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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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\(1^{2}=x^{2}+y^{2}\) \(1 \frac{d l}{d t}=x \frac{d x}{d t}+y \frac{d y}{d t}\) \(y=x^{3 / 2}\) \(\frac{d y}{d t}=\frac{3}{2} x^{1 / 2} \frac{d x}{d t}\) Using (1) \& (2) $11 \sqrt{x^{2}+x^{3}}=x \frac{d x}{d t}+\frac{3}{2} x^{3 / 2} x^{1 / 2} \frac{d x}{d t}$ \(\frac{d x}{d t}=\frac{66}{3+\frac{27}{2}}=\frac{66 \times 2}{33}=4\)
\(\mathrm{h}=\mathrm{ar}+\mathrm{b}\) $\frac{\mathrm{dh}}{\mathrm{dt}}=\mathrm{a} \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \mathrm{a}=3, \mathrm{~b}=3$ \(\mathrm{v}=3 \pi \mathrm{r}^{2}(\mathrm{r}+1)\) $\frac{\mathrm{dv}}{\mathrm{dt}}=3 \pi\left(3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}+2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}\right)$ \(1=3 \pi\left(3-6^{2}+12\right) \frac{\mathrm{dr}}{\mathrm{dt}}\) \(\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{360 \pi}\) Now, when \(\mathrm{r}=36\) $\frac{d v}{d t}=3 \pi\left(3 \cdot(36)^{2}+72\right) \times \frac{1}{360 \pi}$ \(=33\)
\(x^{2}+y^{2}-\frac{10}{3} y+1=0\) \(x^{2}+\left(y-\frac{5}{3}\right)^{2}=\left(\frac{4}{3}\right)^{2}\) \(y^{2}=x^{3}\) \(2 y y_{1}=3 x^{2}\) \(y_{1}=\frac{3 x^{2}}{2 y}\) Normal eqn \(\quad y-y_{1}=-\frac{2 y_{1}}{3 x_{1}^{2}}\left(x-x_{1}\right)\) as it passes through \(\left(0, \frac{5}{3}\right)\) \(\frac{5}{3}-y_{1}=\frac{2 y_{1}}{3 x_{1}}\) \(5-3 y_{1}=2 y^{1 / 3}\) \(5-9 y^{3}-225 y+135 y^{2}=8 y\) \(9 y^{3}+233 y-135 y^{2}-25=0\)
A) As \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R\) \(\Rightarrow a=2 R \sin A\) \(\Rightarrow \frac{d a}{d A}=2 R \cos A\) \(\Rightarrow \frac{d a}{\cos A}=2 R d A\) Similarly, \(\frac{d b}{\cos B}=2 R d B, \frac{d c}{\cos c}=2 R d C\) \(\quad \frac{d a}{\cos A}+\frac{d b}{\cos B}+\frac{d c}{\cos C}+1=\) \(2 R(d A+d B+d C)+1\) \(\quad A s A+B+C=\pi\) \(\Rightarrow d A+d B+d C=0\) using eq (I) \(|\mathbf{m}|=1\) B) $\begin{aligned} & \Rightarrow \mathrm{m}=\pm 1 \\ \mathrm{x}^{2} \mathrm{y}^{2}=& 16 \end{aligned}$ \(2 x^{2} y \frac{d y}{d x}+2 x y^{2}=0\) \(\frac{d y}{d x}=-\frac{y}{x}=1\) Subtangent \(=y / d y / d x=2=|\mathrm{k}|\) \(\Rightarrow \mathrm{k}=\pm 2\) C) \(y=2 e^{2 x}\) \(\frac{d y}{d x}=4 e^{2 x}=4\) for yaxis $\tan ^{-1} 4=\frac{\pi}{2}-\cot ^{-1}\left(\frac{8 n-4}{3}\right)=\tan ^{-1}\left(\frac{8 n+4}{3}\right)$ \(12=|8 n-4|\) \(\mathrm{n}=2\) or \(-1\) D) \(\mathrm{x}=\underline{\mathrm{e}^{\sin y}}\) \(1=e^{\sin y} \cos y \frac{d y}{d x}\) At \((1,0) \Rightarrow \frac{d y}{d x}=1\) eqn of Normal \(\rightarrow y=-(x-1)\) \(y+x-1=0\) Area of \(\Delta=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\) \(\Rightarrow|2 t+1|=3\) \(2 t+1=\pm 3 \Rightarrow t=1\) or \(-2\) $\mathrm{A} \rightarrow(\mathrm{PQ}), \mathrm{B} \rightarrow(\mathrm{RS}), \mathrm{C} \rightarrow(\mathrm{RQ}), \mathrm{D} \rightarrow(\mathrm{PS})$
\(\mathrm{x}=\mathrm{t}^{2}+\mathrm{t}+1\) \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}=2 \mathrm{t}+1\) \(\mathrm{y}=\mathrm{t}^{2}-\mathrm{t}+1\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{t}-1\) \(\frac{d y}{\mathrm{~d} \mathrm{x}}=\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\) $\Rightarrow \mathrm{y}-\left(\mathrm{t}^{2}-\mathrm{t}+1\right)=\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\left(\mathrm{x}-\left(\mathrm{t}^{2}+\mathrm{t}+1\right)\right)$ $\Rightarrow\left(\mathrm{t}-\mathrm{t}^{2}\right)=\left(\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\right)\left(-\left(\mathrm{t}^{2}+\mathrm{t}\right)\right)$ $\Rightarrow 2 \mathrm{t}^{2}+\mathrm{t}-2 \mathrm{t}^{3}-\mathrm{t}^{2}=\mathrm{t}^{2}+\mathrm{t}-2 \mathrm{t}^{3}-2 \mathrm{t}^{2}$ \(\Rightarrow 2 \mathrm{t}^{2}=0\) \(\Rightarrow \mathrm{t}=0\)
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