Related Rates Solution 5

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SOLUTION 5: Draw a cube with edge lengths

$x$ , and assume that

$x$ is a function of time

$t$ .

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$\ \ \ \$ a.) The surface area (Add the areas of 6 square surfaces.) of a cube is

$S=x^2+x^2+x^2+x^2+x^2+x^2 \ \ \ \ \longrightarrow$

$S=6x^2$ GIVEN:

$\ \ \ \displaystyle{ dx \over dt } = -2 \ cm/min.$

FIND:

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

$x =80 \ cm.$

Now differentiate the surface area equation with respect to time

$t$ , getting

$D \{ S \} = D \{ 6x^2 \} \ \ \ \longrightarrow$

$\displaystyle{ dS \over dt } = 6 \cdot 2x \displaystyle{ dx \over dt } \ \ \ \longrightarrow$

$\Big($ Now let

$\displaystyle{ dx \over dt } = -2$ and

$x=80. \Big)$

$\displaystyle{ dS \over dt } = 12(80)(-2) \ \ \ \longrightarrow$

$\displaystyle{ dS \over dt } = -1920 \ cm^2/min.$

$\ \ \ \$ b.) The volume of a cube is

$V = (length)(width)(height) \ \ \ \ \longrightarrow$

$V = x^3$ GIVEN:

$\ \ \ \displaystyle{ dx \over dt } = -2 \ cm/min.$

FIND:

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

$x =80 \ cm.$

Now differentiate the volume equation with respect to time

$t$ , getting

$D \{ V \} = D \{ x^3 \} \ \ \ \longrightarrow$

$\displaystyle{ dV \over dt } = 3 x^2 \displaystyle{ dx \over dt } \ \ \ \longrightarrow$

$\Big($ Now let

$\displaystyle{ dx \over dt } = -2$ and

$x=80. \Big)$

$\displaystyle{ dV \over dt } = 3(80)^2(-2) = -38,400 \ \ cm^3/min.$

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