Question

1. for the function and the vector, $vec{v}$, and the point $p$ given, compute the directional derivative of the given function in the given direction, at the given point, $p$. make sure to normalize the direction vector! (a) $f(x,y)=xy^{2}$, $vec{v}=3hat{i}-4hat{j}$, $p=(3, - 2)$. (b) $g(x,y,z)=xy^{2}z^{3}$, $vec{v}=hat{i}-2hat{j}+2hat{k}$, $p=(3, - 2, - 1)$. (c) $h(x,y)=x^{3}+y$, $vec{v}=langle - 2,5

angle$, $p=(3, - 6)$. (d) $f(x,y)=\frac{x^{4}+xy + y^{2}}{sqrt{x^{2}+y^{2}}}$, $vec{v}=langle3, - 7
angle$, $p=(4, - 3)$.

Explanation:

Step1: Recall the formula for the directional derivative

The directional derivative of a function $z = f(x,y)$ in the direction of a unit - vector $\vec{u}=\langle u_1,u_2
angle$ at the point $(x_0,y_0)$ is given by $D_{\vec{u}}f(x_0,y_0)=
abla f(x_0,y_0)\cdot\vec{u}$, where $
abla f(x,y)=\langle f_x(x,y),f_y(x,y)
angle$. First, we need to find the gradient of the function and then normalize the direction vector.

Step2: For part (a)

1. **Find the gradient of $f(x,y)=xy^2$**:
- Calculate the partial derivatives: $f_x = y^2$ and $f_y = 2xy$.
- At the point $P=(3, - 2)$, $f_x(3,-2)=(-2)^2 = 4$ and $f_y(3,-2)=2\times3\times(-2)=-12$. So, $
abla f(3,-2)=\langle4,-12
angle$.
2. **Normalize the vector $\vec{v}=3\vec{i}-4\vec{j}=\langle3,-4
angle$**:
- The magnitude of $\vec{v}$ is $|\vec{v}|=\sqrt{3^2+( - 4)^2}=\sqrt{9 + 16}=5$.
- The unit - vector $\vec{u}=\frac{\vec{v}}{|\vec{v}|}=\langle\frac{3}{5},-\frac{4}{5}
angle$.
3. **Calculate the directional derivative**:
- $D_{\vec{u}}f(3,-2)=
abla f(3,-2)\cdot\vec{u}=\langle4,-12
angle\cdot\langle\frac{3}{5},-\frac{4}{5}
angle=4\times\frac{3}{5}+(-12)\times(-\frac{4}{5})=\frac{12 + 48}{5}=\frac{60}{5}=12$.

Step3: For part (b)

1. **Find the gradient of $g(x,y,z)=xy^2z^3$**:
- Calculate the partial derivatives: $g_x = y^2z^3$, $g_y = 2xyz^3$, $g_z = 3xy^2z^2$.
- At the point $P=(3,-2,-1)$, $g_x(3,-2,-1)=(-2)^2\times(-1)^3=-4$, $g_y(3,-2,-1)=2\times3\times(-2)\times(-1)^3 = 12$, $g_z(3,-2,-1)=3\times3\times(-2)^2\times(-1)^2=36$. So, $
abla g(3,-2,-1)=\langle - 4,12,36
angle$.
2. **Normalize the vector $\vec{v}=\vec{i}-2\vec{j}+2\vec{k}=\langle1,-2,2
angle$**:
- The magnitude of $\vec{v}$ is $|\vec{v}|=\sqrt{1^2+( - 2)^2+2^2}=\sqrt{1 + 4+4}=3$.
- The unit - vector $\vec{u}=\frac{\vec{v}}{|\vec{v}|}=\langle\frac{1}{3},-\frac{2}{3},\frac{2}{3}
angle$.
3. **Calculate the directional derivative**:
- $D_{\vec{u}}g(3,-2,-1)=
abla g(3,-2,-1)\cdot\vec{u}=(-4)\times\frac{1}{3}+12\times(-\frac{2}{3})+36\times\frac{2}{3}=\frac{-4-24 + 72}{3}=\frac{44}{3}$.

Step4: For part (c)

1. **Find the gradient of $h(x,y)=x^3 + y$**:
- Calculate the partial derivatives: $h_x = 3x^2$ and $h_y = 1$.
- At the point $P=(3,-6)$, $h_x(3,-6)=3\times3^2 = 27$ and $h_y(3,-6)=1$. So, $
abla h(3,-6)=\langle27,1
angle$.
2. **Normalize the vector $\vec{v}=\langle - 2,5
angle$**:
- The magnitude of $\vec{v}$ is $|\vec{v}|=\sqrt{(-2)^2+5^2}=\sqrt{4 + 25}=\sqrt{29}$.
- The unit - vector $\vec{u}=\frac{\vec{v}}{|\vec{v}|}=\langle\frac{-2}{\sqrt{29}},\frac{5}{\sqrt{29}}
angle$.
3. **Calculate the directional derivative**:
- $D_{\vec{u}}h(3,-6)=
abla h(3,-6)\cdot\vec{u}=27\times\frac{-2}{\sqrt{29}}+1\times\frac{5}{\sqrt{29}}=\frac{-54 + 5}{\sqrt{29}}=-\frac{49}{\sqrt{29}}$.

Step5: For part (d)

1. **Find the gradient of $F(x,y)=\frac{x^4+xy + y^2}{\sqrt{x^2 + y^2}}$**:
- First, use the quotient rule. Let $u=x^4+xy + y^2$ and $v=(x^2 + y^2)^{\frac{1}{2}}$.
- $u_x = 4x^3+y$, $u_y=x + 2y$, $v_x=\frac{x}{\sqrt{x^2 + y^2}}$, $v_y=\frac{y}{\sqrt{x^2 + y^2}}$.
- By the quotient rule $F_x=\frac{(4x^3 + y)\sqrt{x^2 + y^2}-\frac{x(x^4+xy + y^2)}{\sqrt{x^2 + y^2}}}{x^2 + y^2}=\frac{(4x^3 + y)(x^2 + y^2)-x(x^4+xy + y^2)}{(x^2 + y^2)^{\frac{3}{2}}}$, $F_y=\frac{(x + 2y)\sqrt{x^2 + y^2}-\frac{y(x^4+xy + y^2)}{\sqrt{x^2 + y^2}}}{x^2 + y^2}=\frac{(x + 2y)(x^2 + y^2)-y(x^4+xy + y^2)}{(x^2 + y^2)^{\frac{3}{2}}}$.
- At the point $P=(4,-3)$, $x = 4$ and $y=-3$.
- First, calculate the denominator $(x^2 + y^2)^{\frac{3}{2}}=(16 + 9)^{\frac{3}{2}}=125$.
- For $F_x$:
- $(4x^3 + y)(x^2 + y^2)-x(x^4+xy + y^2)=(4\times4^3-3)(16 + 9)-4(4^4+4\times(-3)+(-3)^2)=(4\times64-3)\times25-4(256-12 + 9)=(256-3)\times25-4(253)=253\times25-1012=6325-1012 = 5313$.
- For $F_y$:
- $(x + 2y)(x^2 + y^2)-y(x^4+xy + y^2)=(4+2\times(-3))(16 + 9)-(-3)(256-12 + 9)=(-2)\times25 + 3\times253=-50+759 = 709$.
- So, $
abla F(4,-3)=\langle\frac{5313}{125},\frac{709}{125}
angle$.
- Normalize the vector $\vec{v}=\langle3,-7
angle$, $|\vec{v}|=\sqrt{3^2+( - 7)^2}=\sqrt{9 + 49}=\sqrt{58}$. The unit - vector $\vec{u}=\langle\frac{3}{\sqrt{58}},-\frac{7}{\sqrt{58}}
angle$.
- $D_{\vec{u}}F(4,-3)=
abla F(4,-3)\cdot\vec{u}=\frac{5313}{125}\times\frac{3}{\sqrt{58}}+\frac{709}{125}\times(-\frac{7}{\sqrt{58}})=\frac{15939-4963}{125\sqrt{58}}=\frac{10976}{125\sqrt{58}}$.

Answer:

(a) $12$
(b) $\frac{44}{3}$
(c) $-\frac{49}{\sqrt{29}}$
(d) $\frac{10976}{125\sqrt{58}}$