Calculus in physical world Q? (1 Viewer)

[oly1991]

Hey, ive just staterd this topic and having a little bit of trouble with this Q. i derived V to get -6sinpi/x and then subbed it into the formula accordingly but i somehow cant get the right answer.

The volume of water in a tidal pool is given by V=2cos.3pi/x, where x is the depth of water in the pool. Find the rate at which the depth of the pool will be increasing when th volume in the pool is increasing at the rate of 12m^3/h and the depth is 1.2m.

 

[shaon0]

Hey, ive just staterd this topic and having a little bit of trouble with this Q. i derived V to get -6sinpi/x and then subbed it into the formula accordingly but i somehow cant get the right answer.

The volume of water in a tidal pool is given by V=2cos.3pi/x, where x is the depth of water in the pool. Find the rate at which the depth of the pool will be increasing when th volume in the pool is increasing at the rate of 12m^3/h and the depth is 1.2m.

dV/dx= -2cos(2pi)/x^2
dV/dt=dV/dx.dx/dt
12=-2cos(3pi)/(1.2)^2.(dx/dt)
-6(1.44)/cos(3pi)=dx/dt
(dx/dt)=8.64

 

[oly1991]

dV/dx= -2cos(2pi)/x^2
dV/dt=dV/dx.dx/dt
12=-2cos(3pi)/(1.2)^2.(dx/dt)
-6(1.44)/cos(3pi)=dx/dt
(dx/dt)=8.64

thx for the response but im afraid thats still wrong

 

[oly1991]

anyone else want to have a go?

 

[tommykins]

 

[Drongoski]

anyone else want to have a go?

Is answer: 13.417 ... m/h ?

Edit:

I checked again; I forgot the original '2' in formula of V. So Tommy's final formula is correct.

So answer should be half of above: 6.708 ..

 

[oly1991]

dV/dx= -2cos(2pi)/x^2
dV/dt=dV/dx.dx/dt
12=-2cos(3pi)/(1.2)^2.(dx/dt)
-6(1.44)/cos(3pi)=dx/dt
(dx/dt)=8.64

the answer is 0.92 but i think it may be wrong.....

shaon0's actual answer was 8.75... and if you times that by dV/dx u get 12 which means that the formula holds so i think the answer in the book is wrong...on the other hand how come shaon0 didn't change cos to -sin when he derived V.

 

[tommykins]

the answer is 0.92 but i think it may be wrong.....

shaon0's actual answer was 8.75... and if you times that by dV/dx u get 12 which means that the formula holds so i think the answer in the book is wrong...on the other hand how come shaon0 didn't change cos to -sin when he derived V.

Check my solution, it is correct.

 

[Drongoski]

Check my solution, it is correct.

I agree; answer is 6.708 ... m/h I think.

 

[oly1991]

I agree; answer is 6.708 ... m/h I think.

yeh i think thats right.....thx tommy

 

[shaon0]

thx for the response but im afraid thats still wrong

Yeah, i didn't derive it properly. Sorry.

 

[tommykins]

I agree; answer is 6.708 ... m/h I think.

Nah, it's 0.92 as specified in the answers he's given us. They're not wrong.

 

[Drongoski]

Nah, it's 0.92 as specified in the answers he's given us. They're not wrong.

I was wrong again; I forgot to use radians when working out sin( ). I think I should just give up.

 

[oly1991]

Nah, it's 0.92 as specified in the answers he's given us. They're not wrong.

but ur solution gives 6.7...

 

[tommykins]

but ur solution gives 6.7...

no it doesn't. make sure you set your calculator to radians or convert pi to 180

 

[oly1991]

I was wrong again; I forgot to use radians when working out sin( ). I think I should just give up.

lol oh yeah i forgot that aswell....

 

[Drongoski]

but ur solution gives 6.7...

I think u fell into the same trap. haha !

 

[oly1991]

I think u fell into the same trap. haha !

haha and i was revising the radians topic for the past few days aswell..

 

[oly1991]

BUMP, ive got another question

A kite, 20m high, is blown along horizontally at 3m/s by the wind. At what rate is the string being let out when it is 40m long.

 

[tommykins]

Draw a triangle.