A man 6 ft tall walks at a rate of 6 ft/sec away from a lamppost that is 24 ft high. At what rate is the length of his shadow changing when he is 75 ft away from the lamppost

2 ft/s

Step-by-step explanation:

The lamppost is 24 ft. tall, and the man is 6 ft. tall. So, we will use a proportion to find the shadow.

Let s is the length of the base of the lamppost to the shadow while x is the length of the base of the lamppost to the man, so the length of the shadow is s - x.

Using triangular ratio, we have;

24/6 = s/(s - x)

4 = s/(s - x)

We cross multiply and distribute to get;

4s - 4x = s

4s - s = 4x

3s = 4x

s = 4x/3

Taking the derivative of both sides according to time, we have;

ds/dt = (4/3)dx/dt

Now, dx/dt is given as 6 ft/s

So;

ds/dt = (4/3) × 6

ds/dt = 8 ft/s

For us to find the rate of length of the shadow according to time, we recall that the shadow = s - x, so we will just take the derivative of each and subtract. Thus;

d(s - x)/dt = ds/dt - dx/dt

Plugging in the relevant values, we have;

ds/dt - dx/dt = 8 - 6 = 2 ft/s

answer: i realy don't get it that much can you me figer it out pleas you

step-by-step explanation: