## MATH1001 Quizzes

Quiz 5: Partial derivatives and tangent planes
Question 1 Questions
Which option correctly gives the two first order partial derivatives of the following function?
$f\left(x,y\right)={e}^{x}+\frac{x}{y}+{\left(2x+y\right)}^{4}$
Exactly one option must be correct)
 a) ${f}_{x}={e}^{x}+\frac{1}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}=xlny+4{\left(2x+y\right)}^{3}$ b) ${f}_{x}={e}^{x}+\frac{{x}^{2}}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}=-\frac{x}{{y}^{2}}+8{\left(2x+y\right)}^{3}$ c) ${f}_{x}={e}^{x}+\frac{{x}^{2}}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}=xlny+8{\left(2x+y\right)}^{3}$ d) ${f}_{x}={e}^{x}+\frac{1}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}=-\frac{x}{{y}^{2}}+4{\left(2x+y\right)}^{3}$ e) ${f}_{x}={e}^{x}+\frac{{x}^{2}}{y}+8{\left(2x+y\right)}^{3},\phantom{\rule{1em}{0ex}}{f}_{y}={e}^{x}-\frac{x}{{y}^{2}}+8{\left(2x+y\right)}^{3}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect
Find the two first order partial derivatives, with respect to $x$ and $y$, of
$z=cos\left({x}^{2}y\right)+siny.$
Exactly one option must be correct)
 a) $\frac{\partial z}{\partial x}=-2xsin\left({x}^{2}y\right)$ and $\frac{\partial z}{\partial y}=-{x}^{2}ysin\left({x}^{2}y\right)-cosy$ b) $\frac{\partial z}{\partial x}=-2xysin\left({x}^{2}y\right)$ and $\frac{\partial z}{\partial y}={x}^{2}sin\left({x}^{2}y\right)-cosy$ c) $\frac{\partial z}{\partial x}=2xsin\left({x}^{2}y\right)-cosy$ and $\frac{\partial z}{\partial y}={x}^{2}sin\left({x}^{2}y\right)$ d) $\frac{\partial z}{\partial x}=-2xysin\left({x}^{2}y\right)$ and $\frac{\partial z}{\partial y}=-{x}^{2}sin\left({x}^{2}y\right)+cosy$ e) $\frac{\partial z}{\partial x}=-{x}^{2}sin\left({x}^{2}y\right)$ and $\frac{\partial z}{\partial y}=-2xsin\left({x}^{2}y\right)+cosy$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect
Find the first order partial derivative with respect to $y$ of
$f\left(x,y\right)=x{e}^{x{y}^{2}}+ln\left(xy\right).$
Exactly one option must be correct)
 a) ${e}^{x{y}^{2}}+2y{x}^{2}{e}^{x{y}^{2}}+\frac{1}{y}+\frac{1}{x}$ b) $2y{x}^{2}{e}^{x{y}^{2}}+\frac{1}{y}$ c) $x{e}^{x{y}^{2}}+\frac{1}{xy}$ d) ${e}^{x{y}^{2}}+{y}^{2}x{e}^{x{y}^{2}}+\frac{1}{x}$ e) $2y{x}^{2}{e}^{x{y}^{2}}+\frac{1}{xy}$

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Find the first order partial derivative with respect to $x$ of
$f\left(x,y\right)=sin\left(\frac{1}{{x}^{2}+{y}^{2}}\right).$
Exactly one option must be correct)
 a) $2xcos\left(\frac{1}{{x}^{2}+{y}^{2}}\right)$ b) $\frac{2y}{{\left({x}^{2}+{y}^{2}\right)}^{2}}cos\left(\frac{1}{{x}^{2}+{y}^{2}}\right)$ c) $-\frac{2x}{{\left({x}^{2}+{y}^{2}\right)}^{2}}cos\left(\frac{1}{{x}^{2}+{y}^{2}}\right)$ d) $2yln\left({x}^{2}+{y}^{2}\right)cos\left(\frac{1}{{x}^{2}+{y}^{2}}\right)$ e) $-sin\left(\frac{1}{{\left({x}^{2}+{y}^{2}\right)}^{2}}\right)$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect
Find the equation of the tangent line to $y=\sqrt{{e}^{5x}+3}$ at $x=0$. Exactly one option must be correct)
 a) $x-4y+8=0$ b) $x-4y-8=0$ c) $5x-4y+8=0$ d) $5x-4y+4\sqrt{3}=0$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Find the equation of the tangent plane to the surface
$z={x}^{2}-{y}^{2}$
at the point $\left(5,-4,9\right).$ Exactly one option must be correct)
 a) $z-9=10\left(x-5\right)+8\left(y+4\right)$ b) $z-9=10\left(x-5\right)-8\left(y-4\right)$ c) $z-9=8\left(x+4\right)+10\left(y-5\right)$ d) $z-9=8\left(x-5\right)-10\left(y-4\right)$ e) $z-9=5\left(x-8\right)+4\left(y-8\right)$

Choice (a) is correct!
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Find the equation of the tangent plane to the surface
$z={e}^{xy}$
at the point where $x=2$ and $y=0$. Exactly one option must be correct)
 a) $z=-2x+2y+5$ b) $z=-y+2$ c) $z=2y+1$ d) $z=x+2y-1$ e) $z=x-y-1$

