Related Rates D5Bbac

1. **Problem:** Find $\frac{dy}{dt}$ at $x=1$ for $y = x^2 + 3$ given $\frac{dx}{dt} = 4$. 2. **Problem:** Find $\frac{dx}{dt}$ at $x=-2$ for $y = 2x^2 + 1$ given $\frac{dy}{dt} = -1$. 3. **Problem:** Find $\frac{dz}{dt}$ at $(x,y) = (1,3)$ for $z^2 = x^2 + y^2$ given $\frac{dx}{dt} = 4$ and $\frac{dy}{dt} = 3$. 4. **Problem:** Find the rate of change of total resistance $\Omega$ for two resistors in parallel with given rates for $R_1$ and $R_2$. 5. **Problem:** Find how fast the bottom of a 10-ft ladder moves along the ground when the top slides down. --- ### 1. Find $\frac{dy}{dt}$ - Given: $y = x^2 + 3$, $\frac{dx}{dt} = 4$, $x=1$. - Differentiate both sides with respect to $t$: $$\frac{dy}{dt} = 2x \frac{dx}{dt}$$ - Substitute values: $$\frac{dy}{dt} = 2(1)(4) = 8$$ **Answer:** $\frac{dy}{dt} = 8$ --- ### 2. Find $\frac{dx}{dt}$ - Given: $y = 2x^2 + 1$, $\frac{dy}{dt} = -1$, $x = -2$. - Differentiate both sides with respect to $t$: $$\frac{dy}{dt} = 4x \frac{dx}{dt}$$ - Solve for $\frac{dx}{dt}$: $$\frac{dx}{dt} = \frac{\frac{dy}{dt}}{4x}$$ - Substitute values: $$\frac{dx}{dt} = \frac{-1}{4(-2)} = \frac{-1}{-8} = \frac{1}{8}$$ **Answer:** $\frac{dx}{dt} = \frac{1}{8}$ --- ### 3. Find $\frac{dz}{dt}$ - Given: $z^2 = x^2 + y^2$, $\frac{dx}{dt} = 4$, $\frac{dy}{dt} = 3$, at $(x,y) = (1,3)$. - Differentiate implicitly with respect to $t$: $$2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$ - Simplify: $$z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}$$ - Calculate $z$: $$z = \sqrt{x^2 + y^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$$ - Substitute values: $$\sqrt{10} \frac{dz}{dt} = (1)(4) + (3)(3) = 4 + 9 = 13$$ - Solve for $\frac{dz}{dt}$: $$\frac{dz}{dt} = \frac{13}{\sqrt{10}}$$ **Answer:** $\frac{dz}{dt} = \frac{13}{\sqrt{10}}$ --- ### 4. Rate of change of total resistance $\Omega$ - Total resistance for resistors in parallel: $$\frac{1}{\Omega} = \frac{1}{R_1} + \frac{1}{R_2}$$ - Differentiate implicitly with respect to $t$: $$-\frac{1}{\Omega^2} \frac{d\Omega}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt}$$ - Rearrange: $$\frac{d\Omega}{dt} = \Omega^2 \left( \frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2} \frac{dR_2}{dt} \right)$$ - Calculate $\Omega$ when $R_1=20$, $R_2=50$: $$\frac{1}{\Omega} = \frac{1}{20} + \frac{1}{50} = \frac{5}{100} + \frac{2}{100} = \frac{7}{100}$$ $$\Omega = \frac{100}{7} \approx 14.2857$$ - Given rates: $\frac{dR_1}{dt} = 0.5$, $\frac{dR_2}{dt} = -1.1$ - Substitute values: $$\frac{d\Omega}{dt} = (14.2857)^2 \left( \frac{0.5}{20^2} + \frac{-1.1}{50^2} \right)$$ $$= 204.08 \left( \frac{0.5}{400} - \frac{1.1}{2500} \right)$$ $$= 204.08 \left( 0.00125 - 0.00044 \right) = 204.08 \times 0.00081 = 0.165$$ **Answer:** The total resistance increases at approximately $0.165$ $\Omega$/min. --- ### 5. Ladder sliding problem - Ladder length $L = 10$ ft (constant). - Let $x$ be distance from wall to bottom of ladder, $y$ height of ladder top on wall. - Relation by Pythagoras: $$x^2 + y^2 = 10^2 = 100$$ - Differentiate with respect to $t$: $$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$ - Simplify: $$x \frac{dx}{dt} = - y \frac{dy}{dt}$$ - Given $\frac{dy}{dt}$ (top sliding down) is negative; suppose $\frac{dy}{dt} = -v$ ft/sec (value missing in prompt). - When bottom is at $x$, height is: $$y = \sqrt{100 - x^2}$$ - Solve for $\frac{dx}{dt}$: $$\frac{dx}{dt} = - \frac{y}{x} \frac{dy}{dt}$$ **Note:** The exact numerical answer requires the value of $\frac{dy}{dt}$ and the position $x$ or $y$ when calculating. --- **Summary:** - $\frac{dy}{dt} = 8$ - $\frac{dx}{dt} = \frac{1}{8}$ - $\frac{dz}{dt} = \frac{13}{\sqrt{10}}$ - $\frac{d\Omega}{dt} \approx 0.165$ - Ladder problem formula: $\frac{dx}{dt} = - \frac{y}{x} \frac{dy}{dt}$ (needs more data)