1. **Problem:** Air is pumped into a spherical balloon at 5 cm³/min. Find the rate of change of the radius when diameter is 20 cm. Step 1: Volume of sphere is $$V=\frac{4}{3}\pi r^3$$. Step 2: Differentiate w.r.t. time $$t$$: $$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$. Step 3: Given $$\frac{dV}{dt}=5$$, diameter 20 cm implies radius $$r=10$$ cm. Step 4: Solve for $$\frac{dr}{dt}$$: $$5 = 4 \pi (10)^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{5}{400\pi} = \frac{1}{80\pi}$$ cm/min. --- 2. **Problem:** A 15 ft ladder leaning against a wall is pushed toward the wall at 0.25 ft/sec starting 10 ft away. Find how fast the top rises after 12 seconds. Step 1: Use Pythagoras: $$x$$ distance from wall, $$y$$ height; $$x^2 + y^2 = 15^2 = 225$$. Step 2: Initially $$x=10$$ ft, rate $$\frac{dx}{dt} = -0.25$$ ft/sec (toward wall). Step 3: After 12 sec, $$x = 10 - 0.25\times12 = 7$$ ft. Step 4: Solve for $$y$$: $$y = \sqrt{225 - 7^2} = \sqrt{225-49} = \sqrt{176}$$. Step 5: Differentiate: $$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \Rightarrow y \frac{dy}{dt} = -x \frac{dx}{dt}$$. Step 6: Solve $$\frac{dy}{dt}$$: $$\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} = -\frac{7}{\sqrt{176}} (-0.25) = \frac{7 \times 0.25}{\sqrt{176}} = \frac{1.75}{13.266} \approx 0.132$$ ft/sec. --- 3. **Problem:** Cone-shaped tank leaks at 2 ft³/hr. Radius = 5 ft, height = 14 ft. (a) Rate of change of water depth when depth = 6 ft. (b) Rate of change of radius of water surface when depth = 6 ft. Step 1: Volume formula: $$V=\frac{1}{3} \pi r^2 h$$, with $$r$$ radius of water surface, $$h$$ depth. Step 2: Relate $$r$$ and $$h$$ using similar triangles: $$\frac{r}{h} = \frac{5}{14} \Rightarrow r=\frac{5}{14}h$$. Step 3: Substitute: $$V=\frac{1}{3} \pi \left(\frac{5}{14} h\right)^2 h= \frac{1}{3} \pi \frac{25}{196} h^3= \frac{25\pi}{588} h^3$$. Step 4: Differentiate w.r.t. time: $$\frac{dV}{dt} = \frac{25\pi}{588} \times 3 h^2 \frac{dh}{dt} = \frac{75\pi}{588} h^2 \frac{dh}{dt}$$. Step 5: Given $$\frac{dV}{dt} = -2$$ ft³/hr, depth $$h=6$$ ft. Step 6: Solve for $$\frac{dh}{dt}$$: $$-2 = \frac{75\pi}{588} \times 36 \times \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{-2}{\frac{75\pi}{588} \times 36} = \frac{-2}{\frac{75\pi \times 36}{588}} = \frac{-2 \times 588}{75 \pi \times 36} = \frac{-1176}{2700 \pi} = -\frac{14}{225 \pi} \approx -0.0198$$ ft/hr. Step 7: For $$\frac{dr}{dt}$$, recall $$r=\frac{5}{14}h$$, so $$\frac{dr}{dt} = \frac{5}{14} \frac{dh}{dt} = \frac{5}{14} \times (-\frac{14}{225 \pi}) = -\frac{5}{225\pi} = -\frac{1}{45\pi} \approx -0.00707$$ ft/hr. --- 4. **Problem:** Two cyclists start 350 m apart. A rides north at 5 m/sec from start time; B starts 7 min (420 sec) later south at 3 m/sec. Find rate of change of distance between them 25 min after A starts. Step 1: Positions after $$t=1500$$ sec (25 min): A