Question

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!

Answer 1

`(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` represent the area of the rectangle as a function of time `t`, we get `A(t)=sqrt(2t)(2t^2+3t-7)`.
`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` and get
`(dA)/dt==5sqrt(2)t^1.5+4.5sqrt(2)t^0.5-3.5sqrt(2)t^-0.5`.