Directional Derivatives - Multivariable Calculus

1. Find the directional derivative of the function $f(x,y,z) = \sqrt{x^2+y^2+z^2}$ at the point $P(1,2,2)$ in the direction of the vector from $P$ to $Q(3,1,0)$.
   Answer: $-4/9$ The gradient of $f$ is $\nabla f = \frac{1}{2\sqrt{x^2+y^2+z^2}} \langle 2x, 2y, 2z \rangle = \frac{1}{\sqrt{x^2+y^2+z^2}} \langle x, y, z \rangle$. At $P(1,2,2)$, we have $\sqrt{1^2+2^2+2^2} = \sqrt{9} = 3$. So, $\nabla f(1,2,2) = \frac{1}{3} \langle 1, 2, 2 \rangle$. The direction vector is $\vec{v} = Q - P = \langle 3-1, 1-2, 0-2 \rangle = \langle 2, -1, -2 \rangle$. Its magnitude is $||\vec{v}|| = \sqrt{2^2+(-1)^2+(-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$. The unit direction vector is $\vec{u} = \frac{1}{3} \langle 2, -1, -2 \rangle$. The directional derivative is $D_{\vec{u}}f(P) = \nabla f(P) \cdot \vec{u} = \frac{1}{3} \langle 1, 2, 2 \rangle \cdot \frac{1}{3} \langle 2, -1, -2 \rangle = \frac{1}{9}(1(2) + 2(-1) + 2(-2)) = \frac{1}{9}(2-2-4) = -4/9$.
2. Consider $h(x,y) = \ln(x^2 + y^2)$ for $x^2 + y^2 > 0$. If the directional derivative of $h$ at point $(3,4)$ in the direction toward point $(6,8)$ is $\frac{a}{b}$ where $a$ and $b$ are integers in lowest terms, what is $a + b$?
   Answer: $7$ First find the gradient: $h_x = \frac{2x}{x^2+y^2}$ and $h_y = \frac{2y}{x^2+y^2}$. At $(3,4)$: $h_x(3,4) = \frac{2(3)}{9+16} = \frac{6}{25}$ and $h_y(3,4) = \frac{2(4)}{25} = \frac{8}{25}$. So $\nabla h(3,4) = \left\langle \frac{6}{25}, \frac{8}{25} \right\rangle$. The direction vector from $(3,4)$ to $(6,8)$ is $\vec{v} = \langle 6-3, 8-4 \rangle = \langle 3, 4 \rangle$. The unit vector is $\hat{u} = \frac{\langle 3, 4 \rangle}{|\langle 3, 4 \rangle|} = \frac{\langle 3, 4 \rangle}{5}$. The directional derivative is $D_{\hat{u}}h = \nabla h \cdot \hat{u} = \left\langle \frac{6}{25}, \frac{8}{25} \right\rangle \cdot \frac{\langle 3, 4 \rangle}{5} = \frac{1}{5} \left( \frac{6}{25} \cdot 3 + \frac{8}{25} \cdot 4 \right) = \frac{1}{5} \left( \frac{18}{25} + \frac{32}{25} \right) = \frac{1}{5} \cdot \frac{50}{25} = \frac{1}{5} \cdot 2 = \frac{2}{5}$. Since $\frac{2}{5}$ is in lowest terms, $a = 2$ and $b = 5$, so $a + b = 7$.
3. Let $f(x,y) = g(x^2+y^2)$, where $g$ is a differentiable function of a single variable. If $g'(2)=5$, what is the directional derivative of $f$ at the point $(1,1)$ in the direction of the vector $\vec{v} = \langle 3, 4 \rangle$?
   Answer: 14 First, we find the gradient of $f(x,y)$ using the chain rule. Let $u = x^2+y^2$. Then $f(x,y) = g(u)$. The partial derivatives are $\frac{\partial f}{\partial x} = g'(u) \frac{\partial u}{\partial x} = g'(x^2+y^2) \cdot 2x$ and $\frac{\partial f}{\partial y} = g'(u) \frac{\partial u}{\partial y} = g'(x^2+y^2) \cdot 2y$. So, $\nabla f(x,y) = \langle 2x g'(x^2+y^2), 2y g'(x^2+y^2) \rangle$. At the point $(1,1)$, we have $x^2+y^2 = 1^2+1^2=2$. We are given $g'(2)=5$. So, $\nabla f(1,1) = \langle 2(1)g'(2), 2(1)g'(2) \rangle = \langle 2(5), 2(5) \rangle = \langle 10, 10 \rangle$. The direction vector is $\vec{v} = \langle 3, 4 \rangle$. Its magnitude is $||\vec{v}|| = \sqrt{3^2+4^2}=\sqrt{25}=5$. The unit vector is $\vec{u} = \langle 3/5, 4/5 \rangle$. The directional derivative is $D_{\vec{u}}f(1,1) = \nabla f(1,1) \cdot \vec{u} = \langle 10, 10 \rangle \cdot \langle 3/5, 4/5 \rangle = 10(3/5) + 10(4/5) = 6+8=14$.
4. At a certain point $P$, the maximum rate of change of a differentiable function $f$ is $12$. What is the directional derivative of $f$ at $P$ in a direction that makes an angle of $2\pi/3$ with the direction of maximum rate of change?
   Answer: $-6$ The directional derivative of $f$ at a point $P$ in the direction of a unit vector $\vec{u}$ is given by $D_{\vec{u}}f(P) = ||\nabla f(P)|| \cos\theta$, where $\theta$ is the angle between the gradient vector $\nabla f(P)$ and the direction vector $\vec{u}$. The maximum rate of change is the magnitude of the gradient, so $||\nabla f(P)|| = 12$. The direction of maximum rate of change is the direction of the gradient. The problem states that the angle between our direction $\vec{u}$ and the direction of the gradient is $\theta = 2\pi/3$. Therefore, the directional derivative is $D_{\vec{u}}f(P) = 12 \cos(2\pi/3) = 12(-1/2) = -6$.
5. The directional derivative of a differentiable function $f(x,y)$ at $P(1,2)$ in the direction toward $Q(3,3)$ is $2\sqrt{5}$. The directional derivative at $P$ in the direction toward $R(0,4)$ is $0$. What is the gradient of $f$ at $P(1,2)$?
   Answer: $\langle 4, 2 \rangle$ Let $\nabla f(1,2) = \langle a, b \rangle$. The direction from $P$ to $Q$ is $\vec{v}_1 = Q-P = \langle 2, 1 \rangle$. The unit vector is $\vec{u}_1 = \frac{1}{\sqrt{2^2+1^2}} \langle 2, 1 \rangle = \frac{1}{\sqrt{5}} \langle 2, 1 \rangle$. The directional derivative is $D_{\vec{u}_1}f = \nabla f \cdot \vec{u}_1 = \langle a, b \rangle \cdot \frac{1}{\sqrt{5}} \langle 2, 1 \rangle = \frac{2a+b}{\sqrt{5}}$. We are given this is $2\sqrt{5}$, so $\frac{2a+b}{\sqrt{5}} = 2\sqrt{5}$, which implies $2a+b=10$. The direction from $P$ to $R$ is $\vec{v}_2 = R-P = \langle -1, 2 \rangle$. The unit vector is $\vec{u}_2 = \frac{1}{\sqrt{(-1)^2+2^2}} \langle -1, 2 \rangle = \frac{1}{\sqrt{5}} \langle -1, 2 \rangle$. The directional derivative is $D_{\vec{u}_2}f = \nabla f \cdot \vec{u}_2 = \langle a, b \rangle \cdot \frac{1}{\sqrt{5}} \langle -1, 2 \rangle = \frac{-a+2b}{\sqrt{5}}$. We are given this is $0$, so $-a+2b=0$, which means $a=2b$. Substituting this into the first equation gives $2(2b)+b=10$, so $5b=10$, and $b=2$. Then $a=2(2)=4$. Thus, $\nabla f(1,2) = \langle 4, 2 \rangle$.
6. Let $z = f(x,y)$ be a differentiable function of $x$ and $y$ defined implicitly by the equation $xyz + z^3 = 10$. Find the directional derivative of $f$ at the point $(x,y)=(1,1)$ in the direction of the vector $\vec{v} = \langle 1, \sqrt{3} \rangle$.
