Gradient Vector - Multivariable Calculus

1. A surface is defined implicitly by the equation $x\ln(y) + yz^3 + z\sqrt{x} = 11$. Find the value of the partial derivative $\frac{\partial z}{\partial x}$ at the point $(4, 1, 2)$.
   Answer: $-1/28$ When you encounter an implicitly defined surface and need to find a partial derivative, you're dealing with implicit differentiation in multiple variables. The key insight is that since the surface equation equals a constant, differentiating both sides with respect to any variable gives you zero. Starting with $x\ln(y) + yz^3 + z\sqrt{x} = 11$, differentiate both sides with respect to $x$, treating $y$ as constant and $z$ as a function of $x$: $\frac{\partial}{\partial x}[x\ln(y)] + \frac{\partial}{\partial x}[yz^3] + \frac{\partial}{\partial x}[z\sqrt{x}] = 0$ This gives us: $\ln(y) + 3yz^2\frac{\partial z}{\partial x} + \frac{\partial z}{\partial x} \cdot \sqrt{x} + z \cdot \frac{1}{2\sqrt{x}} = 0$ Solving for $\frac{\partial z}{\partial x}$: $\frac{\partial z}{\partial x}(3yz^2 + \sqrt{x}) = -\ln(y) - \frac{z}{2\sqrt{x}}$ At point $(4, 1, 2)$: $\frac{\partial z}{\partial x}(3 \cdot 1 \cdot 4 + 2) = -\ln(1) - \frac{2}{4} = 0 - \frac{1}{2} = -\frac{1}{2}$ So: $\frac{\partial z}{\partial x} \cdot 14 = -\frac{1}{2}$, giving us $\frac{\partial z}{\partial x} = -\frac{1}{28}$ Answer (A) $1/28$ has the wrong sign—likely from a sign error in the implicit differentiation. Answer (C) $-28$ inverts the fraction, suggesting algebraic mistakes. Answer (D) $-6/7$ might result from computational errors when substituting the point values. Strategy tip: Always verify your point satisfies the original equation before substituting, and be extra careful with signs when moving terms across the equation during implicit differentiation.
2. Let $f(x, y) = x^2 e^{xy}$. Which of the following vectors represents the direction of the most rapid decrease of the function $f$ at the point $P(1, 0)$?
   Answer: $\langle -2, -1 \rangle$ The direction of the most rapid increase of a function is given by its gradient vector, $\nabla f$. The direction of the most rapid decrease is in the opposite direction, $-\nabla f$. First, we compute the partial derivatives of $f(x, y) = x^2 e^{xy}$. Using the product rule for $f_x$: $f_x(x,y) = \frac{\partial}{\partial x}(x^2 e^{xy}) = 2xe^{xy} + x^2(ye^{xy}) = (2x+x^2y)e^{xy}$. For $f_y$: $f_y(x,y) = \frac{\partial}{\partial y}(x^2 e^{xy}) = x^2(xe^{xy}) = x^3e^{xy}$. The gradient vector is $\nabla f(x,y) = \langle (2x+x^2y)e^{xy}, x^3e^{xy} \rangle$. Now, we evaluate the gradient at the point $P(1, 0)$: $f_x(1, 0) = (2(1)+1^2(0))e^{1 \cdot 0} = 2e^0 = 2$. $f_y(1, 0) = 1^3 e^{1 \cdot 0} = 1e^0 = 1$. So, $\nabla f(1, 0) = \langle 2, 1 \rangle$. This is the direction of the greatest increase. The direction of the most rapid decrease is $-\nabla f(1, 0) = -\langle 2, 1 \rangle = \langle -2, -1 \rangle$.
3. The electric potential in a region is given by $V(x,y) = \frac{k}{\sqrt{x^2 + y^2}}$ where $k > 0$ is a constant. The electric field is $\vec{E} = -\nabla V$. At the point $(3, 4)$, what is the magnitude of the electric field?
