3 years ago

Please Answer! - Image attached Write the inequality shown by the graph below.

Mathematics

1 answer:

MaRussiya [10]3 years ago

7 0

Answer:

Step-by-step explanation:

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Given the relationship x2 + 3y2 =12, with y > 0 and dx, dt = 2 units/min., find the value of dy, dt at the instant y = 1 unit

matrenka [14]

$\bf x^2+3y^2=12\implies \stackrel{chain~rule}{2x\cfrac{dx}{dt}}+3 \stackrel{chain~rule}{\left(2y\cfrac{dy}{dt}\right)}=0
\\\\\\
2x\cfrac{dx}{dt}+6y\cfrac{dy}{dt}=0
\implies 
6y\cfrac{dy}{dt}=-2x\cfrac{dx}{dt}\implies \cfrac{dy}{dt}=\cfrac{-2x\frac{dx}{dt}}{
6y\frac{dy}{dt}}
\\\\\\
\boxed{\cfrac{dy}{dt}=-\cfrac{x\frac{dx}{dt}}{3y}}\\\\
-------------------------------$

$\bf \textit{now, when y = 1, what is \underline{x}?}\qquad x^2+3y^2=12\implies x^2+3(1)^2=12
\\\\\\
x^2=9\implies x=\sqrt{9}\implies x=3\\\\
-------------------------------\\\\
\begin{cases}
y=1\\
x=3\\
\frac{dx}{dt}=2
\end{cases}\implies \cfrac{dy}{dt}=-\cfrac{3\cdot 2}{3\cdot 1}\implies \cfrac{dy}{dt}=-2$

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