Answer:
[tex]3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex] square inches per seconds.
Step-by-step explanation:
given that a rectangle is growing such that the length of a rectangle is 5t+4 and its height is √t, where t is time in seconds and the dimensions are in inches
Area of rectangle = length * width
A = [tex]\sqrt{t} (5t+4)\\= 5t\sqrt{t} +4\sqrt{t}[/tex]
To find rate of change of A with respect to t, we can find derivative of A with respect to t.
WE get
[tex]\frac{dA}{dt} =5(\frac{3}{2} )\sqrt{t} +\frac{4}{2\sqrt{t} } \\= 3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex]
Hence rate of change of area with respect to time is
[tex]3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex] square inches per seconds.
