Calculus Solutions Ex # 3.4

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Q1. Equation y=3 x+5 (a) Given that \frac{d x}{d t}=2, find \frac{d y}{d t} when x=1 (b) Given that’ \frac{d y}{d t}=-1, find \frac{d x}{d t} when x=0. .
Solution:
(a)

    \[y=3 x+5\]

Differentiating w.r.t’ t ‘

    \[\begin{aligned} & \frac{d y}{d t}=\frac{3 d x}{d t} \\ \\ & \left.\frac{d y}{d t}\right|_{x=1}=\left.3 \frac{d x}{d t}\right|_{x=1}=3(2)=6 \end{aligned}\]

(b)

    \[y=3 x+5\]

Differentiating w.r.t. 't'

    \[\begin{aligned} & \frac{d y}{d t}=3 \frac{d x}{d t} \\ \\ & \left.3 \frac{d x}{d t}\right|_{x=0}=\left.\frac{d y}{d t}\right|_{x=0} \\ \\ & \left.3 \frac{d x}{d t}\right|_{x=0}=-\left.1 \Rightarrow \frac{d x}{d t}\right|_{x=0}=-1 / 3 \end{aligned}\]

 
Q.02 Equation x+4 y=3. a) Given that \frac{d x}{d t}=1 find \frac{d y}{d t} when x=2 (b) Given that d y / d t=4, find d x / d t when x=3
Solution:
(a)

    \[\quad x+4 y=3\]

    \[\begin{aligned} & \text { Differentiating w.r.t. } t \\ \\ & \frac{d x}{d t}+4 \frac{d y}{d t}=0 \quad \text { (1) } \\ \\ & \left.\frac{d x}{d t}\right|_{x=2}+\left.4 \frac{d y}{d t}\right|_{x=2}=0 \\ \\ & 1+ \left.4 \frac{d y}{d t}\right|_{x=2}=0 \\ \\ & \left.4\frac{d y}{d t}\right|_{x=2}=-1 \\ \\ & \left.\frac {d y}{d t}\right|_{x=2}=\frac{-1}{4}. \end{aligned}\]

(b) from (1) in part (a) we have

    \[\left.\frac{d x}{d t}\right|_{x=3}+\left.4 \frac{d y}{d t}\right|_{x=3}=0\]

    \[\begin{aligned} & \left.\frac{d x}{d t}\right|_{x=3}+4(4)=\left.0 \quad \therefore \frac{d y}{d t}\right|_{x=3}=4 \\ \\ & \left.\frac{d x}{d t}\right|_{x=3}=-16 \end{aligned}\]

 
Q.03 Equation 4 x^{2}+9 y^{2}=1 (a) Given that \frac{d x}{d t}=3, find \frac{d y}{d t} when (x, y)=\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right) (b) Given that \frac{d y}{d t}=8, find \frac{d x}{d t} when (x, y)=(1 / 3,-\sqrt{5} / 9)
Solution:
(a)

    \[\begin{aligned} & 4 x^{2}+9 y^{2}=1 ; \quad \frac{d x}{d t}=3, \qaud \frac{d y}{d t}=? \\ \\ & \text {when} \quad (x, y)=\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right) \\ \\ & \text { Differentiating w.r.t. } t \\ \\ & 4 \frac{d}{d t}\left(x^{2}\right)+9 \frac{d}{d t}\left(y^{2}\right)=0 \\ \\ & 4\left(2 x \frac{d x}{d t}\right)+9\left(2 y \frac{d y}{d t}\right=0 \\ \\ & \left.8 x \frac{d x}{d t}+18 y \frac{d y}{d t}=0----(1) \\ \\ & \left.8 x \frac{d x}{d t}\right|_{\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)}+ \left. 18y \frac{d y}{d t}\right|_{\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)}=0 \\ \\ & 8\left(\frac{1}{2 \sqrt{2}}\right)(3)+\left.18\left(\frac{1}{3 \sqrt{2}}\right) \frac{d y}{d t}\right|_{\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)}=0 \\ \\ & \frac{12}{\sqrt(2)}+\left.\frac{6}{\sqrt{2}} \frac{d y}{d t}\right|_{\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)}=0 \end{aligned}\]

    \[\begin{aligned} & \left.\frac{6}{\sqrt{2}} \frac{d y}{d t}\right|_{\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)}=-\frac{12}{\sqrt{2}} \\ \\ & \left. \frac{d y}{d t}\right|_{\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)}=-\frac{12}{\sqrt{2}} \times \frac{\sqrt{2}}{6}=-2. \end{aligned}\]

(b) From (1) part (a) we have

    \[\begin{aligned} & 8 x \frac{d x}{d t}+18 y \frac{d y}{d t}=0 ; \quad \frac{d y}{d t}=8, \quad \frac{d x}{d t}=? \\ \\ & \text { when }(x, y)=(1 / 3,-\sqrt{5} / 9) \\ \\ & \left. 8 x \frac{d x}{d t}\right|_{\left(\frac{1}{ 3}},-\frac{\sqrt{5}}{9}\right)}+\left. 18 \frac{d y}{d t}\right|_{\left(\frac{1}{ 3}},-\frac{\sqrt{5}}{9}\right)}=0 \\ \\ & \left. 8 (\frac{1}{3})\frac{d x}{d t}\right|_{\left(\frac{1}{ 3}},-\frac{\sqrt{5}}{9}\right)} +8 (-\frac{\sqrt{5}}{9}\right)(8)=0 \\ \\ & \left.\frac{8}{3} \frac{d x}{d t}\right|_{\left(\frac{1}{ 3}},-\frac{\sqrt{5}}{9}\right)} +(-16 \sqrt{5})=0 \\ \\ & \left.\frac{8}{3} \frac{d x}{d t}\right|_{\left(\frac{1}{ 3}},-\frac{\sqrt{5}}{9}\right)}=16 \sqrt{5} \\ \\ & \left. \frac{d x}{d t}\right|_{\left(\frac{1}{ 3}},-\frac{\sqrt{5}}{9}\right)}=16 \sqrt{5} \times \frac{3}{8}=6 \sqrt{5} \end{aligned}\]