The rate at which the area of the rectangle changing at the instant is 25in²/s
If a rectangle is inscribed in a circle with a radius of 5inches, the length of diagonal of the rectangle will be 10inches
Given the length to be 6 inches
Get the width using the Pythagoras theorem;
[tex]d^2=l^2+w^2\\10^2=6^2+w^2\\100=36+w^2\\w^2=100-36\\w^2=64\\w=\sqrt{64}\\w=8in[/tex]
Area of the rectangle is expressed as;
[tex]A=lw\\[/tex]
Get the change of rate of the area as shown:
[tex]\frac{dA}{dt}=l\frac{dw}{dt} + w\frac{dl}{dt}[/tex]
Next is to get the rate at which the width is decreasing. Recall that:
[tex]l^2+w^2=100\\2l\frac{dl}{dt}+2w\frac{dw}{dt} =0\\2(6)(-2)+2(8)\frac{dw}{dt} =0\\-24+16\frac{dw}{dt} =0\\-16\frac{dw}{dt}=-24\\\frac{dw}{dt}=\frac{24}{16} \\\frac{dw}{dt}=1.5in/s[/tex]
Get the rate at which the area is increasing:
[tex]\frac{dA}{dt}=l\frac{dw}{dt} + w\frac{dl}{dt}\\\frac{dA}{dt}=6(1.5) + 8(2)\\\frac{dA}{dt}=9+16\\\frac{dA}{dt}=25in^2/s[/tex]
Hence the rate at which the area of the rectangle changing at the instant is 25in²/s
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