A sphere is losing volume at 6cm^3/minute
How fast is the radius of the sphere decreasing when the surface are of the sphere is 24cm^2?
"The volume of a sphere as a function of its radius is V=4pi/3*r^3. The surface area of a sphere as a function of its radius is SA=4pi*r^2."
This is given as a problem in a calculus chapter containing derivatives. I can't get my initial equation started at all. The more I try to even think of it, I just complicate it more and more.
We are talking radius decreasing so its not the total volume. Otherwise I'd just write 6cm^3/minute.
Please help.
So the wording there really confused me. I've gotten so far out of that, which looks like a fool proof plan. I just don't understand where to substitute.
I can get the radius from the surface area. I get that. And the slope will be affected by the 6cm^3 at which its constantly losing per minute.
I think I'm just not used to seeing those ways of talking about derivatives. If I was a calc pro, I would be done here, but I have to ask for clarifications.
The radius for this is going to be sqrt 6 / sqrt pi... I'm still learning how to write it in math language, but now that I have the radius at this point in time, I believe I should take the derivative of that volume formula. I'm sorry but the more I look at that explanation the more I just want to chalk this one up as a loss.
Yes, too far out! Just in case a picture helps...
... where
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to t, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which subject to the chain rule).
Once you've differentiated with respect to r, you'll see you can just sub two values into these blanks, and virtually read off the value of dr/dt...
__________________________________________
Don't integrate - balloontegrate!
Balloon Calculus: Gallery
Balloon Calculus Drawing with LaTeX and Asymptote!
Wow, I mean, that is some nice stuff, the more I look at it the more I feel like I'm levitating! I feel close to it, but i don't see the step by step approach. I've found the radius. When I substitute the radius into the volume formula, I don't have a variable and I receive a constant value at my given radius of sqrt 6 / sqrt of pi .
Are you saying I should differentiate the volume formula first? That would give me a variable. I've done that and I don't get something that looks like its part of my answer. It looks like a step, but I'm missing a good half of everything else I'm supposed to do.
I'm really trying to get this balloon thing. So if I plug in the pieces of the derivative of that function into the ?'s such as
4= ?1 and 2=?2 (the exponent)... Then I can use the chain rule with the formula for volume and differentiate that all together to get the dr/dt?
Or am I just confusing myself with these balloons. I'm weak at implicit differentiation, and we are approaching related rates of change problems this week. Just so you know where I'm at with comprehension.