MX1 help! Rates of change (1 Viewer)

[jeffwu95]

Sorry for shitty MS paint

 

[jeffwu95]

I got 1.03cm/s but im pretty sure im wrong :S

 

[jeffwu95]

bump

 

[mraznboy]

this problem is over 9000

 

[2xL]

i got 0.034 cm/s.

 

[jeffwu95]

how do you get an equation with just r and h

 

[2xL]

how do you get an equation with just r and h

do you know if my solution is right? no point explaining if it turns out im wrong lol

 

[jeffwu95]

Nah the past paper had no answers
wasnt sure how to relate r and h so i just said that 9r=3.6h but its probably wrong

 

[2xL]

Nah the past paper had no answers
wasnt sure how to relate r and h so i just said that 9r=3.6h but its probably wrong

That's what i did.

 

[jeffwu95]

hmm maybe i made a mistake in calculation imma recheck

 

[jeffwu95]

fml i know wat i did wrong, for some reason i changed sqrt(h)/3 to sqrt(h)/10 ye i got ur answer

 

[jeffwu95]

heres the solution for whoever:
dv/dt=sqrt(h)/3
9r=0.4h (as they are both related)
v=(pi)r^2h/3
v=pi(0.4h)^2.h/4
dv/dh=4pih^2/25

dh/dt=dh/dv . dv/dt
dh/dt=25/4pi(h)^2 . sqrt(h)/10
=0.034cm/s

 

[mraznboy]

sure?