position north: $$5\times1500=7500$$ m B position south: Start at 420 sec, so moving for $$1500 - 420=1080$$ sec, distance $$3 \times 1080 = 3240$$ m south. Step 2: Distance separating them: $$s=350 + 7500 + 3240 = 11090$$ m. Step 3: Rate of change of distance: $$\frac{ds}{dt}=5 + 3 = 8$$ m/sec (since moving in opposite directions). --- 5. **Problem:** Two resistors in parallel: $$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$$, with $$R_1=80\Omega$$ increasing at 0.4 $$\Omega/min$$, $$R_2=105\Omega$$ decreasing at 0.7 $$\Omega/min$$. Find rate of change of $$R$$. Step 1: Differentiate implicitly: $$-\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt}$$. Step 2: Rearranged: $$\frac{dR}{dt} = R^2 \left( \frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2} \frac{dR_2}{dt} \right)$$. Step 3: Compute $$R$$ at given values: $$\frac{1}{R} = \frac{1}{80} + \frac{1}{105} = \frac{21}{1680} + \frac{16}{1680} = \frac{37}{1680} \Rightarrow R = \frac{1680}{37} \approx 45.41 \Omega$$. Step 4: Substitute rates: $$\frac{dR}{dt} = (45.41)^2 \left( \frac{0.4}{80^2} + \frac{-0.7}{105^2} \right)$$. Calculate: $$45.41^2 \approx 2062.9$$, $$\frac{0.4}{6400} = 6.25 \times 10^{-5}$$, $$\frac{-0.7}{11025} = -6.35 \times 10^{-5}$$. Sum inside parenthesis: $$6.25 \times 10^{-5} - 6.35 \times 10^{-5} = -0.1 \times 10^{-5} = -1 \times 10^{-6}$$. Step 5: Finally: $$\frac{dR}{dt} = 2062.9 \times (-1 \times 10^{-6}) = -0.00206 \Omega/min$$. --- 6. **Problem:** Money in bank after $$t$$ years is $$A(t) = 2000 - 10 t e^{5 - t^2/8}$$. Find minimum and maximum of $$A(t)$$ in $$[0,10]$$. Step 1: Differentiate: $$\frac{dA}{dt} = -10 \left(e^{5 - t^2/8} + t \cdot e^{5 - t^2/8} \cdot \left(-\frac{t}{4}\right)\right) = -10 e^{5 - t^2/8} \left(1 - \frac{t^2}{4} \right)$$. Step 2: Set derivative zero: $$1 - \frac{t^2}{4} = 0 \Rightarrow t^2 = 4 \Rightarrow t=2$$ (since $$t\ge0$$). Step 3: Evaluate values at $$t=0,2,10$$: - $$A(0) = 2000 -0 = 2000$$ - $$A(2) = 2000 - 20 e^{5 - 4/8} = 2000 - 20 e^{4.5} \approx 2000 - 20 \times 90.017 = 2000 - 1800.34 = 199.66$$ - $$A(10) = 2000 - 100 e^{5 - 100/8} = 2000 - 100 e^{5 - 12.5} = 2000 - 100 e^{-7.5} \approx 2000 - 100 \times 0.0005531 = 1999.94$$ Step 4: Max at $$t=0$$ or approx $$t=10$$, min near $$t=2$$. --- 7. **Problem:** Maximize area of rectangular field fenced with 500 ft of material; one side bordering building no fence. Step 1: Let $$x$$ be length along building (no fence), $$y$$ the width with fence. Step 2: Fence needed for three sides: $$x + 2y = 500$$. Step 3: Area: $$A = x y$$. Step 4: Express $$x$$: $$x = 500 - 2y$$. Step 5: Substitute: $$A(y) = y(500 - 2y) = 500y - 2y^2$$. Step 6: Differentiate: $$\frac{dA}{dy} = 500 - 4y$$. Step 7: Set zero: $$500 - 4y = 0 \Rightarrow y = 125$$ ft. Step 8: Find $$x$$: $$x = 500 - 2\times125 = 250$$ ft. Step 9: Maximum area: $$A = 250 \times 125 = 31250$$ sq ft.