Question

in session 31 : related rates
the worked example how he found that r = (3v/5pi)^(1/3)
and why when i tried to solve the example with can design method h was different !!?

Answer 1

can you describe the problem ? are you asking about the related rates of a sphere with dv/dt= 1000 cc/sec and r= 8 ?

Answer 2

no it's how we build grain silo

Answer 3

ok, I found it.
The argument is a bit involved. We fix the volume V (i.e. treat is as a constant).
We fix the "shape" cylinder with hemisphere top, and wish to minimize the surface area as a function of two variables: radius of the cylinder (and hemisphere), and h the height of the cylinder.

I would start by writing down an expression for the surface area (including the floor):
\[ A = 3 \pi r^2 + 2 \pi r h\]
we could try to eliminate one of the variables r or h using the "constraint equation" (i.e. expression for the volume). Or we can find the derivative of the Area with respect to one of the variables. Pick r:
\[ \frac{dA}{dr} = 6 \pi r \frac{dr}{dr} + 2 \pi \left( r \frac{dh}{dr}+h \frac{dr}{dr} \right)\\
\frac{dA}{dr} = 6 \pi r + 2 \pi \left( r \frac{dh}{dr}+h \right)
\]
notice we have to treat h as a function of r because r and h are related by a fixed volume: if you change r , for example, and we keep the volume fixed, then the height must change to keep the volume constant.

Answer 4

do the "usual thing", set dA/dr = 0
\[ \frac{dA}{dr} = 6 \pi r + 2 \pi \left( r \frac{dh}{dr}+h \right) = 0 \\
3r +r \frac{dh}{dr}+h = 0 \]
now we can solve for r in terms of h if we can get rid of dh/dr

Answer 5

to find dh/dr, we use the constraint equation
\[ V_0 = \pi r^2 h + \frac{2}{3} \pi r^3 \\
\frac{dV_0}{dr} = 0 = 2 \pi r h + \pi r^2 \frac{dh}{dr} + 2 \pi r^2 \]
notice because V0 is a constant, its derivative is 0
solve for dh/dr
\[ 2 \pi r h + \pi r^2 \frac{dh}{dr} + 2 \pi r^2 = 0 \\
2h + r \frac{dh}{dr} +2r = 0\]
continuing
\[ \frac{dh}{dr} = -2 \frac{r+h}{r} \]

Answer 6

use that expression for dh/dr in the expression where we are minimizing the surface area
\[ 3r +r \frac{dh}{dr}+h = 0 \\ 3r +-2 \cancel{r} \frac{r+h}{\cancel{r} }+h = 0 \]
from which we get
\[ 3r -2r -2h+h= 0 \\ r-h= 0 \\ r= h\]

this says to minimize the surface area, we want the radius equal the height.
use this in the volume expression. i.e. replace h with r and solve for r

Answer 7

start with \[ \pi r^2 h + \frac{2}{3} \pi r^3 = V_0 \]
and with h= r (to minimize the surface area, we have
\[ \frac{3}{3}\pi r^3 + \frac{2}{3} \pi r^3 = V_0 \\
\frac{5}{3} \pi r^3 = V_0 \\
r^3 = \frac{3V_0}{5\pi} \\
r= \left( \frac{3V_0}{5\pi} \right)^\frac{1}{3}
\]

Answer 8

because V0 is a constant, r has a specific value, and using that value (and the fact that h=r) we can find the minimum value for the surface area:
\[ A = 3 \pi r^2 + 2 \pi r h \\ A= 5 \pi r^2 \\ = 5 \pi \left( \frac{3V_0}{5\pi} \right)^\frac{2}{3} \]
which can be written in the form
\[ A = \sqrt[3]{45 \pi V_0^2} \]