Question

sand is falling into a conical pile so that the radius of that base of pile is always equal to half of its height.if the sand is falling at the rate of 10cm3/sec,how fast is height of the pile increasing when the pile is 5cm deep?(topic from calculus 1 related rates)

Answer 1

the answer is dh/dt = 8/5pie.

Answer 2

can someone solve it for me..

Answer 3

we need some derivatives to play with; what formulas can you think of that will relate to the problem at hand?

Answer 4

perhaps since a volume is being added at a specified rate, we could work with the formula for the volume of a cone?

Answer 5

Welcome to Open Study and congrats on asking your first question. :)

Answer 6

we would gladly solve it WITH you, but not FOR you .... we would rather you participate in the solution process please.

Answer 7

volume of cone is 3/4(pie)(r)squared(h)

Answer 8

Yes, as @amistre64 said, we can help you, but we will Not just hand you the answers.

Answer 9

cool, cause i cant recall those things :)

\[V=\frac13pi~r^2h\]
that looks about right to me ... http://math.about.com/od/formulas/ss/surfaceareavol_2.htm

Answer 10

the best thing I find to do is just dive in and take an implicit derivative, i see a product rule for the most part

Answer 11

\[V'=\frac13pi(2rhr'+r^2h')\]
and we want to solve for h'

\[\frac{\frac 3{pi}V'-2hr~r'}{r^2}=h'\]
it should look better with the given values :)

Answer 12

ooo

Answer 13

we are given that r = h/2; therefore the rates related are: r' = 1/2 h'

as an after thought lets fill that in and then simplify it

\[V'=\frac13pi(2rhr'+r^2h')\]
\[V'=\frac13pi(2rh\frac12h'+r^2h')\]
\[V'=\frac13pi(rh~h'+r^2h')\]
\[V'=h'\frac13pi(rh+r^2)\]

Answer 14

i'll try with using that

Answer 15

let us know if you get stuck :) i still have difficulties with these things

Answer 16

yeah my lecturer also have prob with this questions

Answer 17

he ask us to solve for him instead =.=

Answer 18

well, as the lecturer he has that right :)

we know the givens now as:

V' = 10
h = 5
r = 5/2
h' = unknown

\[V'=h'\frac13pi(rh+r^2)\]
\[10=h'\frac13pi(5(\frac52)+(\frac52)^2)\]
\[\frac{30}{pi}=h'(\frac{25}2+\frac{25}4)\]
\[\frac{120}{pi}=h'(50+25)\]
\[\frac{120}{75pi}=h'\]
\[h'=\frac{8}{5pi}\]

Answer 19

the key for me is in finding suitable formulas or relationships between what is given, and what we want to find.

Answer 20

wow thx amistre64

Answer 21

look over it, and try to figure out what i did, with practice it should become more clearer :) good luck