Help with Related Rates Calculus.

Question

A tractor is moving at the rate of 11.8 ft/s away from a building 90.8 ft high. How fast is the distance to the top of the building increasing when the tractor is 151 ft away from the base of the building?

Answer 1

The Speed of the Tractor = 11.8 ft/sec.

Height of the Building = 90.8 ft.

Distance travelled by the Tractor from the base of the Building = 151 ft.

Let s be the Distance from top of the building to the tractor when x = 151 ft.

Apply Pythagoras Theorem

S² = X² + h²

S² = (151)² + (90.8)²

S² = 31045.64

S = 176.19 ft.

The Distance from top of the building to the tractor when x = 151 ft is 176.19 ft.

We know that Speed = Distance / time

Time = Distance / Speed

Time taken by Tractor = 151 / 11.8

t = 12.7966 sec.

Then the Change in the distance from top of the building to the tractor is = Distance / time

= 176.19 / 12.7699

= 13.7684 ft/sec

Therefore the Change in the distance from top of the building to the tractor is 13.7684 ft/sec.