Calculus Rate Change 2Ae630

1. **Problem:** Find the rate of change of area of a circle with respect to radius $r$ at $r=6$ cm. Formula: Area $A = \pi r^2$. Rate of change: $\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r$. At $r=6$, $\frac{dA}{dr} = 2\pi \times 6 = 12\pi$. **Answer:** (b) $12\pi$. 2. **Problem:** Find marginal revenue when $x=15$ for $R(x) = 3x^2 + 36x + 5$. Marginal revenue $= \frac{dR}{dx} = 6x + 36$. At $x=15$, $6(15) + 36 = 90 + 36 = 126$. **Answer:** (d) 126. 3. **Problem:** Volume of sphere increasing at 3 cm$^3$/sec. Find rate of radius increase at $r=2$ cm. Volume $V = \frac{4}{3}\pi r^3$. Differentiating w.r.t. $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Given $\frac{dV}{dt} = 3$, $r=2$. $3 = 4\pi (2)^2 \frac{dr}{dt} = 16\pi \frac{dr}{dt}$. $\Rightarrow \frac{dr}{dt} = \frac{3}{16\pi}$. **Answer:** (b) $\frac{3}{16\pi}$ cm/sec. 4. **Problem:** On $y = x^2$, find points where $\frac{dy}{dt} = 6 \frac{dx}{dt}$. Differentiating: $\frac{dy}{dt} = 2x \frac{dx}{dt}$. Given $2x \frac{dx}{dt} = 6 \frac{dx}{dt}$. Cancel $\frac{dx}{dt}$ (assuming nonzero): $2x = 6 \Rightarrow x=3$. At $x=3$, $y=9$. **Answer:** (b) $(3,9)$. 5. **Problem:** Sides of equilateral triangle increase at 3 cm/sec. Find rate of area increase when side $a=8$ cm. Area $A = \frac{\sqrt{3}}{4} a^2$. Differentiating: $\frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2a \frac{da}{dt} = \frac{\sqrt{3}}{2} a \frac{da}{dt}$. Given $\frac{da}{dt} = 3$, $a=8$. $\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 8 \times 3 = 12\sqrt{3}$. **Answer:** (c) $12\sqrt{3}$ cm$^2$/sec. 6. **Problem:** Cylindrical tank radius $r=10$ m filled at $\frac{dV}{dt} = 314$ m$^3$/hr. Find rate of depth increase $\frac{dh}{dt}$. Volume $V = \pi r^2 h = \pi (10)^2 h = 100\pi h$. Differentiating: $\frac{dV}{dt} = 100\pi \frac{dh}{dt}$. Given $\frac{dV}{dt} = 314$. $314 = 100\pi \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{314}{100\pi} = \frac{314}{314} = 1$ m/hr. **Answer:** (a) 1 m/hr. 7. **Problem:** On curve $y^2 = 8x$, find point where $\frac{dx}{dt} = \frac{dy}{dt}$. Differentiate w.r.t. $t$: $2y \frac{dy}{dt} = 8 \frac{dx}{dt}$. Given $\frac{dx}{dt} = \frac{dy}{dt} = v$. $2y v = 8 v \Rightarrow 2y = 8 \Rightarrow y=4$. Substitute in curve: $4^2 = 8x \Rightarrow 16 = 8x \Rightarrow x=2$. **Answer:** Point is $(2,4)$. 8. **Problem:** For $y=5x^2 - 2x^3$, $\frac{dx}{dt} = 2$ units/sec, find rate of change of slope at $x=3$. Slope $m = \frac{dy}{dx} = 10x - 6x^2$. Rate of change of slope w.r.t. $t$: $\frac{dm}{dt} = \frac{d}{dt}(10x - 6x^2) = (10 - 12x) \frac{dx}{dt}$. At $x=3$, $\frac{dm}{dt} = (10 - 36) \times 2 = (-26) \times 2 = -52$. **Answer:** $-52$ units/sec. 9. **Problem:** Wave radius increases at 5 cm/sec. Find rate of area increase when radius $r=8$ cm. Area $A = \pi r^2$. $\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi \times 8 \times 5 = 80\pi$. **Answer:** $80\pi$ cm$^2$/sec. 10. **Problem:** Volume of cube increases at 9 cm$^3$/sec. Find rate of surface area increase when edge $x=10$ cm. Volume $V = x^3$, Surface area $S = 6x^2$. $\frac{dV}{dt} = 3x^2 \frac{dx}{dt} = 9$. At $x=10$, $3(10)^2 \frac{dx}{dt} = 9 \Rightarrow 300 \frac{dx}{dt} = 9 \Rightarrow \frac{dx}{dt} = \frac{9}{300} = \frac{3}{100}$. Rate of surface area change: $\frac{dS}{dt} = 12x \frac{dx}{dt} = 12 \times 10 \times \frac{3}{100} = \frac{360}{100} = 3.6$. **Answer:** $3.6$ cm$^2$/sec. 11. **Problem:** Surface area of sphere increases at 2 cm$^2$/sec. Find rate of volume increase at radius $r=6$ cm. Surface area $S = 4\pi r^2$, volume $V = \frac{4}{3}\pi r^3$. $\frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 2$. At $r=6$, $8\pi \times 6 \frac{dr}{dt} = 2 \Rightarrow 48\pi \frac{dr}{dt} = 2 \Rightarrow \frac{dr}{dt} = \frac{2}{48\pi} = \frac{1}{24\pi}$. Rate of volume change: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi \times 36 \times \frac{1}{24\pi} = \frac{144\pi}{24\pi} = 6$. **Answer:** 6 cm$^3$/sec. 12. **Problem:** For ellipse $16x^2 + 9y^2 = 400$, find points where $\frac{dy}{dt} = -\frac{dx}{dt}$. Differentiate: $32x \frac{dx}{dt} + 18y \frac{dy}{dt} = 0$. Given $\frac{dy}{dt} = -\frac{dx}{dt}$. Substitute: $32x \frac{dx}{dt} + 18y (-\frac{dx}{dt}) = 0 \Rightarrow (32x - 18y) \frac{dx}{dt} = 0$. Assuming $\frac{dx}{dt} \neq 0$, $32x - 18y = 0 \Rightarrow y = \frac{16}{9} x$. Substitute in ellipse: $16x^2 + 9 \left( \frac{16}{9} x \right)^2 = 400$. $16x^2 + 9 \times \frac{256}{81} x^2 = 400$. $16x^2 + \frac{2304}{81} x^2 = 400$. Multiply both sides by 81: $1296 x^2 + 2304 x^2 = 32400$. $3600 x^2 = 32400 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$. Then $y = \frac{16}{9} \times 3 = \frac{48}{9} = \frac{16}{3}$. Points: $(3, \frac{16}{3})$ and $(-3, -\frac{16}{3})$. 13. **Problem:** If area of circle increases uniformly, prove perimeter varies inversely as radius. Area $A = \pi r^2$, perimeter $P = 2\pi r$. Given $\frac{dA}{dt} = k$ (constant). $\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = k$. $\Rightarrow \frac{dr}{dt} = \frac{k}{2\pi r}$. Perimeter $P = 2\pi r$. Express $r$ in terms of $P$: $r = \frac{P}{2\pi}$. Substitute in $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{k}{2\pi r} = \frac{k}{2\pi \times \frac{P}{2\pi}} = \frac{k}{P}$. Since $\frac{dr}{dt} \propto \frac{1}{P}$, perimeter $P$ varies inversely as radius $r$. 14. **Problem:** Find values of $k$ for which $f(x) = kx^3 + 5$ is decreasing. Derivative: $f'(x) = 3kx^2$. For decreasing, $f'(x) \leq 0$ for all $x$. Since $x^2 \geq 0$, sign depends on $k$. For $f'(x) \leq 0$, $k \leq 0$. **Answer:** $k \leq 0$. 15. **Problem:** Find interval where $f(x) = 5x^2 + 18$ is strictly decreasing. Derivative: $f'(x) = 10x$. For decreasing, $f'(x) < 0 \Rightarrow 10x < 0 \Rightarrow x < 0$. **Answer:** $(-\infty, 0)$. 16. **Problem:** Rate of change of area of circle w.r.t. radius at $r=6$ cm. Same as Q1: $\frac{dA}{dr} = 2\pi r = 12\pi$. 17. **Problem:** Interval where $(2 + x)e^{-x}$ is increasing. Let $f(x) = (2 + x)e^{-x}$. Derivative: $f'(x) = e^{-x} (1 - (2 + x)) = e^{-x} (1 - 2 - x) = e^{-x} (-1 - x)$. Since $e^{-x} > 0$, sign depends on $-1 - x$. $f'(x) > 0$ when $-1 - x > 0 \Rightarrow x < -1$. **Answer:** $(-\infty, -1)$. 18. **Problem:** Interval where $f(x) = x - \frac{1}{x}$ is increasing. Derivative: $f'(x) = 1 + \frac{1}{x^2} > 0$ for all $x \neq 0$. So $f(x)$ is increasing on $(-\infty, 0) \cup (0, \infty)$. 19. **Problem:** $f(x) = x^3 + \frac{1}{x^3}$, $x \neq 0$. Find intervals of increase/decrease. Derivative: $f'(x) = 3x^2 - 3x^{-4} = 3x^2 - \frac{3}{x^4}$. Set $f'(x) = 0$: $3x^2 = \frac{3}{x^4} \Rightarrow x^6 = 1 \Rightarrow x = \pm 1$. Test intervals: - For $x > 1$, $f'(x) > 0$ (increasing). - For $0 < x < 1$, $f'(x) < 0$ (decreasing). - For $-1 < x < 0$, $f'(x) < 0$ (decreasing). - For $x < -1$, $f'(x) > 0$ (increasing). **Answer:** Increasing on $(-\infty, -1) \cup (1, \infty)$, decreasing on $(-1, 0) \cup (0, 1)$. 20. **Problem:** $f(x) = \log(\cos x)$ decreasing intervals. Derivative: $f'(x) = \frac{-\sin x}{\cos x} = -\tan x$. $f'(x) < 0$ when $-\tan x < 0 \Rightarrow \tan x > 0$. $\tan x > 0$ in intervals $(0, \frac{\pi}{2}), (\pi, \frac{3\pi}{2}), ...$ So $f(x)$ is decreasing where $\tan x > 0$. 21. **Problem:** $f(x) = x^3 - 6x^2 + 9x + 18$. Find intervals where $f$ is strictly decreasing. Derivative: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$. $f'(x) < 0$ between roots: $(1,3)$. **Answer:** (d) $(1, 3)$.