Answer:
A)
[tex]\displaystyle \frac{dy}{dt}=-\frac{33}{8}[/tex]
B)
[tex]\displaystyle \frac{dx}{dt}=\frac{3}{2}[/tex]
Step-by-step explanation:
x and y are differentiable functions of t, and we are given the equation:
[tex]xy=6[/tex]
First, let's differentiate both sides of the equation with respect to t. So:
[tex]\displaystyle \frac{d}{dt}\left[xy\right]=\frac{d}{dt}[6][/tex]
By the Product Rule and rewriting:
[tex]\displaystyle \frac{d}{dt}[x(t)]y+x\frac{d}{dt}[y(t)]=0[/tex]
Therefore:
[tex]\displaystyle y\frac{dx}{dt}+x\frac{dy}{dt}=0[/tex]
A)
We want to find dy/dt when x = 4 and dx/dt = 11.
Using our original equation, find y when x = 4:
[tex]\displaystyle (4)y=6\Rightarrow y=\frac{3}{2}[/tex]
Therefore:
[tex]\displaystyle \frac{3}{2}\left(11\right)+(4)\frac{dy}{dt}=0[/tex]
Solve for dy/dt:
[tex]\displaystyle \frac{dy}{dt}=-\frac{33}{8}[/tex]
B)
We want to find dx/dt when x = 1 and dy/dt = -9.
Again, using our original equation, find y when x = 1:
[tex](1)y=6\Rightarrow y=6[/tex]
Therefore:
[tex]\displaystyle (6)\frac{dx}{dt}+(1)\left(-9)=0[/tex]
Solve for dx/dt:
[tex]\displaystyle \frac{dx}{dt}=\frac{3}{2}[/tex]
