## Part I ### Problem 1 (This question has already been solved in the lecture) A circular balloon has radius \ cm and volume Its radius is increasing at \ cm/s. Find \ when \ cm. ### Problem 2 (This question has already been solved in the lecture) A 10 m ladder rests against a vertical wall. The foot of the ladder is pulled away from the wall at 1.2 m/s. How fast is the top of the ladder moving when the foot is 6 m from the wall? ### Problem 3 The sides \ and \ of a rectangle change with time. At one instant, Find the rate of change of the area and state whether the area is increasing or decreasing. ### Problem 4 Two cars are approaching the same roundabout on perpendicular roads. Let \ and \ be their distances from the roundabout. At one instant, Car A is 0.6 km from the roundabout and moving toward it at 40 km/h, while Car B is 0.8 km from the roundabout and moving toward it at 30 km/h. How fast is the straight-line distance between the cars changing at that instant? ### Problem 5 For a differentiable function \ , the linearization at \ is (a) Find the linearization of \ at \ . Then use it to approximate \ . (b) Find the linearization of \ at \ . Use it to estimate \ . (c) Find the linearization of \ at \ . Use it to estimate \ . ### Problem 6 (a) For \ at \ , calculate \ and \ when \ . Compare the two values. (b) For \ at \ , calculate \ and \ when \ . Compare the two values.
Answer:
Step-by-step solution
Step 1: State the volume formula and given rates
We are given the volume formula of a spherical balloon . The radius expands at a constant rate of , and we want to find the rate of change of volume specifically when the radius reaches .
Step 2: Differentiate volume with respect to time
Applying the Chain Rule with respect to time , we differentiate to get . The factor of in the numerator cancels with the denominator , simplifying the derivative of volume to , which is the surface area multiplied by the rate of change of the radius.
Step 3: Substitute the given values into the rate equation
Substitute and into our rate equation. Squaring yields . Multiplying gives , leaving the exact rate of increase as , or approximately .