The length of a rectangle is given by 5t+4sqrt(t), and its height is sqrt(t), where t is time in seconds and the dimensions are in centimeters. fine the rate of change of the area of the rectangle with respect to time when the length of the rectangle is 9 centimeters.

11.5 cm²/s

Step-by-step explanation:

We must solve two problems to answer this question:

For the first problem, we have ...

... 9 = 5t +4√t

... 9 -5t = 4√t . . . . . subtract 5t

... 81 -90t +25t² = 16t . . . . square both sides

... 25t² -106t +81 = 0 . . . subtract 16t

... (t -1)(25t -81) = 0 . . . . factor

... t = 1 or 3.24 . . . . . . t=3.24 is an extraneous solution

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For the second problem, we have area (a(t)) is ...

... a(t) = length×height = (5t +4√t)(√t) = 5t^(3/2) +4t

Then the derivative is ...

... a'(t) = (3/2)(5√t) +4

and

... a'(1) = (3/2)(5√1) +4 = 11.5 . . . . . cm²/s

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Comment on the attachment

A graphing calculator can help solve both problems. It can help find the time at which length is 9 cm by solving ... length-9 = 0. (The nice thing here is that there is no extraneous solution.) It can compute the derivative of a function, too.

36

step-by-step explanation: