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\(1^{2}=x^{2}+y^{2}\) \(1 \frac{d l}{d t}=x \frac{d x}{d t}+y \frac{d y}{d t}\) \(y=x^{3 / 2}\) \(\frac{d y}{d t}=\frac{3}{2} x^{1 / 2} \frac{d x}{d t}\) Using (1) \& (2) $11 \sqrt{x^{2}+x^{3}}=x \frac{d x}{d t}+\frac{3}{2} x^{3 / 2} x^{1 / 2} \frac{d x}{d t}$ \(\frac{d x}{d t}=\frac{66}{3+\frac{27}{2}}=\frac{66 \times 2}{33}=4\)

Short Answer

Expert verified
In summary, the rate of change of \(x\) with respect to time, \(\frac{dx}{dt}\), in the given system of related rates, is equal to \(4\).

Step by step solution

01

Identify the given equations and variables

We have the following equations given to us: 1. \(1^2 = x^2 + y^2\) 2. \(1\frac{dl}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}\) 3. \(y = x^{\frac{3}{2}}\) 4. \(\frac{dy}{dt}=\frac{3}{2}x^{\frac{1}{2}}\frac{dx}{dt}\) Our objective is to find \(\frac{dx}{dt}\).
02

Use given equations to solve for y

From equation (3), we have: \(y = x^{\frac{3}{2}}\) Square both sides: \(y^2 = x^3\) Replace \(y^2\) in equation (1) using this result: \(1^2 = x^2 + x^3\) Simplify: \(1 = x^2 + x^3\)
03

Solve for \(\frac{dy}{dt}\)

From equation (4), we have: \(\frac{dy}{dt}=\frac{3}{2}x^{\frac{1}{2}}\frac{dx}{dt}\)
04

Replace \(y\) and \(\frac{dy}{dt}\) in equation (2)

Using equations (3) and (4), we can rewrite equation (2) as: \(1\frac{dl}{dt} = x\frac{dx}{dt} + x^{\frac{3}{2}}\left(\frac{3}{2}x^{\frac{1}{2}}\frac{dx}{dt}\right)\)
05

Solve for \(\frac{dx}{dt}\)

Now, we have a single equation that can be solved for \(\frac{dx}{dt}\): \(1\frac{dl}{dt} = x\frac{dx}{dt} + \frac{3}{2}x^2\frac{dx}{dt}\) Given: \(11\sqrt{x^2+x^3}=x\frac{dx}{dt}+\frac{3}{2}x^{\frac{3}{2}}\frac{dx}{dt}\) Multiply both sides by \(\frac{2}{3}\): \(\frac{22}{3}\sqrt{x^2+x^3}=\frac{2}{3}x\frac{dx}{dt}+\frac{1}{2}x^{\frac{3}{2}}\frac{dx}{dt}\) Combine the terms on right side: \(\frac{22}{3}\sqrt{x^2+x^3}=\left(\frac{2}{3}x+\frac{1}{2}x^{\frac{3}{2}}\right)\frac{dx}{dt}\) Now, divide both sides to find \(\frac{dx}{dt}\): \(\frac{dx}{dt}=\frac{\frac{22}{3}\sqrt{x^2+x^3}}{\frac{2}{3}x+\frac{1}{2}x^{\frac{3}{2}}}\) Given the provided input, \(\frac{dx}{dt}=\frac{66}{3+\frac{27}{2}}=\frac{66\times 2}{33}=4\) The rate of change of \(x\) with respect to time, \(\frac{dx}{dt}\), is equal to \(4\).

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Most popular questions from this chapter

\(y^{2}-2 x^{2}-4 y+8=0\) \(2 y \frac{d y}{d x}-4 x-4 \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{2 x}{y-2}\) Now, \(y^{2}-2 x^{2}-4 y+8=0\) \((y-2)^{2}-2\left(x^{2}-2\right)=0\) eqn of tangent \(y-y_{1}=\frac{2 x_{1}}{y_{1}-2}\left(x-x_{1}\right)\) \(-\left(2-y_{1}\right)^{2}=2 x_{1}\left(1-x_{1}\right)\) Using (II) \(-2\left(x_{1}^{2}-2\right)=2 x_{1}-2 x_{1}^{2}\) \(2 \mathrm{x}_{1}=4\) \(\mathrm{x}_{1}=2\) \& \(\mathrm{x}_{1}=0\) (Horizontal tangent)

\(y=e^{2 x}+x^{2}\) \(\frac{d y}{d x}=2 e^{2 x}+2 x\) at \(x=0, \quad \frac{d y}{d x}=2\) \(-\frac{d x}{d y}=-\frac{1}{2}\) eqn of normal \(\rightarrow y-1=-\frac{1}{2} x\) \(\Rightarrow 2 y+x-2=0\) Distance from \((0,0)=\frac{2}{\sqrt{5}}\)

\(y^{2}-2 y-8 x+17=0\) \(2 y \frac{d y}{d x}-2 \frac{d y}{d x}-8=0\) \(\frac{d y}{d x}=\frac{y}{y-1}=1\) \(\Rightarrow y=5, \quad x=4\)

Volume inhaled while exercising \(=\int_{0}^{4} 1.75 \sin \frac{\pi t}{2}\) dt \(=-1.75\left(\cos \frac{\pi t}{2}\right) \times \frac{2}{\pi}\) \(=\frac{7}{\pi}\) Difference \(=\frac{7}{\pi}-\frac{5.1}{\pi}\) \(=\frac{1.9}{\pi}\) Hence, B is correct Comprehension \(5:\) $\begin{aligned} x=& a(2 \cos t+\cos 2 t), \quad y=a(2 \sin t-\sin 2 t\\\ \frac{d x}{d t} &=a(-2 \sin t-2 \sin 2 t) \\\ &=-2 a[\sin t+2 \sin 2 t] \\ &=-2 a \sin t[1+2 \cos t] \\ \frac{d y}{d t} &=a[2 \cos t-2 \cos 2 t] \\ &=-2 a\left[2 \cos ^{2} t-\cos t-1\right] \\\ &=-2 a(\cos t-1)(2 \cos t+1) \end{aligned}$ $\frac{d y}{d x}=\frac{(\cos t-1)(2 \cos t+1)}{(\sin t)(1+2 \cos t)}=-\tan t / 2$ slope of normal \(=\cot t / 2 .\)

\(y=x \ln x+\sin (\pi \ln x)\) \(\frac{d y}{d x}=1+\ln x+\cos (\pi \ln x) \frac{\pi}{x}\) at \(x=e\) \(\frac{d y}{d x}=2-\frac{\pi}{e}=\frac{2 e-\pi}{e}\) at \(x=e^{2}\) \(\frac{d y}{d x}=3+\frac{\pi}{e^{2}}=\frac{3 e^{2}+\pi}{e^{2}}\)

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