   Answer: $-\frac{1+\sqrt{3}}{13}$ First, find the value of $z$ when $(x,y)=(1,1)$. The equation becomes $(1)(1)z + z^3 = 10$, or $z^3+z-10=0$. By inspection, $z=2$ is a solution since $2^3+2-10 = 8+2-10=0$. Let $F(x,y,z) = xyz+z^3-10=0$. We can find the partial derivatives of $f(x,y)$ using implicit differentiation: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ and $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$. We have $F_x=yz$, $F_y=xz$, and $F_z=xy+3z^2$. At $(1,1,2)$, $F_x=2$, $F_y=2$, and $F_z=1(1)+3(2^2)=13$. So, $\frac{\partial z}{\partial x} = -2/13$ and $\frac{\partial z}{\partial y} = -2/13$. The gradient of $f$ at $(1,1)$ is $\nabla f(1,1) = \langle -2/13, -2/13 \rangle$. The direction vector is $\vec{v} = \langle 1, \sqrt{3} \rangle$, with magnitude $||\vec{v}|| = \sqrt{1^2+(\sqrt{3})^2} = \sqrt{4}=2$. The unit vector is $\vec{u} = \langle 1/2, \sqrt{3}/2 \rangle$. The directional derivative is $D_{\vec{u}}f(1,1) = \nabla f(1,1) \cdot \vec{u} = \langle -2/13, -2/13 \rangle \cdot \langle 1/2, \sqrt{3}/2 \rangle = (-2/13)(1/2) + (-2/13)(\sqrt{3}/2) = -\frac{1}{13} - \frac{\sqrt{3}}{13} = -\frac{1+\sqrt{3}}{13}$.
7. At a point $P$, the directional derivative of a function $f$ is $4\sqrt{2}$ in the direction of $\vec{v}_1 = \langle 1, 1 \rangle$ and $4$ in the direction of $\vec{v}_2 = \langle \sqrt{3}, 1 \rangle$. What is the maximum rate of change of $f$ at $P$?
   Answer: $8$ Let the gradient at $P$ be $\nabla f = \langle a, b \rangle$. The maximum rate of change is $||\nabla f|| = \sqrt{a^2+b^2}$. The unit vector for $\vec{v}_1$ is $\vec{u}_1 = \frac{1}{\sqrt{2}}\langle 1, 1 \rangle$. The directional derivative is $D_{\vec{u}_1}f = \langle a,b \rangle \cdot \frac{1}{\sqrt{2}}\langle 1, 1 \rangle = \frac{a+b}{\sqrt{2}} = 4\sqrt{2}$, so $a+b=8$. The unit vector for $\vec{v}_2$ is $\vec{u}_2 = \frac{1}{2}\langle \sqrt{3}, 1 \rangle$. The directional derivative is $D_{\vec{u}_2}f = \langle a,b \rangle \cdot \frac{1}{2}\langle \sqrt{3}, 1 \rangle = \frac{a\sqrt{3}+b}{2} = 4$, so $a\sqrt{3}+b=8$. We have a system of two equations: $a+b=8$ and $a\sqrt{3}+b=8$. Since both expressions equal 8, we have $a+b = a\sqrt{3}+b$, which simplifies to $a = a\sqrt{3}$, or $a(\sqrt{3}-1)=0$. This implies $a=0$. Substituting $a=0$ into $a+b=8$ gives $b=8$. So, the gradient is $\nabla f = \langle 0, 8 \rangle$. The maximum rate of change is the magnitude of the gradient: $||\nabla f|| = ||\langle 0, 8 \rangle|| = \sqrt{0^2+8^2}=8$.
8. The altitude of a mountain (in meters) is modeled by $h(x,y) = 2000 - 0.01x^2 - 0.02y^2$, where $x$ and $y$ are horizontal coordinates. A climber is at the point $(10, 10)$. If the climber moves in the direction of the vector $\vec{v} = \langle 1, -2 \rangle$, what is the initial rate of change of their altitude?