   Answer: $\frac{k}{25}$ First, compute the gradient of $V(x,y) = k(x^2 + y^2)^{-1/2}$. Using the chain rule: $V_x = k \cdot (-\frac{1}{2})(x^2 + y^2)^{-3/2} \cdot 2x = -\frac{kx}{(x^2 + y^2)^{3/2}}$ and $V_y = -\frac{ky}{(x^2 + y^2)^{3/2}}$. So $\nabla V = \langle -\frac{kx}{(x^2 + y^2)^{3/2}}, -\frac{ky}{(x^2 + y^2)^{3/2}} \rangle$. At $(3, 4)$: $x^2 + y^2 = 9 + 16 = 25$, so $(x^2 + y^2)^{3/2} = 25^{3/2} = 125$. Therefore: $\nabla V(3, 4) = \langle -\frac{3k}{125}, -\frac{4k}{125} \rangle$. The electric field is $\vec{E} = -\nabla V = \langle \frac{3k}{125}, \frac{4k}{125} \rangle$. The magnitude is: $||\vec{E}|| = \sqrt{\left(\frac{3k}{125}\right)^2 + \left(\frac{4k}{125}\right)^2} = \sqrt{\frac{9k^2 + 16k^2}{125^2}} = \sqrt{\frac{25k^2}{125^2}} = \frac{5k}{125} = \frac{k}{25}$.
4. The directional derivative of a function $f(x,y)$ at the point $P(1,2)$ in the direction of the vector $\vec{v}_1 = \langle 3, 4 \rangle$ is $10$. The directional derivative at $P(1,2)$ in the direction of the vector $\vec{v}_2 = \langle -4, 3 \rangle$ is $5$. What is the gradient vector $\nabla f(1,2)$?
   Answer: $\langle 2, 11 \rangle$ When you encounter directional derivative problems with multiple directions, you're working with the fundamental relationship: $D_{\vec{u}}f = \nabla f \cdot \vec{u}$, where $\vec{u}$ is a unit vector. This creates a system of equations you can solve for the gradient components. First, convert the given vectors to unit vectors. For $\vec{v}_1 = \langle 3, 4 \rangle$: $|\vec{v}_1| = \sqrt{3^2 + 4^2} = 5$, so $\vec{u}_1 = \langle 3/5, 4/5 \rangle$. For $\vec{v}_2 = \langle -4, 3 \rangle$: $|\vec{v}_2| = \sqrt{(-4)^2 + 3^2} = 5$, so $\vec{u}_2 = \langle -4/5, 3/5 \rangle$. Let $\nabla f(1,2) = \langle a, b \rangle$. The given information creates two equations: - $\langle a, b \rangle \cdot \langle 3/5, 4/5 \rangle = 10$, which gives $\frac{3a + 4b}{5} = 10$, so $3a + 4b = 50$ - $\langle a, b \rangle \cdot \langle -4/5, 3/5 \rangle = 5$, which gives $\frac{-4a + 3b}{5} = 5$, so $-4a + 3b = 25$ Solving this system: multiply the first equation by 3 and the second by 4, then add: $9a + 12b - 16a + 12b = 150 + 100$, giving $-7a + 24b = 250$. From the second equation, $b = \frac{25 + 4a}{3}$. Substituting and solving yields $a = 2$ and $b = 11$. Answer B $\langle 2, 11 \rangle$ is correct. Answer A $\langle 2/5, 11/5 \rangle$ incorrectly assumes you don't need to convert to unit vectors. Answers C and D represent calculation errors in solving the linear system. Always remember: directional derivatives require unit vectors, creating a solvable system when you have enough directional information.
5. Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a differentiable function and let $P$ be a point in its domain. Which of the following statements about the gradient $\nabla f(P)$ is necessarily true?
   Answer: If $\vec{u}$ and $\vec{v}$ are any two orthogonal unit vectors, then $(D_{\vec{u}}f(P))^2 + (D_{\vec{v}}f(P))^2 = |\nabla f(P)|^2$. (A) is false. If $\nabla f(P) = \vec{0}$, $P$ is a critical point, but it could be a saddle point, which is not a local extremum. (B) is false. The gradient vector $\nabla f(P)$ is orthogonal (normal) to the level curve of $f$ through $P$, not parallel to its tangent line. (C) is true. Let $\nabla f(P) = \vec{g}$. The directional derivatives are $D_{\vec{u}}f(P) = \vec{g} \cdot \vec{u}$ and $D_{\vec{v}}f(P) = \vec{g} \cdot \vec{v}$. Since $\vec{u}$ and $\vec{v}$ are orthogonal unit vectors, they form an orthonormal basis for $\mathbb{R}^2$. The vector $\vec{g}$ can be expressed in this basis as $\vec{g} = (\vec{g} \cdot \vec{u})\vec{u} + (\vec{g} \cdot \vec{v})\vec{v}$. The squared magnitude of $\vec{g}$ is given by the sum of the squares of its components in this orthonormal basis: $|\vec{g}|^2 = (\vec{g} \cdot \vec{u})^2 + (\vec{g} \cdot \vec{v})^2$. Substituting the definitions, we get $|\nabla f(P)|^2 = (D_{\vec{u}}f(P))^2 + (D_{\vec{v}}f(P))^2$. This is the 2D version of Parseval's identity. (D) is false. The directional derivative is $D_{\vec{u}}f(P) = |\nabla f(P)|\cos\theta$. If $\vec{u}$ points in the same direction as $\nabla f(P)$, then $\theta=0$ and $\cos\theta=1$, making $D_{\vec{u}}f(P) = |\nabla f(P)|$. The relationship is $D_{\vec{u}}f(P) \le |\nabla f(P)|$, not strictly less than.
