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# C4 CONNECTED RATES OF CHANGE- 5 QUESTIONS watch

1. See Attachment(s)

Found these 5 to be slightly tricky, would appreciate any help.

Thanks
Attached Files
2. Differential Equations.doc (78.0 KB, 504 views)
3. Differential Equations 2.doc (46.5 KB, 302 views)
4. (Original post by melbourne)
See Attachment(s)

Found these 5 to be slightly tricky, would appreciate any help.

Thanks
Question 10.

Time in seconds = t
Surface area of cube = A
Volume of cube = V

2t = A³
hence .. dt/dA = ½A³, This is the rate of change in area due to time.
A³ = V
V = (2t)^1/3
re-arrange to get > (½V)^1/3
5. Question 1:

C = f(x) = y
P = (x,y) on curve C

dy/dx = xy / z, because the rate of change is proportional to the product of x and y, there will be a division of a constant, let the constant be z.

When P = (4, 2) dy/dx = ½

So dy/dx = (4 x 2) / z = ½

rearrange, 8/z = ½

z = 16, substituite this in previous formula: dy/dx = xy / z,

dy/dx = xy / 16
6. (Original post by Vijay1)
Question 1:

C = f(x) = y
P = (x,y) on curve C

dy/dx = xy / z, because the rate of change is proportional to the product of x and y, there will be a division of a constant, let the constant be z.

When P = (4, 2) dy/dx = ½

So dy/dx = (4 x 2) / z = ½

rearrange, 8/z = ½

z = 16, substituite this in previous formula: dy/dx = xy / z,

dy/dx = xy / 16
7. Question 3:

Volume into container = 30cm³ per second
Volume out of container = 2/15Vcm³ per second
Time in seconds = t

dV/dt = 30t, is rate of liquid transferred into container at time in seconds
dV/dt = 2/15V, is rate of liquid leaving container, at time t seconds.

Putting these together, to get volume in container at a given time:

dV/dt = 30t - 2/15V

(dV/ dt = - 2/15V + 30t) x -15

-15(dV/dt) = 2V - 450t
8. (Original post by melbourne)
Oohh thank you very much

I think the last question, I've made a mistake somewhere, its not meant to be a t in the equation! oh well
9. (Original post by Vijay1)
Question 3:

Volume into container = 30cm³ per second
Volume out of container = 2/15Vcm³ per second
Time in seconds = t

dV/dt = 30t, is rate of liquid transferred into container at time in seconds
dV/dt = 2/15V, is rate of liquid leaving container, at time t seconds.

Putting these together, to get volume in container at a given time:

dV/dt = 30t - 2/15V

(dV/ dt = - 2/15V + 30t) x -15

-15(dV/dt) = 2V - 450t

Question 3 (attempt 2 )

Volume into container = 30cm³ per second
Volume out of container = 2/15Vcm³ per second
Time in seconds = t

dV/dt = 30cm, is rate of liquid transferred into container at time in seconds
dV/dt = 2/15V, is rate of liquid leaving container, at time t seconds.

Putting these together, to get volume in container at a given time:

dV/dt = 30 - 2/15V

(dV/ dt = - 2/15V + 30t) x -15

-15(dV/dt) = 2V - 450
10. QUESTION ELEVEN:

Model the conical flask as a cone.

Volume of cone = V = 1/3Ah

Where A = Area of base = pi x r2

A slice through the center of the cone is a triangle with angle 30 degrees at the tip.

tan30 = r/h

1/√3 = r/h

r = h/√3

But A = pi x r2 = pi x (h/√3)2 = pi(h2/3)

Now V = 1/3Ah = 1/3(pi(h2/3))h = 1/3(pi(h3/3))

=> V = 1/9pih3

dV/dt = -6

dV/dh = 1/3pih2

dh/dt = dh/dV x dV/dt

dh/dt = 1/(dV/dh) x dV/dt

dh/dt = -6/(1/3pih2) = -18/(pih2)

Let -18/pi = k = constant

So: dh/dt = k/h2

So rate of change of h is inversely proportional to h2
11. Q 9)

we are given that dV/dt = -k√V , where k is a +ve constant

Vol of cylinder is V = πr²h
dV/dh = πr²

dh/dt = dh/dV*dV/dt
dh/dt = (1/πr²)(-k√V)

and √V = r√(πh)

giving

dh/dt = -kr√(πh)/(πr²)
dh/dt = (-k/r√π)√h
dh/dt = -K√h
==========
12. (Original post by Vijay1)
Question 10.

Time in seconds = t
Surface area of cube = A
Volume of cube = V

2t = A³
hence .. dt/dA = ½A³, This is the rate of change in area due to time.
A³ = V
V = (2t)^1/3
re-arrange to get > (½V)^1/3
Sorry, VJ, but I don't really follow that.

Time in seconds = t
Surface area of cube = A = 6x²
Volume of cube = V = x³
where x = length of a side of the cube.

dA/dx = 12x

dV/dA = dV/dx*dx/dA
dV/dA = (3x²)/(12x)
dV/dA = x/4
=========

we are given that dA/dt = 2

dV/dt = dV/dA*dA/dt
dV/dt = (x/4)*2 = x/2

from V = x³
x = V1/3
so,
dV/dt = (1/2)V1/3
=============
13. (Original post by Fermat)
Sorry, VJ, but I don't really follow that.

Time in seconds = t
Surface area of cube = A = 6x²
Volume of cube = V = x³
where x = length of a side of the cube.

dA/dx = 12x

dV/dA = dV/dx*dx/dA
dV/dA = (3x²)/(12x)
dV/dA = x/4
=========

we are given that dA/dt = 2

dV/dt = dV/dA*dA/dt
dV/dt = (x/4)*2 = x/3

from V = x³
x = V1/3
so,
dV/dt = (1/2)V1/3
=============
Oh right I see, I wasn't thinking logically there.

You've got the correct answer, but I think you've made a mistake her:

dV/dt = (x/4)*2 = x/3

should be:

dV/dt = (x/4)*2 = x/2

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