## MATH1001 Quizzes

Quiz 9: Directional derivatives and the gradient
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 $\left(1,-2\right)$ in the direction $3\mathbf{i}-4\mathbf{j}$ is Exactly one option must be correct)
 a) $\frac{\mathbf{i}}{4}+\frac{\mathbf{j}}{2}$ 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 $\left(3,-6\right)$. 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
In which directions is the directional derivative of $f\left(x,y\right)=xy+cos\phantom{\rule{0.3em}{0ex}}\left({x}^{2}\right)+2y$ at $\left(0,1\right)$ equal to $1$ ? Exactly one option must be correct)
 a) $-3\mathbf{i}+4\mathbf{j}$ and $\mathbf{i}$ b) $±\left(\mathbf{i}-2∕5\mathbf{j}\right)$ c) $±\left(2\mathbf{i}-\mathbf{j}\right)$ d) $2\mathbf{i}+\mathbf{j}$ and $\mathbf{i}$ e) $3\mathbf{i}-4\mathbf{j}$ and $\mathbf{j}$

Choice (a) is correct!
The gradient vector at $\left(x,y\right)$ is $\nabla f\left(x,y\right)=\left(y-2xsin\phantom{\rule{0.3em}{0ex}}{x}^{2}\right)\mathbf{i}+\left(x+2\right)\mathbf{j}$ and so at $\left(0,1\right)$ we have ${D}_{\mathbf{u}}f\left(0,1\right)=\nabla f\left(0,1\right)\cdot \stackrel{̂}{\mathbf{u}}=\left(\mathbf{i}+2\mathbf{j}\right)\cdot \stackrel{̂}{\mathbf{u}}={u}_{1}+2{u}_{2},$ where $\stackrel{̂}{\mathbf{u}}={u}_{1}\mathbf{i}+{u}_{2}\mathbf{j}$ is a unit vector. Thus ${D}_{\mathbf{u}}f\left(0,1\right)=1$ if ${u}_{1}=1-2{u}_{2}$. Since ${u}_{1}^{2}+{u}_{2}^{2}=1$, we obtain the conditions ${u}_{2}=0$ or ${u}_{2}=4∕5$. The corresponding values of ${u}_{1}$ are ${u}_{1}=1$ and ${u}_{1}=-3∕5$. This gives two directions, $\mathbf{i}$ and $-\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Find the maximum rate of change of $f\left(x,y\right)={y}^{2}∕x$ at $\left(2,3\right),$ and the direction in which it occurs. Exactly one option must be correct)
 a) $21∕4$, in the direction of $\left(-9∕4\right)\mathbf{i}-3\mathbf{j}$ b) $21∕4$, in the direction of $\left(-9∕4\right)\mathbf{i}+3\mathbf{j}$ c) $15∕4$, in the direction of $\left(9∕4\right)\mathbf{i}+3\mathbf{j}$ d) $15∕4$, in the direction of $\left(-9∕4\right)\mathbf{i}+3\mathbf{j}$ e) $9∕4$, in the direction of $\left(9∕4\right)\mathbf{i}+3\mathbf{j}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The gradient vector at $\left(x,y\right)$ is $\nabla f\left(x,y\right)=\left(-{y}^{2}∕{x}^{2}\right)\mathbf{i}+\left(2y∕x\right)\mathbf{j}$ and so at $\left(2,3\right)$ we have $\nabla f\left(2,3\right)=\left(-9∕4\right)\phantom{\rule{0.3em}{0ex}}\mathbf{i}+3\phantom{\rule{0.3em}{0ex}}\mathbf{j}.$ This is the direction of maximum rate of change of $f$ at the point $\left(2,3\right)$. The actual maximum rate of change is $|\nabla f\left(2,3\right)|=|\left(-9∕4\right)\phantom{\rule{0.3em}{0ex}}\mathbf{i}+3\phantom{\rule{0.3em}{0ex}}\mathbf{j}|=\sqrt{{\left(-9∕4\right)}^{2}+{3}^{2}}=15∕4.$
Choice (e) is incorrect
Find the direction in which the function $g\left(x,y\right)={x}^{4}y-{x}^{2}{y}^{3}$ decreases fastest at the point $\left(2,-3\right)$. Exactly one option must be correct)
 a) $12\mathbf{i}-92\mathbf{j}$ b) $12\mathbf{i}+92\mathbf{j}$ c) $-12\mathbf{i}-92\mathbf{j}$ d) $-12\mathbf{i}+92\mathbf{j}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The gradient vector at $\left(x,y\right)$ is $\nabla g\left(x,y\right)=\left(4{x}^{3}y-2x{y}^{3}\right)\mathbf{i}+\left({x}^{4}-3{x}^{2}{y}^{2}\right)\mathbf{j}$ and so at $\left(2,-3\right)$ we have $\nabla g\left(2,-3\right)=12\phantom{\rule{0.3em}{0ex}}\mathbf{i}-92\phantom{\rule{0.3em}{0ex}}\mathbf{j}.$ This is the direction of fastest increase of $g$ at the point $\left(2,-3\right)$. The direction of fastest decrease is the opposite direction, namely $-12\phantom{\rule{0.3em}{0ex}}\mathbf{i}+92\phantom{\rule{0.3em}{0ex}}\mathbf{j}.$