}$

Therefore: . $\frac{dx}{dt} \:=\:\tfrac{3}{2}(3) \:=\:\frac{9}{2}$ ft/sec.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Some explanation . . .

In part (a), they ask for the rate of change of the length of the shadow.
That is, how fast is the tip of his shadow $(E)$ moving towards his feet $(B)$?
. . We found this to be: . $1\tfrac{1}{2}$ ft/sec.

In part (b), they ask for the rate of change of point $E$ relative to the world.
Relative to the streetlight, the distance is $x = DE.$
. . We found that: . $\frac{dx}{dt} \:=\:4\tfrac{1}{2}$ ft/sec.

To put it yet another way:

The tip of his shadow is moving towards his feet at 1½ ft/sec.

But his feet are moving at 3 ft/sec.

So, relative to world, the tip of his shadow is moving at 4½ ft/sec.

very helpful. sorry to rebump such an old thread by im a little shaky on one thing

i just want to get something cleared up for me.. would both answers be the same if he was moving AWAY from the lamp post, based on your explanation, it seems that way. please correct me if im wrong

5. If intersted for an animation and solution relating to this pblm see

Optimization and Related Rates

,

,

,

### a man 6ft tall is walking at the rate of 3 ft/s toward a streetlight

Click on a term to search for related topics.