## MATH1901 Quizzes

Quiz 12: Directional derivatives and the gradient vector
Question 1 Questions
Let $f\left(x,y\right)={e}^{{x}^{2}}cosy$. What is $\nabla f\left(x,y\right)?$ Exactly one option must be correct)
 a) ${e}^{{x}^{2}}\mathbf{i}+cosy\mathbf{j}$ b) ${e}^{{x}^{2}}cosy\mathbf{i}-{e}^{{x}^{2}}siny\mathbf{j}$ c) $2x{e}^{{x}^{2}}cosy-{e}^{{x}^{2}}siny$ d) ${e}^{{x}^{2}}+cosy$ e) $2x{e}^{{x}^{2}}cosy\mathbf{i}-{e}^{{x}^{2}}siny\mathbf{j}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
Let $f\left(x,y\right)=\frac{1}{x+{y}^{2}}$. Find the gradient vector $\nabla f\left(1,1\right)$ at the point (1,1). Exactly one option must be correct)
 a) $-\frac{1}{4}\mathbf{i}-\frac{1}{2}\mathbf{j}$ b) $-\mathbf{i}-\frac{1}{2}\mathbf{j}$ c) $-\frac{1}{2}\mathbf{i}-\mathbf{j}$ d) $\frac{1}{4}\mathbf{i}+\frac{1}{2}\mathbf{j}$ e) $\mathbf{i}+\frac{1}{2}\mathbf{j}$

Choice (a) is correct!
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
The directional derivative of $f\left(x,y\right)={x}^{2}{y}^{3}+2{x}^{4}y$ at the point (1,-2) in the direction $3\mathbf{i}-4\mathbf{j}$ is Exactly one option must be correct)
 a) $\frac{1}{4}\mathbf{i}+\frac{1}{2}\mathbf{j}$ b) $-96\mathbf{i}-56\mathbf{j}$ c) -152 d) -30.4 e) $-32\mathbf{i}+14\mathbf{j}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
${D}_{\mathbf{u}}f\left(1,-2\right)=\nabla f\left(1,-2\right)\cdot \stackrel{̂}{\mathbf{u}}$. Here $\stackrel{̂}{\mathbf{u}}=\frac{3}{5}\mathbf{i}-\frac{4}{5}\mathbf{j}$ and $\nabla f\left(1,-2\right)=-32\mathbf{i}+14\mathbf{j}$. The directional derivative is always a scalar as it is the dot product of the two vectors.
Choice (e) is incorrect
Find the direction where the directional derivative is greatest for the function
$f\left(x,y\right)=3{x}^{2}{y}^{2}-{x}^{4}-{y}^{4}$
at the point (1,2). Exactly one option must be correct)
 a) $\frac{1}{\sqrt{2}}\left(-\mathbf{i}+\mathbf{j}\right)$ b) $\frac{1}{\sqrt{2}}\left(\mathbf{i}-\mathbf{j}\right)$ c) $\frac{1}{\sqrt{2}}\left(\mathbf{i}+\mathbf{j}\right)$ d) $\frac{1}{\sqrt{5}}\left(2\mathbf{i}+\mathbf{j}\right)$ e) $-\frac{1}{\sqrt{5}}\left(\mathbf{i}-\mathbf{j}\right)$

Choice (a) is incorrect
Choice (b) is correct!
The gradient vector gives the direction where the directional derivative is steepest. $\nabla f\left(1,-2\right)=20\mathbf{i}-20\mathbf{j}$, so any positive scalar multiple of this vector would provide an answer to this question. One such vector is the unit vector in this direction, $\frac{1}{\sqrt{2}}\left(\mathbf{i}-\mathbf{j}\right)$.
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Find the maximum directional derivative of the function
$f\left(x,y\right)=xlny+{x}^{2}{y}^{2}$
at the point (-1,1). Exactly one option must be correct)
 a) $-2\mathbf{i}+\mathbf{j}$ b) $\frac{1}{\sqrt{5}}\left(-2\mathbf{i}+\mathbf{j}\right)$ c) $1$ d) $\sqrt{5}$ e) $\frac{1}{\sqrt{5}}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The maximum directional derivative is equal to the magnitude of the gradient vector. Here
$|\nabla f|=|-2\mathbf{i}+\mathbf{j}|=|\sqrt{5}|.$
Choice (e) is incorrect
Let the temperature at the point $\left(x,y\right)$ in a flat plate be given by the function
$T\left(x,y\right)=3{x}^{2}+2xy.$
A tub of margarine is placed at (3,-6) in what direction should it be moved to cool most quickly? Exactly one option must be correct)
 a) $6\mathbf{i}+6\mathbf{j}$ b) $\mathbf{i}+\mathbf{j}$ c) $-\mathbf{i}-\mathbf{j}$ d) $6\mathbf{i}-12\mathbf{j}$ e) (3,-6) is already the coolest point on the plate.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
$\nabla T\left(3,-6\right)=6\mathbf{i}+6\mathbf{j}$ so the direction of most rapid increase in $T$ is $\mathbf{i}+\mathbf{j}$. For the most rapid decrease the tub of margarine must be moved in the opposite direction, $-\mathbf{i}-\mathbf{j}$.
Choice (d) is incorrect
Choice (e) is incorrect
Find a vector normal to the curve
${x}^{2}y+lny-2x=0$
at the point (2,1). Exactly one option must be correct)
 a) $2\mathbf{i}+5\mathbf{j}$ b) $5\mathbf{i}-2\mathbf{j}$ c) $-\frac{2}{5}\mathbf{i}+\mathbf{j}$ d) $2\mathbf{i}+\mathbf{j}$ e) None of the above

Choice (a) is correct!
If a planar curve in the $xy$ plane is defined implicitly by $f\left(x,y\right)=c$ then the vector $\nabla f$ is normal to the curve. Here $\nabla f=\left(2xy-2\right)\mathbf{i}+\left({x}^{2}+\frac{1}{y}\right)\mathbf{j}$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect