A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet of the ground (see figure). (a) When he is 10 feet from the base of the light, at what rate is the tip of his shadow moving?

(b) When he is 10 feet from the base of the light, at what rate is the length of his shadow changing?

Thank you in advance!

a) ds/dt = 25/3 ft/s

b) 10/3 ft/s

Step-by-step explanation:

Using the triangles and the ratios given

15 ft            s

=

6 ft           s-x

where s is the length from the lamppost  to the shadow and x  is the length from the lamppost to the man

Using cross products

15( s-x) = 6s

15s -15x = 6s

15s-6s = 15x

9s = 15x

s = 15/9 x

s = 5/3 x

Taking the derivative of each side with respect to time

ds/dt = 5/3 dx/dt

We know dx/dt = 5 ft/s

ds /dt = 5/3 * 5 = 25/3 ft/s

a) ds/dt = 25/3 ft/s

The length of the shadow is ( s-x)

d/dt ( s-x)

ds/dt - dx/dt

25/3 - 5

25/3 - 15/3

10/3 ft/s

b) 10/3 ft/s

ii6tyhj7 juj3yyj333ju jmyhthhtk7h jg hajj jiijh

9 sandwiches per hour

step-by-step explanation:

when we have a rate, we want a 1 in the denominator

27/3 = 9

27 sandwiches/ 3 hours = 9 sandwiches/hour