Answer
(a)
In third and fourth quadrants, $\dfrac{dx}{dt}>0$.
(b)
$18x\dfrac{dx}{dt}+32y\dfrac{dy}{dt}=0$
(c)
The x-coordinate is changing at a rate of $\dfrac{-32}{3}$m/s.
(d)
Both at the top and the bottom of the ellipse $\dfrac{dy}{dt}=0$.
Work Step by Step
(a)
If $\dfrac{dx}{dt}>0$, then $x$ is increasing with respect to time $t$ in those intervals.
Since, $x$ is increasing in the third and fourth quadrants.
$\dfrac{dx}{dt}>0$ in the third and fourth quadrants.
(b)
Differentiate $9x^2+16y^2=25$ with respect $t$ using product rule and chain rule.
We get, $18x\dfrac{dx}{dt}+32y\dfrac{dy}{dt}=0$
Which is the required relation.
(c)
Since particle passing through the point $(1,1)$.
$x=1$ and $y=1$
Substitute $\dfrac{dy}{dt}=6$m/s, $x=1$ and $y=1$ in $18x\dfrac{dx}{dt}+32y\dfrac{dy}{dt}=0$.
$18\dfrac{dx}{dt}+32\cdot6=0$
$\implies \dfrac{dx}{dt}=\dfrac{-32}{3}$
So, the x-coordinate is changing at a rate of $\dfrac{-32}{3}$m/s.
(d)
At top, $y=\dfrac{5}{4}$ and $x=0$.
Substitute the values in $18x\dfrac{dx}{dt}+32y\dfrac{dy}{dt}=0$.
We get, $32\cdot\dfrac{5}{4}\dfrac{dy}{dt}=0$
$\implies \dfrac{dy}{dt}=0$
At bottom, $y=\dfrac{-5}{4}$ and $x=0$.
Substitute the values in $18x\dfrac{dx}{dt}+32y\dfrac{dy}{dt}=0$.
We get, $32\cdot\dfrac{-5}{4}\dfrac{dy}{dt}=0$
$\implies \dfrac{dy}{dt}=0$
