Mathematics-I (3110014)

$\mathrm{We} \mathrm{have}, \mathrm{f}={\mathrm{xy}}^2+{\mathrm{yz}}^2 \mathrm{and} \mathrm{u}¯=\mathrm{i}+2\mathrm{j}+2\mathrm{k}=(1,2,2).$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}={\mathrm{y}}^2 ⟹{\left(\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\right)}_{(2,-1,1)}=(-1{)}^2=1$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=2\mathrm{xy}+{\mathrm{z}}^2 ⟹{\left(\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\right)}_{(2,-1,1)}=2\left(2\right)(-1)+(1{)}^2=-4+1=-3$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=2\mathrm{yz} ⟹{\left(\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\right)}_{(2,-1,1)}=2(-1)\left(1\right)=-2$

$\mathrm{Now},$

$\nabla \mathrm{f}=\frac{\partial \mathrm{f}}{\partial \mathrm{x}} \hat{\mathrm{i}}+\frac{\partial \mathrm{f}}{\partial \mathrm{y}} \hat{\mathrm{j}}+\frac{\partial \mathrm{f}}{\partial \mathrm{z}} \hat{\mathrm{k}}$

$⟹(\nabla \mathrm{f}{)}_{(2,-1,1)}={\left(\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\right)}_{(2,-1,1)}\hat{\mathrm{i}}+{\left(\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\right)}_{(2,-1,1)}\mathrm{j}+{\left(\frac{\partial \mathrm{f}}{\partial \mathrm{z}}\right)}_{(2,-1,1)}\hat{\mathrm{k}}$

$⟹(\nabla \mathrm{f}{)}_{(2,-1,1)}=\left(1\right) \hat{\mathrm{i}}+\left(-3\right) \mathrm{j}+\left(-2\right) \hat{\mathrm{k}}$

$⟹(\nabla \mathrm{f}{)}_{(2,-1,1)}=\left(1, -3, -2\right)$

$\mathrm{Now}, \hat{\mathrm{u}}=\frac{\overline{\mathrm{u}}}{\left|\overline{\mathrm{u}}\right|}$

$\mathrm{Now}, \hat{\mathrm{u}}=\frac{(1,2,2)}{\sqrt{1^2+2^2+2^2}}$

$\mathrm{Now}, \hat{\mathrm{u}}=\frac{(1,2,2)}{\sqrt{9}}$

$\mathrm{Now}, \hat{\mathrm{u}}=\frac{(1,2,2)}{3}$

$\mathrm{For} \mathrm{Directional} \mathrm{Derivative},$

$(\nabla \mathrm{f}{)}_{(2,-1,1)} \cdot \hat{\mathrm{u}}=(1,-3,-2) \cdot \frac{(1,2,2)}{3}$

$(\nabla \mathrm{f}{)}_{(2,-1,1)} \cdot \hat{\mathrm{u}}=\frac{\left(1\right)\cdot \left(1\right)+(-3)\cdot \left(2\right)+(-2)\cdot \left(2\right)}{3}$

$(\nabla \mathrm{f}{)}_{(2,-1,1)} \cdot \hat{\mathrm{u}}=\frac{1-6-4}{3}$

$(\nabla \mathrm{f}{)}_{(2,-1,1)} \cdot \hat{\mathrm{u}}=-\frac{9}{3}$

$(\nabla \mathrm{f}{)}_{(2,-1,1)} \cdot \hat{\mathrm{u}}=-3$

$\mathrm{So}, \mathrm{Directional} \mathrm{Derivative} \mathrm{is} -3.$