- #1
mikky05v
- 53
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1. The problem statement, all variables and given/known
This is a group project for differential equations. I ended up without a group, lucky me. I've been trying to work through this on my own but I am stuck. Sorry about the pictures, typing it all out would of taken ages.
http://imgur.com/qeOkl3n
http://m.imgur.com/uLZWRCX
http://imgur.com/z3SC4st
He sent us the email
Project C: you can get an approximate answer for part (d) by using the formula r = 3h/5 in part (c) [so A(h) = Pi*(3h/5)^2] you really should be able to solve that diff eq in (c) with that info. (note that the "actual" time is a bit longer if you solve it the "correct" way using r = .59h + .5). For part (e)--you first need the formula r = -3/5 h +30 which of course gives you the radius at any given height. This makes your A(h) = Pi* (-3/5 h + 30)^2. When you put this into the differential equation and divide by the square root of h and then integrate you should get an equation with h^5/2 and h^3/2 and h^1/2 and t (and of course coefficients on all of those!). Once you find your constant by inputting your initial condition [h(0)=50] you can solve the resulting implicit equation BUT you can't do it by hand, you have to use technology! (I used desmos by inputting "x" for my "t" and "y" for my "h") the graph then showed me the time to drain the other tank is around 10 minutes (600 seconds). I'll leave it to you to find the exact value.
l did part a by integrating twice.
b. I am not entirely sure what this section wanted. I put A (h) dh/dt= -a(sqrt(2gh))
C. r=3h/5 according to email
A (h) = pi (3h/5)^2
a= pi (1/2) ^2 = pi/4
g= 98.1 cm/s^2
Giving the seperable differential equation
Pi (3h/5)^2 dh/dt = -pi/4 sqrt (2×98.1h)
That simplifies to .0257012h^(3/2) dh=dt
D. Integrating both sides I got t= .0102805h^(5/2) + c
I thought to solve for c by taking t=0 and h=50 but I get c=-181.7352816 so I know I did something wrong.
Thats as far as I've gotten.
This is a group project for differential equations. I ended up without a group, lucky me. I've been trying to work through this on my own but I am stuck. Sorry about the pictures, typing it all out would of taken ages.
http://imgur.com/qeOkl3n
http://m.imgur.com/uLZWRCX
http://imgur.com/z3SC4st
Homework Equations
He sent us the email
Project C: you can get an approximate answer for part (d) by using the formula r = 3h/5 in part (c) [so A(h) = Pi*(3h/5)^2] you really should be able to solve that diff eq in (c) with that info. (note that the "actual" time is a bit longer if you solve it the "correct" way using r = .59h + .5). For part (e)--you first need the formula r = -3/5 h +30 which of course gives you the radius at any given height. This makes your A(h) = Pi* (-3/5 h + 30)^2. When you put this into the differential equation and divide by the square root of h and then integrate you should get an equation with h^5/2 and h^3/2 and h^1/2 and t (and of course coefficients on all of those!). Once you find your constant by inputting your initial condition [h(0)=50] you can solve the resulting implicit equation BUT you can't do it by hand, you have to use technology! (I used desmos by inputting "x" for my "t" and "y" for my "h") the graph then showed me the time to drain the other tank is around 10 minutes (600 seconds). I'll leave it to you to find the exact value.
The Attempt at a Solution
l did part a by integrating twice.
b. I am not entirely sure what this section wanted. I put A (h) dh/dt= -a(sqrt(2gh))
C. r=3h/5 according to email
A (h) = pi (3h/5)^2
a= pi (1/2) ^2 = pi/4
g= 98.1 cm/s^2
Giving the seperable differential equation
Pi (3h/5)^2 dh/dt = -pi/4 sqrt (2×98.1h)
That simplifies to .0257012h^(3/2) dh=dt
D. Integrating both sides I got t= .0102805h^(5/2) + c
I thought to solve for c by taking t=0 and h=50 but I get c=-181.7352816 so I know I did something wrong.
Thats as far as I've gotten.