   Answer: Ascending at a rate of $\frac{3\sqrt{5}}{25}$ m per horizontal unit The rate of change of altitude is the directional derivative of $h(x,y)$. First, find the gradient: $\nabla h = \langle -0.02x, -0.04y \rangle$. At the point $(10,10)$, the gradient is $\nabla h(10,10) = \langle -0.02(10), -0.04(10) \rangle = \langle -0.2, -0.4 \rangle$. The direction of movement is given by $\vec{v} = \langle 1, -2 \rangle$. We need to use a unit vector, so we normalize $\vec{v}$: $||\vec{v}|| = \sqrt{1^2+(-2)^2} = \sqrt{5}$. The unit vector is $\vec{u} = \frac{1}{\sqrt{5}} \langle 1, -2 \rangle$. The directional derivative is $D_{\vec{u}}h(10,10) = \nabla h(10,10) \cdot \vec{u} = \langle -0.2, -0.4 \rangle \cdot \frac{1}{\sqrt{5}} \langle 1, -2 \rangle = \frac{1}{\sqrt{5}}(-0.2(1) - 0.4(-2)) = \frac{-0.2+0.8}{\sqrt{5}} = \frac{0.6}{\sqrt{5}}$. This simplifies to $\frac{0.6\sqrt{5}}{5} = \frac{6\sqrt{5}}{50} = \frac{3\sqrt{5}}{25}$. Since the value is positive, the climber is ascending.
9. Consider the function $f(x,y) = x^3 - xy$. Which of the following is a unit vector describing a direction in which the rate of change of $f$ at the point $P(1,2)$ is zero?
   Answer: $\frac{1}{\sqrt{2}} \langle 1, 1 \rangle$ The rate of change of a function is zero in directions orthogonal to its gradient vector. First, we compute the gradient of $f(x,y)$: $\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle = \langle 3x^2 - y, -x \rangle$. At the point $P(1,2)$, the gradient is $\nabla f(1,2) = \langle 3(1)^2 - 2, -1 \rangle = \langle 1, -1 \rangle$. A direction of zero change is given by a vector $\vec{v} = \langle a, b \rangle$ that is orthogonal to $\nabla f(1,2)$, meaning their dot product is zero: $\langle 1, -1 \rangle \cdot \langle a, b \rangle = a - b = 0$, so $a=b$. A simple choice for such a vector is $\langle 1, 1 \rangle$. To make this a unit vector, we divide by its magnitude, which is $\sqrt{1^2+1^2} = \sqrt{2}$. Thus, a unit vector for a direction of zero change is $\vec{u} = \frac{1}{\sqrt{2}} \langle 1, 1 \rangle$.
10. The temperature at any point $(x,y,z)$ in space is given by $T(x,y,z) = x^2 + 2y^2 - z^2$. A particle is at the point $P(1,1,1)$. Which of the following vectors represents a direction in which the particle should move to experience no initial temperature change?
    Answer: $\langle 1, 1, 3 \rangle$ To experience no initial temperature change, the particle must move in a direction where the directional derivative of the temperature function is zero. This occurs in any direction orthogonal to the gradient vector $\nabla T$. The gradient is $\nabla T = \langle 2x, 4y, -2z \rangle$. At the point $P(1,1,1)$, the gradient is $\nabla T(1,1,1) = \langle 2(1), 4(1), -2(1) \rangle = \langle 2, 4, -2 \rangle$. We need to find which of the given vectors has a dot product of zero with $\langle 2, 4, -2 \rangle$. Let's check the options: A) $\langle 1, 2, -1 \rangle \cdot \langle 2, 4, -2 \rangle = 2+8+2 = 12 \neq 0$. B) $\langle -1, -2, 1 \rangle \cdot \langle 2, 4, -2 \rangle = -2-8-2 = -12 \neq 0$. C) $\langle 1, 1, 3 \rangle \cdot \langle 2, 4, -2 \rangle = 2+4-6=0$. D) $\langle 1, 1, 1 \rangle \cdot \langle 2, 4, -2 \rangle = 2+4-2=4 \neq 0$. The correct direction is given by the vector in option C.
