Rate of change: area and perimeter
The side of rectangle $x = 20m$ increases at the rate of $5m/s$, the other side $y=30m$ decreases at $4m/sec$. What is the rate of change o the perimeter and area of the retangle?
If we put this rectangle in the cartesian plan (starting on the origin) , we can calculate it's perimeter and area by $P = 2x + 2y$ and $A=x.y$(maybe I should use $\triangle x$ and $\triangle y$ instead of x and y ). Since my perimeter is lnearly increasing in x and y, we can find $\frac{dP}{dt} = \frac{\partial P}{\partial x} \frac{\partial x}{ \partial t} + \frac{ \partial P }{\partial y} \frac{\partial y}{\partial t} $ = $2 m/sec$
I couldn't do the same with my area (i don't even know if with I did with the perimeter is correct).
Thanks for your help!
$\endgroup$1 Answer
$\begingroup$We have $\displaystyle\frac{dx}{dt}=5$ and $\displaystyle\frac{dy}{dt}=-4$
Now the rate of change of area $$=\frac{d(xy)}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}$$
Now the rate of change of perimeter $$=\frac{d\{2(x+y)\}}{dt}=\cdots$$
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