Related Rates Solution 2

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SOLUTION 2: Draw a rectangle with length

$x$ and width

$y$ , and assume each edge of the rectangle is a function of time

$t$ .

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$\ \ \ \$ a.) The perimeter of the rectangle is

$P=x+x+y+y \ \ \ \ \longrightarrow$

$P = 2x+2y$ GIVEN:

$\ \ \ \displaystyle{ dx \over dt } = 4 \ ft/min. \$ and

$\ \displaystyle{ dy \over dt } = -3 \ ft/min.$

FIND:

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

$x=8 \ ft.$ and

$y=5 \ ft.$

Now differentiate the perimeter equation with respect to time

$t$ getting

$D \{ P \} = D \{2x+2y \} \ \ \ \longrightarrow$

$\displaystyle{ dP \over dt } = 2 \displaystyle{ dx \over dt } + 2 \displaystyle{ dy \over dt } \ \ \ \longrightarrow$

$\Big($ Now let

$\displaystyle{ dx \over dt } = 4$ and

$\ \displaystyle{ dy \over dt } = -3. \Big)$

$\displaystyle{ dP \over dt } = 2 (4) + 2 (-3 ) = 2 \ ft/min.$

$\ \ \ \$ b.) The area of the rectangle is

$A= (length)(width) \ \ \ \ \longrightarrow$

$A= xy$ GIVEN:

$\ \ \ \displaystyle{ dx \over dt } = 4 \ ft/min.$ and

$\displaystyle{ dy \over dt } = - 3 \ ft/min.$

FIND:

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

$x=8 \ ft.$ and

$y=5 \ ft.$

Now differentiate the area equation with repect to time

$t$ using the product rule getting

$D \{ A \} = D \{xy \} \ \ \ \longrightarrow$

$\displaystyle{ dA \over dt } = x \displaystyle{ dy \over dt } + \displaystyle{ dx \over dt } y \ \ \ \longrightarrow$

$\Big($ Now let

$\displaystyle{ dx \over dt } = 4$ ,

$\ \displaystyle{ dy \over dt } = -3 , x=8, \$ and

$y=5. \Big)$

$\displaystyle{ dA \over dt } = (8)(-3) + (4)(5) = -4 \ ft^2/min.$

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