Exercise 8.6-Additional Problems - Chapter 8 - Differentials and Partial Derivatives - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
Suppose that $\mathrm{z}=y e^{x^2}$ where $\mathrm{x}=2 \mathrm{t}$ and $\mathrm{y}=1-\mathrm{t}$ then find $\frac{d z}{d t}$.
Solution:
$
\begin{aligned}
\frac{d z}{d t} & =\frac{\partial z}{\partial x} \frac{d x}{d t}+\frac{\partial z}{\partial y} \frac{d y}{d t} \\
\frac{\partial z}{\partial x} & =y e^{x^2} 2 x ; \frac{\partial z}{\partial y}=e^{x^2} ; \frac{d x}{d t}=2 ; \frac{d y}{d t}=-1 \\
\frac{d z}{d t} & =y 2 x e^{x^2}(2)+e^{x^2}(-1) \\
& =4 x y e^{x^2}-e^{x^2}=e^{4 t^2}\left[(8 t(1-t)-1]=e^{4 t^2}\left(8 t-8 t^2-1\right)\right.
\end{aligned}
$
$($ Since $x=2 t$ and $y=1-t)$

Question 2.
If $\mathrm{w}=\mathrm{x}+2 \mathrm{y}+\mathrm{z}^2$ and $\mathrm{x}=\cos \mathrm{t} ; \mathrm{y}=\sin \mathrm{t} ; \mathrm{z}=\mathrm{t}$. Find $\frac{d w}{d t}$.
Solution:
We know
$
\begin{aligned}
\frac{\partial w}{\partial x} & =1 ; \frac{d x}{d t}=-\sin t \\
\frac{\partial w}{\partial y} & =2 ; \frac{d y}{d t}=\cos t \\
\frac{\partial w}{\partial z} & =2 z ; \frac{d z}{d t}=1 \\
\therefore \quad \frac{d w}{d t} & =1(-\sin t)+2 \cos t+2 z=-\sin t+2 \cos t+2 t
\end{aligned}
$