## MATH1001 Quizzes

Quiz 8: More on the partial derivative
Question 1 Questions
What is the differential $dz$ of the function given by $z=\frac{1}{1-x-y-{x}^{2}+{y}^{2}}$, at the point $\left(1,-2\right)$ ? Exactly one option must be correct)
 a) $dz=-\frac{3}{5}\phantom{\rule{0.3em}{0ex}}dx-dy$ b) $dz=\frac{3}{25}\phantom{\rule{0.3em}{0ex}}dx+\frac{5}{25}\phantom{\rule{0.3em}{0ex}}dy$ c) $dz=\frac{3}{5}\phantom{\rule{0.3em}{0ex}}dx+\frac{5}{5}\phantom{\rule{0.3em}{0ex}}dy$ d) $dz=\frac{3}{25}\phantom{\rule{0.3em}{0ex}}dx-\frac{5}{25}\phantom{\rule{0.3em}{0ex}}dy$ e) $dz=-\frac{3}{25}\phantom{\rule{0.3em}{0ex}}dx+\frac{5}{25}\phantom{\rule{0.3em}{0ex}}dy$

Choice (a) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{1+2x}{{\left(1-x-y-{x}^{2}+{y}^{2}\right)}^{2}}$ and $\frac{\partial z}{\partial y}=\frac{1-2y}{{\left(1-x-y-{x}^{2}+{y}^{2}\right)}^{2}}$.
Choice (b) is correct!
Choice (c) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{1+2x}{{\left(1-x-y-{x}^{2}+{y}^{2}\right)}^{2}}$ and $\frac{\partial z}{\partial y}=\frac{1-2y}{{\left(1-x-y-{x}^{2}+{y}^{2}\right)}^{2}}$.
Choice (d) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{1+2x}{{\left(1-x-y-{x}^{2}+{y}^{2}\right)}^{2}}$ and $\frac{\partial z}{\partial y}=\frac{1-2y}{{\left(1-x-y-{x}^{2}+{y}^{2}\right)}^{2}}$.
Choice (e) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{1+2x}{{\left(1-x-y-{x}^{2}+{y}^{2}\right)}^{2}}$ and $\frac{\partial z}{\partial y}=\frac{1-2y}{{\left(1-x-y-{x}^{2}+{y}^{2}\right)}^{2}}$.
You are given that $z=ln\left(1+\sqrt{1+x+y}\right)$. Use differentials to find the approximate change in $z$ when $\left(x,y\right)$ changes from $\left(1,7\right)$ to $\left(0.9,6.9\right)$. Exactly one option must be correct)
 a) An increase of approximately $\frac{1}{12}$ b) A decrease of approximately $\frac{1}{24}$ c) An increase of approximately $\frac{1}{120}$ d) An increase of approximately $\frac{1}{24}$ e) A decrease of approximately $\frac{1}{120}$

Choice (a) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{1+x+y}\left(1+\sqrt{1+x+y}\right)}$.
Choice (b) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{1+x+y}\left(1+\sqrt{1+x+y}\right)}$.
Choice (c) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{1+x+y}\left(1+\sqrt{1+x+y}\right)}$.
Choice (d) is incorrect
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{1+x+y}\left(1+\sqrt{1+x+y}\right)}$.
Choice (e) is correct!
The first partial derivatives are $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{1+x+y}\left(1+\sqrt{1+x+y}\right)}$. They equal $\frac{1}{24}$ at $\left(1,7\right)$, and so $dz=\frac{1}{24}\left(dx+dy\right)=\frac{-0.2}{24}=-\frac{1}{120}$. Thus $z$ has decreased by approximately $\frac{1}{120}$.
If $z=sinxcosy,\phantom{\rule{1em}{0ex}}x=\pi t$ and $y=\sqrt{t}$ then $\frac{dz}{dt}$ is Exactly one option must be correct)
 a) $\pi cos\left(\pi t\right)cos\left(\sqrt{t}\right)-\frac{1}{2\sqrt{t}}sin\left(\pi t\right)sin\left(\sqrt{t}\right)$ b) $-\frac{1}{2\sqrt{t}}cos\left(\pi t\right)cos\left(\sqrt{t}\right)+\pi sin\left(\pi t\right)sin\left(\sqrt{t}\right)$ c) $\pi sin\left(\pi t\right)cos\left(\sqrt{t}\right)-\frac{1}{2\sqrt{t}}cos\left(\pi t\right)sin\left(\sqrt{t}\right)$ d) $\pi -\frac{1}{2\sqrt{t}}$ e) $\pi sinxcosy-\frac{1}{2\sqrt{t}}cosxsiny$

Choice (a) is correct!
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
If $z=lny\left(2x+y\right),\phantom{\rule{1em}{0ex}}x=sint$ and $y=cost$ then $\frac{dz}{dt}$ is Exactly one option must be correct)
 a) $-\frac{2cost}{2x+y}+\frac{2sint\left(x+y\right)}{y\left(2x+y\right)}$ b) $-\frac{2{y}^{2}cost}{2x+y}-\frac{2sint\left(x+y\right)}{y\left(2x+y\right)}$ c) $\frac{2\left({y}^{2}-{x}^{2}-xy\right)}{y\left(2x+y\right)}$ d) $\frac{2cost}{2x+y}-\frac{sint}{2x+y}$ e) None of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
$z=lny\left(2x+y\right)$ is easier to differentiate when it is rewritten as $z=lny+ln\left(2x+y\right)$ using log laws. Here $\frac{dx}{dt}=cost=y$ and $\frac{dy}{dt}=-sint=-x$ which makes it easy to write $\frac{dz}{dt}$ in terms of $x$ and $y$ only.
Choice (d) is incorrect
Choice (e) is incorrect
Let $z=x{y}^{2}+{x}^{3}y$ and let $x$ and $y$ be functions of $t$ with $x\left(1\right)=1$ , $y\left(1\right)=2$ , ${x}^{\prime }\left(1\right)=3$ and ${y}^{\prime }\left(1\right)=4$. The value of $\frac{dz}{dt}$ when $t=1$ is

Correct!
We have
$\frac{dz}{dt}=\frac{\partial z}{\partial x}\phantom{\rule{0.3em}{0ex}}\frac{dx}{dt}+\frac{\partial z}{\partial y}\phantom{\rule{0.3em}{0ex}}\frac{dy}{dt}.$

Now, $\frac{\partial z}{\partial x}={y}^{2}+3{x}^{2}y$ and $\frac{\partial z}{\partial y}=2xy+{x}^{3}$, so $\frac{\partial z}{\partial x}=10$ and $\frac{\partial z}{\partial y}=5$ when $t=1$ . Therefore, when $t=0$ we have $\frac{dz}{dt}=10\cdot 3+5\cdot 4=50.$

Recall that $\frac{dz}{dt}=\frac{\partial z}{\partial x}{x}^{\prime }\left(t\right)+\frac{\partial z}{\partial y}{y}^{\prime }\left(t\right)$.
Let $z=\frac{x}{y}$, $x=s{e}^{t},$ $y=1+s{e}^{-t}$. Which of the following alternatives are equal to $\frac{\partial z}{\partial t}$ ? (Zero or more options can be correct)
 a) $\frac{s{e}^{t}x}{y}\left(1-\frac{1}{y}\right)$ b) $\frac{x}{y}+\frac{x\left(1+y\right)}{{y}^{2}}$ c) $\frac{x}{y}-\frac{x\left(1-y\right)}{{y}^{2}}$ d) $\frac{s{e}^{t}-2{s}^{2}}{{\left(1+s{e}^{-t}\right)}^{2}}$ e) $\frac{s{e}^{t}+2{s}^{2}}{{\left(1+s{e}^{-t}\right)}^{2}}$

There is at least one mistake.
For example, choice (a) should be False.
There is at least one mistake.
For example, choice (b) should be False.
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
There is at least one mistake.
For example, choice (e) should be True.
Since $\frac{\partial x}{\partial t}=s{e}^{t}=x$ and $\frac{\partial y}{\partial t}=-s{e}^{-t}=1-y$ it is possible to write $\frac{\partial z}{\partial t}$ in terms of $x$ and $y$ only as well as $s$ and $t$ only.
Correct!
1. False
2. False
3. True
4. False
5. True Since $\frac{\partial x}{\partial t}=s{e}^{t}=x$ and $\frac{\partial y}{\partial t}=-s{e}^{-t}=1-y$ it is possible to write $\frac{\partial z}{\partial t}$ in terms of $x$ and $y$ only as well as $s$ and $t$ only.
Let $z={e}^{x}siny$ and let $x$ and $y$ be functions of $s$ and $t$ with $x\left(0,0\right)=0$, $y\left(0,0\right)=2\pi$, $\frac{\partial x}{\partial s}=3$ and $\frac{\partial y}{\partial s}=4$ at $s=t=0$. What is the value of $\frac{\partial z}{\partial s}$ at $s=t=0$.

