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.

h df tyjgtyj ttd y ytkgfty ty

step-by-step explanation:

tyty jry j y yj yttj yhgtyj y ghyj ytthj

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)