Choice (a) is incorrect
We have $\frac{\partial z}{\partial x}=y{e}^{xy}$ and $\frac{\partial z}{\partial x}=x{e}^{xy}$. Now evaluate these at $x=2$ and $y=0$.
Choice (b) is incorrect
We have $\frac{\partial z}{\partial x}=y{e}^{xy}$ and $\frac{\partial z}{\partial x}=x{e}^{xy}$. Now evaluate these at $x=2$ and $y=0$.
Choice (c) is correct!
At $\left(2,0\right)$ we have $\frac{\partial z}{\partial x}=0,\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial y}=2$ and $z=1$. So the tangent plane has equation $z-1=0\left(x-2\right)+2\left(y-0\right)$, that is, $z=2y+1$.
Choice (d) is incorrect
We have $\frac{\partial z}{\partial x}=y{e}^{xy}$ and $\frac{\partial z}{\partial x}=x{e}^{xy}$. Now evaluate these at $x=2$ and $y=0$.
Choice (e) is incorrect
We have $\frac{\partial z}{\partial x}=y{e}^{xy}$ and $\frac{\partial z}{\partial x}=x{e}^{xy}$. Now evaluate these at $x=2$ and $y=0$.
Find the equation of the tangent plane to the surface
$z=3{x}^{2}y+5{y}^{2}cosx$
at the point where $x=0$ and $y=2$. Exactly one option must be correct)
 a) $z=6x-10y+40$ b) $z=10x+20y+22$ c) $z=20y+12$ d) $z=20y-20$ e) $z=5x-y-18$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect
Which of the options satisfies the following equation? $x\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial x}+y\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial y}=z$ Exactly one option must be correct)
 a) $z=x{e}^{y}$ b) $z=\frac{x+y}{x-y}$ c) $z={x}^{2}+y$ d) $z=\sqrt{{x}^{2}+{y}^{2}}$ e) $z=ln\left({x}^{2}-y\right)$

Choice (a) is incorrect
Since $\frac{\partial z}{\partial x}={e}^{y}$ and $\frac{\partial z}{\partial y}=x{e}^{y}$, we see that in this case, $x\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial x}+y\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial y}=x{e}^{y}+xy{e}^{y}\ne z$
Choice (b) is incorrect
Since $\frac{\partial z}{\partial x}=\frac{-2y}{{\left(x-y\right)}^{2}}$ and $\frac{\partial z}{\partial y}=\frac{2x}{{\left(x-y\right)}^{2}}$, we see that in this case, $x\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial x}+y\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial y}=0\ne z$
Choice (c) is incorrect
Since $\frac{\partial z}{\partial x}=2x$ and $\frac{\partial z}{\partial y}=1$, we see that in this case, $x\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial x}+y\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial y}=2{x}^{2}+y\ne z$
Choice (d) is correct!
Here we have $\frac{\partial z}{\partial x}=\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}$ and $\frac{\partial z}{\partial y}=\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}$. In this case, $x\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial x}+y\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial y}=\frac{{x}^{2}+{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}=\sqrt{{x}^{2}+{y}^{2}}=z$
Choice (e) is incorrect
Since $\frac{\partial z}{\partial x}=\frac{2x}{{x}^{2}-y}$ and $\frac{\partial z}{\partial y}=\frac{-1}{{x}^{2}-y}$, we see that in this case, $x\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial x}+y\phantom{\rule{0.3em}{0ex}}\frac{\partial z}{\partial y}=\frac{2{x}^{2}-y}{{x}^{2}-y}\ne z$
Find the equation of the tangent plane to the surface
$z=ln\left({x}^{2}+{y}^{2}\right)$
at the point where $x=1$ and $y=-2$. Exactly one option must be correct)
 a) $5z=2\left(x-1\right)-4\left(y+2\right)$ b) $5z=ln5+2x-4y+10$ c) $5z=5ln5+4y-8$ d) $5z=5ln5-2x+4y+4$ e) $5z=5ln5+2x-4y-10$

Choice (a) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{2x}{{x}^{2}+{y}^{2}}$ and $\frac{\partial z}{\partial x}=\frac{2y}{{x}^{2}+{y}^{2}}$. Now evaluate these at $x=1$ and $y=-2$.
Choice (b) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{2x}{{x}^{2}+{y}^{2}}$ and $\frac{\partial z}{\partial x}=\frac{2y}{{x}^{2}+{y}^{2}}$. Now evaluate these at $x=1$ and $y=-2$.
Choice (c) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{2x}{{x}^{2}+{y}^{2}}$ and $\frac{\partial z}{\partial x}=\frac{2y}{{x}^{2}+{y}^{2}}$. Now evaluate these at $x=1$ and $y=-2$.
Choice (d) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{2x}{{x}^{2}+{y}^{2}}$ and $\frac{\partial z}{\partial x}=\frac{2y}{{x}^{2}+{y}^{2}}$. Now evaluate these at $x=1$ and $y=-2$.
Choice (e) is correct!