Correct!
We have that

$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\phantom{\rule{0.3em}{0ex}}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\phantom{\rule{0.3em}{0ex}}\frac{\partial y}{\partial s}={e}^{x}siny\frac{\partial x}{\partial s}+{e}^{x}cosy\frac{\partial y}{\partial s}.$

So when $\left(s,t\right)=\left(0,0\right)$ we find

$\frac{\partial z}{\partial s}{\left|\right}_{\left(s,t\right)=\left(0,0\right)}={e}^{0}\cdot 0\cdot 3+{e}^{0}\cdot 1\cdot 4=0$

Recall that $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\phantom{\rule{0.3em}{0ex}}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\phantom{\rule{0.3em}{0ex}}\frac{\partial y}{\partial s}$
The point $3,1$ lies on the level curve $f\left(x,y\right)=10$ of the function $f\left(x,y\right)$. The partial derivatives at this point are ${f}_{x}\left(3,1\right)=2$ and ${f}_{y}\left(3,1\right)=-5$. What is the slope of the tangent line to the level curve at the point $\left(3,1\right)$ ? Exactly one option must be correct)
 a) $-2$ b) $\frac{2}{5}$ c) $-\frac{1}{2}$ d) $5$ e) There is not enough information to be able to decide.

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
A function $y=f\left(x\right)$ is defined implicitly by $xcosy+ycosx=1$. Find $\frac{dy}{dx}$. Exactly one option must be correct)
 a) $\frac{cosy-ysinx}{cosx-xsiny}$ b) $\frac{ysinx-cosy}{-xsiny+cosx}$ c) $\frac{1-cosy+ysinx}{cosx-xsiny}$ d) $\frac{1-ysinx+cosy}{cosx-xsiny}$ e) $\frac{dy}{dx}$ is not defined because $cosx-xcosy$ can be zero.

Choice (a) is incorrect
Choice (b) is correct!
Let $g\left(x,y\right)=xcosy+ycosx-1=0$ then

$\frac{dy}{dx}=-\frac{{g}_{x}}{{g}_{y}}=-\frac{cosy-ysinx}{cosx-xsiny}.$
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
A function $y=f\left(x\right)$ is defined implicitly by $ln\left(x+y\right)=1+x{y}^{2}$. Find $\frac{dy}{dx}$.
Exactly one option must be correct)
 a) $\frac{dy}{dx}=\frac{1+x{y}^{2}}{1-2{x}^{2}y-2x{y}^{2}}$ b) $\frac{dy}{dx}=\frac{1-x{y}^{2}-{y}^{2}}{1+2{x}^{2}y-2x{y}^{2}}$ c) $\frac{dy}{dx}=\frac{1-x{y}^{2}-2{y}^{3}}{1+x{y}^{2}+2{y}^{3}}$ d) $\frac{dy}{dx}=-\frac{1-x{y}^{2}-{y}^{3}}{1-2{x}^{2}y-2x{y}^{2}}$ e) None of the above

Choice (a) is incorrect
Hint: The partial derivatives of the function $g\left(x,y\right)=ln\left(x+y\right)-1-x{y}^{2}$ are ${g}_{x}\left(x,y\right)=\frac{1}{x+y}-{y}^{2}$ and ${g}_{y}\left(x,y\right)=\frac{1}{x+y}-2xy$.
Choice (b) is incorrect
Hint: The partial derivatives of the function $g\left(x,y\right)=ln\left(x+y\right)-1-x{y}^{2}$ are ${g}_{x}\left(x,y\right)=\frac{1}{x+y}-{y}^{2}$ and ${g}_{y}\left(x,y\right)=\frac{1}{x+y}-2xy$.
Choice (c) is incorrect
Hint: The partial derivatives of the function $g\left(x,y\right)=ln\left(x+y\right)-1-x{y}^{2}$ are ${g}_{x}\left(x,y\right)=\frac{1}{x+y}-{y}^{2}$ and ${g}_{y}\left(x,y\right)=\frac{1}{x+y}-2xy$.
Choice (d) is correct!
Since the partial derivatives of the function $g\left(x,y\right)=ln\left(x+y\right)-1-x{y}^{2}$ are ${g}_{x}\left(x,y\right)=\frac{1}{x+y}-{y}^{2}$ and ${g}_{y}\left(x,y\right)=\frac{1}{x+y}-2xy$, we have $\frac{dy}{dx}=-\frac{{g}_{x}}{{g}_{y}}=-\frac{1-x{y}^{2}-{y}^{3}}{1-2{x}^{2}y-2x{y}^{2}}$.
Choice (e) is incorrect
Hint: The partial derivatives of the function $g\left(x,y\right)=ln\left(x+y\right)-1-x{y}^{2}$ are ${g}_{x}\left(x,y\right)=\frac{1}{x+y}-{y}^{2}$ and ${g}_{y}\left(x,y\right)=\frac{1}{x+y}-2xy$.