Related Rates Solution 8

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SOLUTION 8: Draw a closed right circular cylinder of base radius

$r$ and height

$h$ , and assume that

$r$ and

$h$ are functions of time

$t$ .

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$\ \ \ \$ a.) The surface area of a closed right circular cylinder of radius

$r$ is the sum of the areas of two circles and a rectangle:

$S= \pi r^2 + \pi r^2 + (base)(height) \ \ \ \ \longrightarrow$

$S= 2\pi r^2 + (circumference \ of \ circle)(height) \ \ \ \ \longrightarrow$

$S= 2\pi r^2 + (2\pi r)(h) \ \ \ \ \longrightarrow$

$S= 2\pi r^2 + 2\pi r h$

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GIVEN:

$\ \ \ \displaystyle{ dr \over dt } = 5 \ cm/hr.$ and

$\displaystyle{ dh \over dt } = -4 \ cm/hr.$

FIND:

$\ \ \ \displaystyle{ dS \over dt }$ when

$r=20 \ cm.$ and

$h=12 \ cm.$

Now differentiate the surface area equation with respect to time

$t$ using the product rule, getting

$D \{ S \} = D \{ 2\pi r^2 + 2\pi rh \} \ \ \ \longrightarrow$

$\displaystyle{ dS \over dt } = 2 \pi \cdot 2r \displaystyle{ dr \over dt } + ( 2 \pi r \cdot \displaystyle{ dh \over dt } + 2 \pi \displaystyle{ dr \over dt } \cdot h ) \ \ \ \longrightarrow$

$\Big($ Now let

$\displaystyle{ dr \over dt } = 5, \displaystyle{ dh \over dt } = -4, r=20,$ and

$h =12. \Big)$

$\displaystyle{ dS \over dt } = 4 \pi (20) (5) + 2 \pi (20)(-4) + 2 \pi (5) (12) \ \ \ \longrightarrow$

$\displaystyle{ dS \over dt } = 400 \pi -160 \pi + 120 \pi \ \ \ \longrightarrow$

$\displaystyle{ dS \over dt } = 360 \pi \ cm^2/hr. \approx 1130.97 \ cm^2/hr.$

$\ \ \ \$ b.) The volume of a closed right circular cylinder of radius

$r$ is

$V = (area \ of \ circular \ base)(height) \ \ \ \ \longrightarrow$

$V = \pi r^2 h$ GIVEN:

$\ \ \ \displaystyle{ dr \over dt } = 5 \ cm/hr.$ and

$\displaystyle{ dh \over dt } = -4 \ cm/hr.$

FIND:

$\ \ \ \displaystyle{ dV \over dt }$ when

$r=20 \ cm.$ and

$h=12 \ cm.$

Now differentiate the volume equation with respect to time

$t$ using the product rule, getting

$D \{ V \} = D \{ \pi r^2 h \} \ \ \ \longrightarrow$

$\displaystyle{ dV \over dt } = \pi r^2 \cdot \displaystyle{ dh \over dt } + 2\pi r \displaystyle{ dr \over dt } \cdot h \ \ \ \longrightarrow$

$\Big($ Now let

$\displaystyle{ dr \over dt } = 5, \displaystyle{ dh \over dt } = -4, r=20,$ and

$h =12. \Big)$

$\displaystyle{ dV \over dt } = \pi (20)^2 (-4) + 2 \pi (20) (5)(12) \ \ \ \longrightarrow$

$\displaystyle{ dV \over dt } = -1600 \pi + 2400 \pi \ \ \ \longrightarrow$

$\displaystyle{ dV \over dt } = 800 \pi \ cm^3/hr. \approx 2513.27 \ cm^3/hr.$

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