• Jun 19th 2009, 01:26 PM
mvho
If:

$\displaystyle V=(6-\frac{r^2}{3})*h$

and

$\displaystyle \frac{h}{2-r}$=$\displaystyle \frac{5}{2}$

write V as a function of h . Simplify the function so that you can express it as a polynomial in h .

My answer: 1st, I solved for R in the equation:

$\displaystyle \frac{h}{2-r}$=$\displaystyle \frac{5}{2}$ and figured it to be: $\displaystyle \frac{4h^2-40h+100}{75}$=$\displaystyle \frac{r^2}{3}$

I then, plugged it into the equation $\displaystyle V=(6-\frac{\pi^2}{3})*h$
since I now know R. I am positive my answer is correct but the computer is rejecting it.

$\displaystyle V=\frac{4h^3+40h^2}{75}$ +350
• Jun 19th 2009, 01:34 PM
yeongil
V is already a function of h... :confused:

01
• Jun 19th 2009, 05:15 PM
VonNemo19
Quote:

Originally Posted by mvho
If $\displaystyle V=(6-\frac{\pi^2}{3})*h$ and $\displaystyle \frac{h}{2-r}$ =$\displaystyle \frac{5}{2}$

Write V as a function of "h". Simplify the function so you can express it as a polynomial in "h".

My Answer: I solved for r first and plugged in R in the original equation where v=(6-etc...). I solved for R in the equation: h/2-r=5/2 which I found as: 4h^2-40h+100/75=r^2.

My algebra is not the best so if you catch any mistakes let me know!

Answer (I think): (4*h^3/75) + (40*h^2)+350h

TIA

It seems like you've already solved the problem.

You've already got V in terms of h. If V was in terms of r and h, then to write it as a function of h by substituting for r.

This is a strange question altogether because I've never seen quadratic volume.
• Jun 19th 2009, 05:46 PM
mvho
To be honest, I am quite lost as well. Our TA just told us to solve for R first and that was the "clue".
• Jun 19th 2009, 07:43 PM
VonNemo19
Quote:

Originally Posted by mvho
To be honest, I am quite lost as well. Our TA just told us to solve for R first and that was the "clue".

Now that you've corrected the problem, I can help.

$\displaystyle \frac{h}{2-r}=\frac{5}{2}$
cross multiply
$\displaystyle 2h=5(2-r)$
multiply out parentheses
$\displaystyle 2h=10-5r$
subtract and divide
$\displaystyle \frac{2h-10}{-5}=r$

NOW you have r in terms of h.

So substitute

$\displaystyle V(h)=[6-\frac{(\frac{2h-10}{-5})^2}{3}]*h$

now simplify

$\displaystyle V(h)=[6-\frac{(\frac{4h^2-40h+100}{25})}{3}]*h=6h-\frac{h(4h^2-40h+100)}{75}$

If you need to simplify further I'm sure you can do it. You may not be inputting correctly into your computer.
• Jun 19th 2009, 07:45 PM
yeongil
Quote:

Originally Posted by mvho
If:

$\displaystyle V=(6-\frac{{\color{red}r^2}}{3})*h$

and

$\displaystyle \frac{h}{2-r}$=$\displaystyle \frac{5}{2}$

write V as a function of h . Simplify the function so that you can express it as a polynomial in h .

Oh, now that's different. (Giggle)

\displaystyle \begin{aligned} \frac{h}{2-r} &= \frac{5}{2} \\ 10 - 5r &= 2h \\ -5r &= 2h - 10 \\ r &= \frac{10 - 2h}{5} \end{aligned}

\displaystyle \begin{aligned} V &= \left(6-\frac{\left(\frac{10 - 2h}{5}\right)^2}{3}\right)h \\ &= \left(6-\frac{\left(\frac{100 - 40h + 4h^2}{25}\right)}{3}\right)h \\ &= \left(6-\frac{4h^2 - 40h + 100}{75}\right)h \\ &= \left(\frac{450 - (4h^2 - 40h + 100)}{75}\right)h \\ \end{aligned}
\displaystyle \begin{aligned} &= \left(\frac{-4h^2 + 40h + 350}{75}\right)h \\ &= \frac{-h(4h^2 - 40h - 350)}{75} \end{aligned}

I'm not getting what you're getting, mvho...

01
• Jun 19th 2009, 07:51 PM
VonNemo19
Quote:

Originally Posted by yeongil

I'm not getting what you're getting, mvho...

01

We've all got the same thing.
• Jun 19th 2009, 08:15 PM
mvho
ok Thanks guys! Ill have to review this again. The hardest part about calculus is algebra.