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 2 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?

dy/dt= 0.6m/s

Explanation:

Ok, so we hace two right triangles ABC and ADE which are similar triangles, so we have their corresponding sides like:

If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12-x

Now we have to take the derivatives of both sides

We are now going to divide the hole equation by dt

and y=3

Let´s sustitute the data on the equation

So we have that

answer: cesare beccaria developed the classical school of criminology with from philosopher jeremy bentham

explanation:

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s