A spherical balloon is inflated so that its volume is increasing at the rate of 3.7 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.8 feet? The diameter is increasing at______ft/min.
What is the formula for volume of a sphere? From here you take the derivative of it with respect to time. Try it out, I'll help you.
kainui such a good boyy
yup
ok
The idea is that the volume and radius are both changing. So that means you can take the rate of change of the volume and radius right? In fact, if we look at the formula for volume of a sphere it looks like this: \[\Large V=\frac{4}{3} \pi r^3\] So what does the question want us to find? How the diameter is changing. Well if we plug into this formula, we know that diamter is just twice the radius, so r=d/2 \[\Large V = \frac{4}{3} \pi (\frac{d}{2})^3 \\ \large V= \frac{\pi}{6} d^3 \] How do we find out how much each thing is changing over time? Take the derivative with respect to time. This is an implicit derivative for both V and d. So try to take the implicit derivative here. Looking at our problem, we have been given some things that we should try to understand as more mathematical terms. \[\large 3.7 \frac{ft^3}{\min}=\frac{dV}{dt}\] See how the units match up? -- Look this might not make sense to you, but this is where people start to actually understand calculus and I can help you understand it. What's important is that you try to understand and take the first step by trying to take the derivative of that volume function implicitly.
i follow in what you say, the other thing is that when i type 3.7 im getting it wrong
3.7 is not the answer. It's part of the question.
ohh ok
Until you attempt to take the derivative of that function, I will not offer any more help.
ohh my bad i didnt see that, ill do it right now
i get (rd^2)/2
so from there i just plug in?
A solution using Mathematica 9 Home Edition and the Total Derivative is attached.
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