# A Derative Question

1. Apr 26, 2004

### stuckie27

Smiple Equation.

What is the derative of 3pi*r^2+ ((2,000,000-(2/3pi*r^3))/r)

Do you use the quotient rule

Last edited: Apr 26, 2004
2. Apr 26, 2004

It'd work, but it's more work than you need to do.

That same expression can be written as

$$\frac{7}{3}\pi r^2 + 2\times 10^6 r^{-1}$$

which can be differentiated easily.

Last edited: Apr 26, 2004
3. Apr 26, 2004

### JonF

Yup you use the quotient rule. This is what I got when I did (warning, im prone to make arithmetic mistakes)

$6\pir + \frac{ r (\frac{18/pir^2}{3/pir^5}) –(2,000,000 – (\frac{2}{3\pir^3}){r^2}$

4. Apr 26, 2004

I agree, which is why you should simplify the expression before differentiating it.

5. Apr 26, 2004

### stuckie27

how did you simplify the equation?

6. Apr 26, 2004

### Ebolamonk3y

sigh... waste of good time on these trivial problems...

7. Apr 26, 2004

Distributed the r^(-1) in the second term and then combined coefficients of like powers of r.

And a trivial problem to you, Ebolamonk3y, may not be trivial to somebody else.

8. Apr 26, 2004

### stuckie27

6\pir + \frac{ r (\frac{18/pir^2}{3/pir^5}) –(2,000,000 – (\frac{2}{3\pir^3}){r^2}

What is this? \frac{ r (\frac{18/pir^2}{3/pir^5}) Its not in my origional equation

9. Apr 26, 2004

use $$........ 10. Apr 26, 2004 ### stuckie27 Any reason for this comment? Is this not a calc help forum? I can write you the full prolbem if you would like to solve the whole thing. Here It is A metal storage tank with a volume of 1000 liters is to be constructed in the shape of a right cylinder surmounted by a hemisphere. What dimentions will require the least ammount of metal? 11. Apr 26, 2004 ### cookiemonster It's supposed to be the expression for the derivative of your original expression. I think. I can't read it well enough to really know, though. cookiemonster 12. Apr 26, 2004 ### cookiemonster Let's not turn this into a flame war, okay? cookiemonster 13. Apr 26, 2004 ### Ebolamonk3y Well... long time ago I proposed this huge problem that I made up for a friend of mine... by the end of it I noticed that the problem didn't led to anywhere but a handful of arithemetic exercises... And that doesn't serve any purpose, I am not udnersetanding more just because I did some rote problems... It was some derivative involving the chainrule and this stuff where there is so many layers and functions of functions that one get lost it in and finally one wonders, what is this for? Waste of time for me... So thats why I said that... Take no offense, that is how I felt about my experience... Like... if one wants to really find antiderivated to e^(x^2)... because one didn't read about it, one this they can use some stuff on it and then they realize.... :( Sorry about that. 14. Apr 26, 2004 ### JonF [sarcasm] But… but… my brain can beat up your brain [/sarcasm] 15. Apr 26, 2004 ### Ebolamonk3y heh. perhaps. my mind is afflicted with many things like depression to easily succumb to an incoming invasion of anothers mind. :p 16. Apr 27, 2004 ### Integral Staff Emeritus Lets start from scratch. The volume is the sum of the volume of a hemisphere and a cylinder. [tex] V= \frac 2 3 \pi r^3 + \pi r^2 h$$
The surface area is
$$S= 2 \pi r h + 2 \pi r^2$$
Isolate h in the Volumn equation.
$$h = \frac {V- \frac 2 3 \pi r^2} {\pi r^2}$$
plug into the Surface area equation.
$$S= 2 \pi r \frac {V- \frac 2 3 \pi r^2} {\pi r^2} + 2 \pi r^2$$

$$S= \frac {2 V} r + \frac 2 3 \pi r^2$$
Compute the derivative.
$$\frac {dS} {dr} = - \frac {2V} {r^2} + \frac 4 3 \pi r = 0$$
Solve for r
$$-2V + \frac 4 3 \pi r^3 = 0$$

$$r^3 = \frac {3V} {2 \pi}$$

Complete by solving for h.

Last edited: Apr 27, 2004