## Part I ### Problem 1 (This question has already been solved in the lecture) A circular balloon has radius \ rr cm and volume V=43πr3V=\frac{4}{3}\pi r^3 Its radius is increasing at \ 0.500.50 cm/s. Find \ dVdt\frac{dV}{dt} when \ r=6r=6 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 \ xx and \ yy of a rectangle change with time. At one instant, x=8 cm,y=5 cmx=8\text{ cm},\quad y=5\text{ cm} dxdt=0.7 cm/s,dydt=−0.2 cm/s\frac{dx}{dt}=0.7\text{ cm/s},\quad \frac{dy}{dt}=-0.2\text{ cm/s} 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 \ xx and \ yy 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 \ ff, the linearization at \ x=ax=a is L(x)=f(a)+f′(a)(x−a)L(x)=f(a)+f'(a)(x-a) (a) Find the linearization of \ f(x)=x+5f(x)=\sqrt{x+5} at \ a=4a=4. Then use it to approximate \ 9.2\sqrt{9.2}. (b) Find the linearization of \ f(x)=cos⁡xf(x)=\cos x at \ a=0a=0. Use it to estimate \ cos⁡(0.06)\cos(0.06). (c) Find the linearization of \ f(x)=exf(x)=e^x at \ a=0a=0. Use it to estimate \ e0.03e^{0.03}. ### Problem 6 (a) For \ y=x3−2x+1y=x^3-2x+1 at \ x=2x=2, calculate \ Δy\Delta y and \ dydy when \ dx=0.05dx=0.05. Compare the two values. (b) For \ y=x3+2x2−5x+4y=x^3+2x^2-5x+4 at \ x=1x=1, calculate \ Δy\Delta y and \ dydy when \ dx=0.04dx=0.04. Compare the two values.

Answer: dVdt=72π cm3/s≈226.19 cm3/s\frac{dV}{dt} = 72\pi\text{ cm}^3\text{/s} \approx 226.19\text{ cm}^3\text{/s}

Step-by-step solution

Step 1: State the volume formula and given rates

We are given the volume formula of a spherical balloon V=43πr3V = \frac{4}{3}\pi r^3. The radius expands at a constant rate of drdt=0.50 cm/s\frac{dr}{dt} = 0.50\text{ cm/s}, and we want to find the rate of change of volume dVdt\frac{dV}{dt} specifically when the radius reaches r=6 cmr = 6\text{ cm}.

Step 2: Differentiate volume with respect to time

Applying the Chain Rule with respect to time tt, we differentiate r3r^3 to get 3r2drdt3r^2 \frac{dr}{dt}. The factor of 33 in the numerator cancels with the denominator 33, simplifying the derivative of volume to 4πr2drdt4\pi r^2 \frac{dr}{dt}, 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 r=6r = 6 and drdt=0.50\frac{dr}{dt} = 0.50 into our rate equation. Squaring 66 yields 3636. Multiplying 4×36×0.504 \times 36 \times 0.50 gives 7272, leaving the exact rate of increase as 72π cm3/s72\pi\text{ cm}^3\text{/s}, or approximately 226.19 cm3/s226.19\text{ cm}^3\text{/s}.

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