6. Let $f(x, y) = x^2 + y^2$. At which point $P$ on the line $y = x - 1$ is the gradient vector $\nabla f(P)$ normal to the line?
   Answer: $(1/2, -1/2)$ When you see a question about gradient vectors being normal to curves, you're dealing with the fundamental relationship between gradients and perpendicularity. The gradient vector $\nabla f$ at any point is always perpendicular to the level curves of the function, but this question asks when it's perpendicular to a different line entirely. First, find the gradient: $\nabla f(x,y) = (2x, 2y)$. For any point $(x, x-1)$ on the line $y = x - 1$, the gradient is $\nabla f = (2x, 2(x-1)) = (2x, 2x-2)$. The line $y = x - 1$ has direction vector $(1, 1)$ (since its slope is 1). For the gradient to be normal (perpendicular) to this line, their dot product must equal zero: $(2x, 2x-2) \cdot (1, 1) = 2x + (2x-2) = 4x - 2 = 0$ Solving: $x = \frac{1}{2}$, so $y = \frac{1}{2} - 1 = -\frac{1}{2}$. The point is $(\frac{1}{2}, -\frac{1}{2})$. Looking at the wrong answers: A) $(-\frac{1}{2}, \frac{1}{2})$ isn't even on the line since $\frac{1}{2} \neq -\frac{1}{2} - 1$. B) $(1, -1)$ also fails the line equation since $-1 \neq 1 - 1 = 0$. C) $(1, 0)$ is on the line, but when you check the perpendicularity condition, the dot product gives $4(1) - 2 = 2 \neq 0$. Remember: always verify that candidate points actually lie on the given constraint, then check the geometric condition. Many wrong answers in these problems fail the basic constraint test.
7. Let $f(x,y)$ be a differentiable function with gradient $\nabla f$ at a point $P$. The directional derivative of $f$ at $P$ in the direction of a unit vector $\vec{u}$ is $D_{\vec{u}}f(P)$. If $\theta$ is the angle between $\nabla f(P)$ and $\vec{u}$, for which value of $\theta$ is $D_{\vec{u}}f(P)$ equal to exactly half of its maximum possible value?
   Answer: $\pi/3$ The formula for the directional derivative is $D_{\vec{u}}f(P) = \nabla f(P) \cdot \vec{u}$. Using the geometric definition of the dot product, this can be written as $D_{\vec{u}}f(P) = |\nabla f(P)| |\vec{u}| \cos\theta$, where $\theta$ is the angle between the gradient vector and the direction vector $\vec{u}$. Since $\vec{u}$ is a unit vector, $|\vec{u}| = 1$. So, the formula simplifies to $D_{\vec{u}}f(P) = |\nabla f(P)| \cos\theta$. The maximum value of $\cos\theta$ is 1, which occurs when $\theta = 0$. Therefore, the maximum possible value of the directional derivative is $|\nabla f(P)|$. We want to find the angle $\theta$ for which the directional derivative is half of this maximum value: $D_{\vec{u}}f(P) = \frac{1}{2} |\nabla f(P)|$. Substituting the formula for the directional derivative gives: $|\nabla f(P)| \cos\theta = \frac{1}{2} |\nabla f(P)|$. Assuming $\nabla f(P)$ is not the zero vector, we can divide both sides by $|\nabla f(P)|$ to get: $\cos\theta = 1/2$. The angle $\theta$ in the range $[0, \pi]$ for which $\cos\theta = 1/2$ is $\theta = \pi/3$. (A) $\pi/6$ would be correct if $\cos\theta = \sqrt{3}/2$. (B) $\pi/4$ corresponds to $\cos\theta = \sqrt{2}/2$. (D) $2\pi/3$ corresponds to $\cos\theta = -1/2$, which would give a value that is half the minimum possible value.
