Respuesta :
Question: We want to know the value of the rate of change of the perimiter of the square:
[tex]\frac{dP}{dt}=\text{?}[/tex]1) We know that the perimiter of a square is:
[tex]P=4L[/tex]We find the rate of change of the perimiter (dL/dt) taking its derivative:
[tex]\frac{dP}{dt}=4\frac{dL}{dt}[/tex]From the last equation we see that we need the rate of change of the side (dL/dt) in order to calculate the rate of change of the perimeter (dP/dt).
2) Now, according to the question the area of the square is changing at a rate:
[tex]\frac{dA}{dt}=-89\frac{m^2}{\min}[/tex]We know that the area of square is:
[tex]A=L^2[/tex]We can calculate the rate of change of the side (dL/dt) taking the derivative of the area:
[tex]\frac{dA}{dt}=2L\cdot\frac{dL}{dt}[/tex]Now, taking the data of the rate of change of the area and the last formula we see that:
[tex]\begin{gathered} 2L\cdot\frac{dL}{dt}=-89\frac{m^2}{\min} \\ \frac{dL}{dt}=-\frac{89}{2L}\frac{m^2}{\min} \end{gathered}[/tex]3) Replacing the value of the rate of change of the side (dL/dt) in the formula of the rate of change of the perimeter (dP/dt) we find that:
[tex]\frac{dP}{dt}=4\frac{dL}{dt}=-4\cdot\frac{89}{2L}\cdot\frac{m^2}{\min}[/tex]Finally, replacing the value of the side L = 7 m we find that te rate of change of the perimeter is:
[tex]\frac{dP}{dt}=-4\cdot\frac{89}{2\cdot7m}\cdot\frac{m^2}{\min}\cong-25.429\frac{m^{}}{\min}[/tex]So the perimeter of the square is decreasing at a rate of 25.429 m/min.
