#Partial differentiation
hot spadeBOT
storm bridge
can anyone see if they get a different answer, if so can they point me in the right direction from there
final ice
- You can write both $x$ and $y$ in terms of $t$, ultimately you can write $f(x,y)$ in terms of $t$.
Let $f(t)=x(t)\sqrt{x(t)-3[y(t)]^2}$. You can then derive $f$ with respect to $t$.
lament walrusBOT
Crystopher
final ice
$\frac{df}{dt}$ gave me
$$f'(t)=x'(t)\sqrt{x(t)-3[y(t)]^2}-$$
$$\frac{3x(t)x'(t)y(t)y'(t)}{\sqrt{x(t)-3[y(t)]^2}}$$
lament walrusBOT
Crystopher
final ice
If I am not mistaken with the derivation, then set $x(t)=20,x'(t)=-5,y(t)=1,y'(t)=2$
lament walrusBOT
Crystopher
final ice
This gives approximately 124.91
- First order Taylor approximation of $f(\vec{x})$, where $\vec{x}= \begin{pmatrix}x \ y\end{pmatrix}$ is given by:
$$f(\vec{x}) \approx f(\vec{x_0})+ J(\vec{x})\cdot (\vec{x}-\vec{x_0})$$
where $\vec{x_0}=\begin{pmatrix}0.5 \ 0.6\end{pmatrix}$ and $J(\vec{x_0})$ is the Jacobian of $f$ evaluated on $\vec{x_0}$
lament walrusBOT
Crystopher
storm bridge
you've done it differently i see
not sure if its teh sme
hold on let me write out what i did
final ice
sure
storm bridge
thats for the first question
i admit i was wrong on the first one since i forgot the negative sign on the second part
(-30/rt7)
but im not even sure if this is correct
this is what i was taught to do though
final ice
I can try evaluating $\frac{df}{dx}\cdot\frac{dx}{dt} + \frac{df}{dy}\cdot\frac{dy}{dt}$ and see if I get the same. The way I did it seemed the most intuitive to me.
lament walrusBOT
Crystopher
storm bridge
needs to be partial derivative x and y
jtluk
This is my second answer, I actually got this different as well 😂
ill brb just making hotdogs 😎
im back
final ice
I got
$$\frac{df}{dx}=(x-3y^2)^{\frac{1}{2}}+\frac{x}{2}(x-3y^2)^{-\frac{1}{2}}=$$
$$(20-3)^{\frac{1}{2}}+\frac{20}{2}(20-3)^{-\frac{1}{2}}=$$
$$(17)^{\frac{1}{2}}+10 \cdot (17)^{-\frac{1}{2}}=$$
$$\frac{27}{\sqrt{17}}$$
lament walrusBOT
Crystopher
storm bridge
i forgot the -^1/2
i put 1/2
nvm
i didnt
ah i see what i did
i halfed 20 twice
into 5
yh i thouht x=10
for some reason
final ice
$$\frac{df}{dy}=-\frac{6xy(x-3y^2)}{2}-3xy (x-3y^2)^{-\frac{1}{2}}=$$
$$-\frac{3xy}{\sqrt{x-3y^2}}=-\frac{60}{\sqrt{17}}$$
lament walrusBOT
Crystopher
storm bridge
yea, i get those equation
equations*
just subbed in wrong
i think ill try it again and see if its wrnog after eating
thank you
solid vine
.solved