Question

b. for ( x = x(t)>0 ) and ( y = y(t)>0 ), if ( xy = 10 ) and ( \frac{dx}{dt}=3 ), find ( \frac{dy}{dt} ) 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 xy = 10 with respect to t

Using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x\) and \(v = y\).
We get \(x\frac{dy}{dt}+y\frac{dx}{dt}=0\).

Step2: Find the value of y when x = 2

Since \(xy = 10\) and \(x = 2\), then \(y=\frac{10}{x}=\frac{10}{2} = 5\).

Step3: Substitute the known values into the differentiated equation

We know that \(\frac{dx}{dt}=3\), \(x = 2\), and \(y = 5\).
Substitute into \(x\frac{dy}{dt}+y\frac{dx}{dt}=0\), so \(2\frac{dy}{dt}+5\times3 = 0\).

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

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(for part c):

Step1: Recall the volume formula for a cylinder

The volume of a cylinder \(V=\pi r^{2}h\), where \(r\) is the radius and \(h\) is the height of the water - level. Here \(r = 5\) (constant), so \(V = 25\pi h\).

Step2: Differentiate V with respect to t

Using the chain - rule \(\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}\). Since \(\frac{dV}{dt}=3\) (given) and \(\frac{dV}{dh}=25\pi\) (derivative of \(V = 25\pi h\) with respect to \(h\)).

Step3: Solve for \(\frac{dh}{dt}\)

From \(\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}\), we have \(\frac{dh}{dt}=\frac{\frac{dV}{dt}}{\frac{dV}{dh}}\).
Substitute \(\frac{dV}{dt}=3\) and \(\frac{dV}{dh}=25\pi\) into the formula, \(\frac{dh}{dt}=\frac{3}{25\pi}\approx\frac{3}{25\times3.14}=\frac{3}{78.5}\approx0.038\) m/min.

Answer:

\(\frac{dy}{dt}=-7.5\)