11. At a point $P$, the directional derivative of a differentiable function $f(x,y)$ in the direction of the unit vector $\vec{u} = \langle 1/\sqrt{2}, 1/\sqrt{2} \rangle$ is $3\sqrt{2}$. The directional derivative in the direction of the unit vector $\vec{w} = \langle 1/\sqrt{5}, 2/\sqrt{5} \rangle$ is $11/\sqrt{5}$. What is the directional derivative at $P$ in the direction of the vector $\vec{v} = \langle 0, -1 \rangle$?
    Answer: $-5$ Let the gradient of $f$ at $P$ be $\nabla f = \langle a, b \rangle$. From the given information, we can set up a system of linear equations. First, $D_{\vec{u}}f = \nabla f \cdot \vec{u} = \langle a, b \rangle \cdot \langle 1/\sqrt{2}, 1/\sqrt{2} \rangle = \frac{a+b}{\sqrt{2}} = 3\sqrt{2}$, which simplifies to $a+b=6$. Second, $D_{\vec{w}}f = \nabla f \cdot \vec{w} = \langle a, b \rangle \cdot \langle 1/\sqrt{5}, 2/\sqrt{5} \rangle = \frac{a+2b}{\sqrt{5}} = \frac{11}{\sqrt{5}}$, which simplifies to $a+2b=11$. Now we solve the system: $(a+2b) - (a+b) = 11-6$, which gives $b=5$. Substituting $b=5$ into $a+b=6$ gives $a=1$. So, the gradient is $\nabla f = \langle 1, 5 \rangle$. We need to find the directional derivative in the direction of $\vec{v} = \langle 0, -1 \rangle$. This is already a unit vector. The directional derivative is $D_{\vec{v}}f = \nabla f \cdot \vec{v} = \langle 1, 5 \rangle \cdot \langle 0, -1 \rangle = 1(0) + 5(-1) = -5$.
12. At a point $P$, a function $f$ has a directional derivative of 10 in the direction of steepest ascent and a directional derivative of $-5$ in another direction $\vec{u}$. What is the angle between the direction of steepest ascent and the direction $\vec{u}$?
    Answer: $2\pi/3$ The direction of steepest ascent is the direction of the gradient vector, $\nabla f$. The directional derivative in this direction is the magnitude of the gradient, $||\nabla f||$. We are given that this maximum rate of change is 10, so $||\nabla f|| = 10$. The formula for the directional derivative in the direction of a unit vector $\vec{u}$ is $D_{\vec{u}}f = ||\nabla f|| \cos\theta$, where $\theta$ is the angle between $\nabla f$ and $\vec{u}$. We are given that $D_{\vec{u}}f = -5$. Plugging in the known values, we get $-5 = 10 \cos\theta$. Solving for $\cos\theta$, we find $\cos\theta = -5/10 = -1/2$. The angle $\theta$ in the interval $[0, \pi]$ for which $\cos\theta = -1/2$ is $\theta = 2\pi/3$.
13. Let $f(x,y) = x^2 e^y$. What is the directional derivative of $f$ at the point $P(2,0)$ in the direction of the vector $\vec{v} = \langle 3, -4 \rangle$?
    Answer: $-4/5$ First, compute the gradient of $f$. $\nabla f(x,y) = \langle f_x, f_y \rangle = \langle 2xe^y, x^2e^y \rangle$. At the point $(2,0)$, the gradient is $\nabla f(2,0) = \langle 2(2)e^0, 2^2 e^0 \rangle = \langle 4, 4 \rangle$. Next, find the unit vector $\vec{u}$ in the direction of $\vec{v} = \langle 3, -4 \rangle$. The magnitude of $\vec{v}$ is $||\vec{v}|| = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5$. So, the unit vector is $\vec{u} = \frac{\vec{v}}{||\vec{v}||} = \langle 3/5, -4/5 \rangle$. The directional derivative is $D_{\vec{u}}f(2,0) = \nabla f(2,0) \cdot \vec{u} = \langle 4, 4 \rangle \cdot \langle 3/5, -4/5 \rangle = 4(3/5) + 4(-4/5) = 12/5 - 16/5 = -4/5$.

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