8. A particle's position is described by the vector function $\vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t \rangle$. The particle moves through a region where the temperature is given by $T(x,y,z) = xy + z^2$. What is the rate of change of temperature with respect to time that the particle experiences at $t=1/2$?
   Answer: $1 - \pi$ When a particle moves through a temperature field, you need to find how temperature changes along the particle's path using the chain rule. This requires computing $\frac{dT}{dt} = \nabla T \cdot \vec{r}'(t)$, where you take the dot product of the temperature gradient with the particle's velocity vector. First, find the gradient of $T(x,y,z) = xy + z^2$: $\nabla T = \langle y, x, 2z \rangle$ Next, find the velocity vector by differentiating $\vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t \rangle$: $\vec{r}'(t) = \langle -\pi\sin(\pi t), \pi\cos(\pi t), 1 \rangle$ At $t = 1/2$, evaluate both vectors: - Position: $\vec{r}(1/2) = \langle \cos(\pi/2), \sin(\pi/2), 1/2 \rangle = \langle 0, 1, 1/2 \rangle$ - Velocity: $\vec{r}'(1/2) = \langle -\pi\sin(\pi/2), \pi\cos(\pi/2), 1 \rangle = \langle -\pi, 0, 1 \rangle$ - Gradient at this point: $\nabla T(0,1,1/2) = \langle 1, 0, 1 \rangle$ The rate of change is: $\frac{dT}{dt} = \langle 1, 0, 1 \rangle \cdot \langle -\pi, 0, 1 \rangle = -\pi + 1 = 1 - \pi$ This confirms answer D. A) $1 - \pi/2$ likely comes from incorrectly using $z = 1/2$ instead of $2z = 1$ in the gradient calculation. B) $1 + \pi$ results from a sign error when computing $-\pi\sin(\pi/2)$. C) $1$ ignores the velocity component entirely, using only the $z$-direction contribution. Remember: for particle motion through scalar fields, always compute $\nabla f \cdot \vec{v}$ to find the directional rate of change along the path.
9. The temperature distribution in a metal plate is given by $T(x,y) = 100 - x^2 - 2y^2 + xy$. An ant at position $(2, 1)$ wants to move in the direction of steepest temperature increase. If the ant moves a small distance $\epsilon$ in this optimal direction, approximately how much will the temperature increase?
   Answer: $\epsilon \sqrt{13}$ The direction of steepest increase is given by the gradient vector $\nabla T$. First, compute: $T_x = -2x + y$ and $T_y = -4y + x$. At $(2,1)$: $T_x(2,1) = -4 + 1 = -3$ and $T_y(2,1) = -4 + 2 = -2$. So $\nabla T(2,1) = \langle -3, -2 \rangle$. The magnitude is $||\nabla T(2,1)|| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$. When moving a small distance $\epsilon$ in the direction of the gradient, the temperature change is approximately $\epsilon \cdot ||\nabla T|| = \epsilon\sqrt{13}$. Note that the gradient points in the direction of steepest increase, and its magnitude gives the rate of change per unit distance in that direction.
10. Let $S$ be the level surface of the function $f(x, y, z) = x^2 - y^2 + 2z^2$ that passes through the point $P(1, 1, 1)$. Which of the following is a vector equation for the line normal to the surface $S$ at point $P$?
    Answer: $L(t) = \langle 1, 1, 1 \rangle + t \langle 2, -2, 4 \rangle$ The gradient vector $\nabla f$ at a point $P$ is normal to the level surface of $f$ that passes through $P$. The normal line to the surface at $P$ is the line that passes through $P$ and has the direction of the gradient vector $\nabla f(P)$. First, compute the gradient of $f(x, y, z) = x^2 - y^2 + 2z^2$: $\nabla f(x, y, z) = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle = \langle 2x, -2y, 4z \rangle$. Next, evaluate the gradient at the point $P(1, 1, 1)$: $\nabla f(1, 1, 1) = \langle 2(1), -2(1), 4(1) \rangle = \langle 2, -2, 4 \rangle$. This vector is the direction vector for the normal line. The line passes through the point $P(1, 1, 1)$. The vector equation of a line is $\vec{r}(t) = \vec{p}_0 + t\vec{d}$, where $\vec{p}_0$ is a point on the line and $\vec{d}$ is the direction vector. Thus, the equation is $L(t) = \langle 1, 1, 1 \rangle + t \langle 2, -2, 4 \rangle$. (B) incorrectly swaps the point and the direction vector. (C) is the equation of the tangent plane to the surface at $P$, not the normal line. (D) makes a sign error in the partial derivative with respect to $y$.

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