The function V=s^3 describes the volume of a cube, V, in cubic inches, whose length, width, and height each measures s inches. Find the instantaneous rate of change of the volume with respect to s when s = 3 inches.
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So I'm assuming I find the derivative of V=s^3 and then plug 3 in for s? Which would give me 27. correct?
Yes, that is correct. Find the derivative of \(V(s)\) with respect to \(s\) and evaluate it at \(s=3\).
You can check your answer to see if it is in the right ballpark by evaluating \(V(3)\) and \(V(3+0.001)\). What is the slope of the line between \((3,V(3))\) and \((3+0.001,V(3+0.001))\)? It should be very close to your answer. You are approximating the tangent line to the function at \(s=3\) by using a straight line that passes through \(3,V(3))\) and has the slope of the derivative of \(V(3)\). Try the experiment with various values for the second point and see how it behaves as you get closer and closer.