Question

Question No: 5 Marks: 3+3+3 +4 a) Assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 6 ft/s. i) Find the rate at which the circumference is increasing ii) How fast is the area of the spill increasing when the radius of the spill is 30 ft? b) The three different balls are thrown vertically upwards from the top of a building, the following functions express the height above the ground of an object after it is thrown (height in meters, time in seconds) Basketball: h(t) = -167 +48 + 20 Football : h(t) = -16° +50t+36 Volleyball : h(t) = -167 +30t +25 i) Find the acceleration of the Football. ii) Find the velocity of the Volleyball at t = 3 seconds. iii) How long does it take for the Basketball to hit the ground? c) Let f(x)=-1+ with initial condition f(4) = 2. Does y=f(x) have a local minimum, a local maximum, or neither at x=4? Justify your answer. х

Answer 1

a)

i)

Let r be the radius of the oil spilled and C be its circumference.

As the oil spreads in a circular pattern,

Differentiating both sides of the above equation with respect to time t, we get

The radius increases at a constant rate of 6 ft/s which means

The rate of change of circumference is

Hence, the circumference is increasing at the rate of 12 ft/s.

ii)

Let r be the radius of the oil spilled and A be its area.

As the oil spreads in a circular pattern,

Differentiating both sides of the above equation with respect to time t, we get

The radius increases at a constant rate of 6 ft/s which means

The rate of change of area at r = 30 ft is

Hence, the area is increasing at a rate of 360/s when the radius of the spill is 30 ft.

b)

i)

The height(in meters) of the Football from the ground after t seconds of throwing is given by

The acceleration of the Football is the second derivative of the height function h(t).

Hence, the acceleration of the Football is -32 m/.

ii)

The height(in meters) of the Volleyball from the ground after t seconds of throwing is given by

The velocity of the Volleyball is the first derivative of the height function h(t).

At t = 3 seconds, the velocity is

Hence, the velocity of the Volleyball at t = 3 seconds is -66 m/s.

Note - I have answered 4 sub-parts as HOMEWORKLIB guidelines.