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An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
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@terenzreignz
okay... if we let x be the length of one edge of the cube, we get that the volume of the cube is equal to... \[\huge V = x^3\] Now... we're looking for \[\huge \left.\frac{dV}{dt}\right.\]
Differentiating seems simplest... \[\huge \frac{dV}{dt}=\frac{dV}{dx}\times \frac{dx}{dt}\] And we do have a value for the dx/dt... it's just 3. So once this is done, plug in x = 10, and you're all set.
Thankyou sir
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