# Rate of how fast a shadow grows as you walk

#### daigo

##### Member There is a lamp post 15 feet tall casting a shadow 'B' ft. long of a person that is 6 feet tall standing 'A' ft. from the lamp post. If the person moves away from the lamp post at 5 feet per second, how fast does the shadow lengthen?
So here I thought it might be something to formulate a triangle, or actually two: So the problem is asking me to find the derivative of the shadow's length with respect to time: $$\frac{dB}{dt}$$

I think the triangles are similar:

$$\frac{15}{A+B} = \frac{6}{B} \\ \\ \cdots \\ B = \frac{2}{3}A$$

So we have B and I will attempt to take the derivative of this:

$$\frac{dB}{dt} = [\frac{2}{3}A]'$$

Constant/Chain rules:

$$\frac{dB}{dt} = \frac{2}{3} \cdot [A'] \\ = \frac{2}{3} \cdot \frac{dA}{dt}$$

Now how do I find the derivative of 'A'? It must have something to do with the movement velocity because it was given information and I haven't used it yet. Unless I am being tricked and it is extraneous/unnecessary information.

#### HallsofIvy

##### Well-known member
MHB Math Helper
You've done very well. Now, with your "A" and "B", the total distance from the light pole to the person is A+ B. You are told that the person is walking away from the light pole at 5 ft/s so that (A+ B)'= A'+ B'= 5 and therefore A'= 5- B'. Replace A' in your last equation with that and solve for B'.

#### Reckoner

##### Member
You've done very well. Now, with your "A" and "B", the total distance from the light pole to the person is A+ B. You are told that the person is walking away from the light pole at 5 ft/s so that (A+ B)'= A'+ B'= 5 and therefore A'= 5- B'. Replace A' in your last equation with that and solve for B'.
The distance from the pole to the person is actually $$A$$, according to the original poster's drawing.

So, daigo, if the person is moving 5 feet per second away from the post, then how fast is $$A$$ increasing?

#### daigo

##### Member
The distance from the pole to the person is actually $$A$$, according to the original poster's drawing.

So, daigo, if the person is moving 5 feet per second away from the post, then how fast is $$A$$ increasing?
By 5 ft./sec.? This is a constant velocity, so...this is just a slope of 0 on the y-axis since it's a horizontal line. If function 'A' is the distance travelled while going at a constant 5 ft./sec (f(A) = 5t), then the slope of a line is always going to be the same so taking the derivative of a line like '5t' would just be a constant 5, I think?

#### Reckoner

##### Member
By 5 ft./sec.?
Yes. $$A$$ is increasing by 5 ft every second. Therefore,

$\frac{dA}{dt} = 5\ \mathrm{ft}/\mathrm{s}.$

Now substitute this into your equation.

• daigo