Someone help me with this please... The length of a rectangle is given by 2t^2 + 3t – 7 and the height is sq. root of 2t, where t is time in seconds and the dimensions are in feet. Find the rate of change of the area with respect to time? Thank you!

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Someone help me with this please... The length of a rectangle is given by 2t^2 + 3t – 7 and the height is sq. root of 2t, where t is time in seconds and the dimensions are in feet. Find the rate of change of the area with respect to time? Thank you!

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Answer 1 · 2018-06-19T15:58:05

$\frac{d A}{d t} = 5 \sqrt{2} t^{1.5} + 4.5 \sqrt{2} t^{0.5} - 3.5 \sqrt{2} t^{- 0.5}$ $(dA)/dt=5sqrt(2)t^1.5+4.5sqrt(2)t^0.5-3.5sqrt(2)t^-0.5$

Explanation:

The area of a rectangle is defined as the product of its length and height. Letting $A$ $A$ represent the area of the rectangle as a function of time $t$ $t$ , we get $A (t) = \sqrt{2 t} (2 t^{2} + 3 t - 7)$ $A(t)=sqrt(2t)(2t^2+3t-7)$ .
$A (t) = 2 \sqrt{2} t^{2.5} + 3 \sqrt{2} t^{1.5} - 7 \sqrt{2} t^{0.5}$ $A(t)=2sqrt(2)t^2.5+3sqrt(2)t^1.5-7sqrt(2)t^0.5$ .
To find the rate of change of the area with respect to time, we differentiate both sides with respect to $t$ $t$ and get
$\frac{d A}{d t} = = 5 \sqrt{2} t^{1.5} + 4.5 \sqrt{2} t^{0.5} - 3.5 \sqrt{2} t^{- 0.5}$ $(dA)/dt==5sqrt(2)t^1.5+4.5sqrt(2)t^0.5-3.5sqrt(2)t^-0.5$ .

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Someone help me with this please... The length of a rectangle is given by 2t^2 + 3t – 7 and the height is sq. root of 2t, where t is time in seconds and the dimensions are in feet. Find the rate of change of the area with respect to time? Thank you!

1 Answer

Explanation:

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