3. for the function and the vector, $vec{v}$, given, compute the directional derivative of the given…

Explanation:

Step1: Recall the formula for directional derivative

The directional derivative of a function $z = f(x,y)$ in the direction of a unit - vector $\vec{u}=\langle u_1,u_2\rangle$ is given by $D_{\vec{u}}f=\nabla f\cdot\vec{u}$, where $\nabla f=\langle f_x,f_y\rangle$ and for a function $w = f(x,y,z)$ $\nabla f=\langle f_x,f_y,f_z\rangle$. First, we need to normalize the given vector $\vec{v}$. The magnitude of a vector $\vec{v}=\langle a,b\rangle$ is $\vert\vec{v}\vert=\sqrt{a^{2}+b^{2}}$ and for $\vec{v}=\langle a,b,c\rangle$ is $\vert\vec{v}\vert=\sqrt{a^{2}+b^{2}+c^{2}}$. The unit - vector $\vec{u}=\frac{\vec{v}}{\vert\vec{v}\vert}$.

Step2: Calculate the unit - vector for part (a)

Given $\vec{v}=3\vec{i}-4\vec{j}$, $\vert\vec{v}\vert=\sqrt{3^{2}+(- 4)^{2}}=\sqrt{9 + 16}=5$. So, $\vec{u}=\frac{3}{5}\vec{i}-\frac{4}{5}\vec{j}=\langle\frac{3}{5},-\frac{4}{5}\rangle$. For $f(x,y)=xy^{2}$, $f_x = y^{2}$ and $f_y = 2xy$. Then $\nabla f=\langle y^{2},2xy\rangle$. The directional derivative $D_{\vec{u}}f=\nabla f\cdot\vec{u}=y^{2}\times\frac{3}{5}+2xy\times(-\frac{4}{5})=\frac{3y^{2}-8xy}{5}$.

Step3: Calculate the unit - vector for part (b)

Given $\vec{v}=\vec{i}-2\vec{j}+2\vec{k}$, $\vert\vec{v}\vert=\sqrt{1^{2}+(-2)^{2}+2^{2}}=\sqrt{1 + 4+4}=3$. So, $\vec{u}=\frac{1}{3}\vec{i}-\frac{2}{3}\vec{j}+\frac{2}{3}\vec{k}=\langle\frac{1}{3},-\frac{2}{3},\frac{2}{3}\rangle$. For $g(x,y,z)=xy^{2}z^{3}$, $g_x=y^{2}z^{3}$, $g_y = 2xyz^{3}$, $g_z=3xy^{2}z^{2}$. Then $\nabla g=\langle y^{2}z^{3},2xyz^{3},3xy^{2}z^{2}\rangle$. The directional derivative $D_{\vec{u}}g=\nabla g\cdot\vec{u}=\frac{y^{2}z^{3}}{3}-\frac{4xyz^{3}}{3}+2xy^{2}z^{2}=\frac{y^{2}z^{3}-4xyz^{3}+6xy^{2}z^{2}}{3}$.

Step4: Calculate the unit - vector for part (c)

Given $\vec{v}=\langle - 2,5\rangle$, $\vert\vec{v}\vert=\sqrt{(-2)^{2}+5^{2}}=\sqrt{4 + 25}=\sqrt{29}$. So, $\vec{u}=\langle-\frac{2}{\sqrt{29}},\frac{5}{\sqrt{29}}\rangle$. For $h(x,y)=x^{3}+y$, $h_x = 3x^{2}$ and $h_y = 1$. Then $\nabla h=\langle 3x^{2},1\rangle$. The directional derivative $D_{\vec{u}}h=\nabla h\cdot\vec{u}=3x^{2}\times(-\frac{2}{\sqrt{29}})+1\times\frac{5}{\sqrt{29}}=\frac{-6x^{2}+5}{\sqrt{29}}$.

Step5: Calculate the unit - vector for part (d)

Given $\vec{v}=\langle 3,-7\rangle$, $\vert\vec{v}\vert=\sqrt{3^{2}+(-7)^{2}}=\sqrt{9 + 49}=\sqrt{58}$. So, $\vec{u}=\langle\frac{3}{\sqrt{58}},-\frac{7}{\sqrt{58}}\rangle$. For $F(x,y)=\frac{x^{4}+xy + y^{2}}{\sqrt{x^{2}+y^{2}}}$, using the quotient rule: Let $u=x^{4}+xy + y^{2}$ and $v=(x^{2}+y^{2})^{\frac{1}{2}}$. $u_x = 4x^{3}+y$, $u_y=x + 2y$, $v_x=\frac{x}{\sqrt{x^{2}+y^{2}}}$, $v_y=\frac{y}{\sqrt{x^{2}+y^{2}}}$. $F_x=\frac{(4x^{3}+y)\sqrt{x^{2}+y^{2}}-\frac{x(x^{4}+xy + y^{2})}{\sqrt{x^{2}+y^{2}}}}{x^{2}+y^{2}}=\frac{(4x^{3}+y)(x^{2}+y^{2})-x(x^{4}+xy + y^{2})}{(x^{2}+y^{2})^{\frac{3}{2}}}=\frac{4x^{5}+4x^{3}y^{2}+x^{2}y+y^{3}-x^{5}-x^{2}y-xy^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}=\frac{3x^{5}+4x^{3}y^{2}+y^{3}-xy^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}$. $F_y=\frac{(x + 2y)\sqrt{x^{2}+y^{2}}-\frac{y(x^{4}+xy + y^{2})}{\sqrt{x^{2}+y^{2}}}}{x^{2}+y^{2}}=\frac{(x + 2y)(x^{2}+y^{2})-y(x^{4}+xy + y^{2})}{(x^{2}+y^{2})^{\frac{3}{2}}}$. $D_{\vec{u}}F=\nabla F\cdot\vec{u}=F_x\times\frac{3}{\sqrt{58}}+F_y\times(-\frac{7}{\sqrt{58}})$.

Answer:

For part (a): $D_{\vec{u}}f=\frac{3y^{2}-8xy}{5}$ For part (b): $D_{\vec{u}}g=\frac{y^{2}z^{3}-4xyz^{3}+6xy^{2}z^{2}}{3}$ For part (c): $D_{\vec{u}}h=\frac{-6x^{2}+5}{\sqrt{29}}$ For part (d): $D_{\vec{u}}F=F_x\times\frac{3}{\sqrt{58}}+F_y\times(-\frac{7}{\sqrt{58}})$ where $F_x=\frac{3x^{5}+4x^{3}y^{2}+y^{3}-xy^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}$ and $F_y=\frac{(x + 2y)(x^{2}+y^{2})-y(x^{4}+xy + y^{2})}{(x^{2}+y^{2})^{\frac{3}{2}}}$