- #1
a_skier
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oops I meant "Rate of change of area of a square with respect to its side length"
Ok I have to use this annoying Stewart textbook for my Calc class in college. Most of the questions require what I like to call "Monkey Math," where you just memorize a set of steps and then follow them rigidly for each and every problem.
However, this problem I found has me really thinking, here it is:
Show that the rate of change of the area of a square with respect to its side length is half its perimeter. Try to explain geometrically why this is true.
So I can easily "show" why this is true.
A(x)=x2
A'(x)=2x
P(x)=4x
so (A'(x))/P(x)=1/2
But I am drawing this out on my white board and I can't conceptually understand why the rate of change of the area would be 1/2 the perimeter at the exact same moment...
I can understand why this is true algebraically, but I guess I can't visualize what the derivative of a geometric shape's area is.
Can anyone shed some light on this for me?
Ok I have to use this annoying Stewart textbook for my Calc class in college. Most of the questions require what I like to call "Monkey Math," where you just memorize a set of steps and then follow them rigidly for each and every problem.
However, this problem I found has me really thinking, here it is:
Show that the rate of change of the area of a square with respect to its side length is half its perimeter. Try to explain geometrically why this is true.
So I can easily "show" why this is true.
A(x)=x2
A'(x)=2x
P(x)=4x
so (A'(x))/P(x)=1/2
But I am drawing this out on my white board and I can't conceptually understand why the rate of change of the area would be 1/2 the perimeter at the exact same moment...
I can understand why this is true algebraically, but I guess I can't visualize what the derivative of a geometric shape's area is.
Can anyone shed some light on this for me?
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