Question

1. (4 points; 2 points each) solve the related rates problem. choose 2.

a. for ( x = x(t) > 0 ) and ( y = y(t) > 0 ), if ( x^{2}+y^{2}=25 ) and ( \frac{d x}{d t}=6 ), find ( \frac{d y}{d t} ) when ( x = 3 ).
b. for ( x = x(t) > 0 ) and ( y = y(t) > 0 ), if ( x y = 10 ) and ( \frac{d x}{d t}=3 ), find ( \frac{d y}{d t} ) when ( x = 2 ).
c. water is being pumped into a vertical cylinder of radius 5 meters and height 20 meters at a rate of ( 3 mathrm{~m}^{3} / mathrm{min} ). how fast is the water level rising when the cylinder is half full?

Explanation:

Step1: Differentiate the equation with respect to \(t\)

For \(x^{2}+y^{2}=25\), using the chain - rule \(\frac{d}{dt}(x^{2}+y^{2})=\frac{d}{dt}(25)\).
By the chain - rule \(\frac{d}{dt}(x^{2}) = 2x\frac{dx}{dt}\) and \(\frac{d}{dt}(y^{2})=2y\frac{dy}{dt}\), and \(\frac{d}{dt}(25) = 0\). So, \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\), which simplifies to \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\).

Step2: Find the value of \(y\) when \(x = 3\)

Substitute \(x = 3\) into \(x^{2}+y^{2}=25\). We get \(3^{2}+y^{2}=25\), so \(y^{2}=25 - 9=16\). Since \(y>0\), \(y = 4\).

Step3: Substitute the known values into the differentiated equation

We know that \(x = 3\), \(\frac{dx}{dt}=6\), and \(y = 4\). Substitute these into \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\).
We have \(3\times6+4\times\frac{dy}{dt}=0\).
\(18 + 4\frac{dy}{dt}=0\).

Step4: Solve for \(\frac{dy}{dt}\)

Subtract 18 from both sides: \(4\frac{dy}{dt}=-18\).
Divide both sides by 4: \(\frac{dy}{dt}=-\frac{18}{4}=-\frac{9}{2}\).

Answer:

\(\frac{dy}{dt}=-\frac{9}{2}\)