## MATH1901 Quizzes

Quiz 11: Chain rules for functions of two variables
Question 1 Questions
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 $z=\frac{x}{y}$, $x=s{e}^{t}$ and $y=1+s{e}^{-t}$. 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 $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\phantom{\rule{0.3em}{0ex}}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\phantom{\rule{0.3em}{0ex}}\frac{\partial y}{\partial s}$
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
( Harder ) A function $z=f\left(x,y\right)$ is defined implicitly by $ln\left(x+yz\right)=1+x{y}^{2}{z}^{3}$. Find $\frac{\partial z}{\partial y}$.
Exactly one option must be correct)
 a) $\frac{-\left(1-{y}^{2}{z}^{3}\left(x+yz\right)\right)}{y-3x{y}^{2}{z}^{2}\left(x+yz\right)}$ b) $\frac{-\left(y-2xy{z}^{3}\left(x+yz\right)\right)}{z-3x{y}^{2}{z}^{2}\left(x+yz\right)}$ c) $\frac{y-3x{y}^{2}{z}^{2}\left(x+yz\right)}{z-2xy{z}^{3}\left(x+yz\right)}$ d) $\frac{-\left(z-2xy{z}^{3}\left(x+yz\right)\right)}{y-3x{y}^{2}{z}^{2}\left(x+yz\right)}$ e) $\frac{z-2xy{z}^{3}\left(x+yz\right)}{y-3x{y}^{2}{z}^{2}\left(x+yz\right)}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect