Chapter 5: Problem 91
At exercise, respiratory cycle has a period of 4 . Hence difference in frequency is \(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)
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\(x y^{2}=1\) \(y^{2}+2 x y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-y}{2 x}\) \(-\frac{d x}{d y}=\frac{2 x}{y}=\frac{2}{y^{3}}\) \(y-y_{1}=\frac{2}{y_{1}^{3}}\left(x-x_{1}\right)\) \(+y_{1}^{4}=2 x_{1}\) \(y_{1}^{6}=2\) \(y_{1}=\pm 2^{1 / 6}\) \(x_{1}=\pm 2^{-\sqrt{3}}\)
\(P(t)=60 t^{2}-t^{3}\) \(P^{\prime}(t)=120 t-3 t^{2}=900\) \(\Rightarrow t^{2}-40 t+300=0\) \(t=10,30\)
\(x^{3 / 2}+y^{3 / 2}=2 a^{3 / 2}\) \(\frac{3}{2} x^{1 / 2}+\frac{3}{2} y^{1 / 2} \frac{d y}{d x}=0\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\sqrt{\frac{\mathrm{x}}{\mathrm{y}}}=-1\) \(\mathrm{y}=\mathrm{x}\) Putting in eqn \(x^{3 / 2}=a^{3 / 2}\) \(\Rightarrow \mathrm{x}=\mathrm{a}, \mathrm{y}=\mathrm{a}\)
\(\mathrm{l}_{1}=\mathrm{l}_{2}+\mathrm{l}_{2}^{3}+6\) $\frac{\mathrm{d} \mathrm{l}_{1}}{\mathrm{dt}}=\frac{\mathrm{dl}_{2}}{\mathrm{dt}}+3 \mathrm{l}_{2}^{2} \frac{\mathrm{dl}_{2}}{\mathrm{dt}}$ $\frac{\mathrm{d} \mathrm{S}_{2}}{\mathrm{dS}_{1}}=\frac{\mathrm{dS}_{2}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dS}_{1}}$ $=\frac{2 \mathrm{l}_{2} \mathrm{~d}_{2} / \mathrm{dt}}{2 \mathrm{l}_{1} \mathrm{dl}_{1} / \mathrm{dt}}$ \(=\frac{1}{4} \times \frac{1}{8}\)
\(\mathrm{xy}=1\) \(\mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}}{\mathrm{x}}\) \(-\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\mathrm{x}_{1}^{2}\) \(\Rightarrow \mathrm{x}_{1}^{2}=\frac{\mathrm{a}-3}{\mathrm{a}} \geq 0\) \(\mathrm{a} \in(-\infty, 0) \mathrm{U}(3, \infty)\)
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