Aspotlight on the ground shines on a wall 12 m away. if a man 2 m tall walks along the x-axis from the spotlight toward the building at a speed of 1.8 m/s, which is taken as the given dx/dt, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man=

Therefore,

Differentiate w.r.t t

When the man is 4 m from  the building

Then, we have x=12-4=8 m

Substitute the values in above equation then, we get

Substitute the values then we get

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23