Rate of Change - Any help Greatly appreciated (1 Viewer)

[tristambrown]

Can anyone please help me with Rate of change … I am getting half the q’s with no problem, but the other half seem to be working out horribly wrong …
I have tried to work backwards from the answers but am only ending up more confused trying to compare “what looks different” across questions.
I am having NO problems with questions like
4) find rate of change of V with respect to t when v = 4/3PiR^3, r=10 and dr/dt = 0.1
Method used
Find rate (find derivative of volume function) …
ie Dv/Dt(4/3PiR^3) = 4PiR^2
Dr/DT = 0.1 (given)
Link functions, substituting r=10
Dv/Dt * Dr/Dt (with r value substituted)
=4Pi(10)^2 * 0.1
=40Pi units

However I am having horribly problems with the following 2 questions whilst trying to apply the same method.
5) A spherical balloon is being inflated so that the radius increases at a constant rate of 20 mm/s. Calculate the rate of change of volume when the radius of the balloon is 500mm
My working for this:
Dr/Dt = 20mm/s (Given)
V =4/3PiR^3
Dr/Dt = 4PiR^2
Link together = Dr/Dt * Dv/Dt with R = 500 substituted in
4Pi(500)^2 * 20mm/s = 20000000Pi mm^3/s
This answer unfortunately is incorrect..
Correct answer is 20000Pi mm^3/s
To get the answer from my working I had to modify the eqn to read:
4Pi(500)^2 * 20/100 mm/s = 20000Pi mm^3/s (correct answer according to book)

6)A vessel is of such a shape that when the depth of water is x cm the volume is V cm^3, where v=30x = X^3 . Water is poured in at a rate of 25cm^3/s. At what rate is the level of water rising when the depth is 5cm ?
My working for this:
Dv/Dt (30x + X^3) = 30 + 3x^2
Dr/Dt = 25cm^3/s (given)
Link Dv/Dt * Dr/DT sub x=5 for at 5cm
=30+3(5^2) *25 = 105 *25 = 2625cm/s
This too is incorrect, however the previous adjustment of does not work for this q.
To get the correct answer i have had to rearrange the eqn again to read
Dr/dt / Dv/dt = answer
IE
25cm/s / 105 = 0.2380952381 =~0.24cm/s (which IS the correct answer according to book)

I cant seem to find a standardised adjustment to the method to apply across all questions … I cannot see why the method used in q4 (above) works

Can anybody post worked solutions for these questions or hint at what it is that I need to add to my method to make this work ?
Thanks all! 
Cheers
Tristam

 

[Riviet]

4) dr/dt=0.1
v=4pi.r³/3
dv/dr=4pi.r²
When r=10,
dv/dt=(dv/dr) x (dr/dt)
=4pi.r² x 0.1
=4.pi.10² x 0.1
=40 pi

5) dr/dt=20 mm/s
v=4pi.r³/3
dv/dr=4pi.r²
When r=500
dv/dt=(dv/dr) x (dr/dt) = 4pi.r² x 20
=4pi.500² x 20
=2 x 10⁷ mm³/s

6) dv/dt = 25cm³/s
v=30x + x³
dv/dx=30 + 3x²
dx/dv=1/(30 + 3x²)
When x=5,
dx/dt=(dx/dv) x (dv/dt)
=1/(30 + 3x²) x 25
=25/{30 + 3(5²)}
=25/105
=0.24 cm/s (2 decimal places)

 

[Riviet]

is that 2U maths, am i seriously expected to know that next year, ahhh

It's HSC 2 unit level, which you won't be encountering for another 1-2 years.

 

[SoulSearcher]

It's HSC 2 unit level, which you won't be encountering for another 1-2 years.

Hmm, I thought this was only dealt with in the extension 1 course, certainly didn't see this last year.

 

[Riviet]

Hmm, you're probably right SS. However, I do recall doing these types of questions in the prelim extension 1 course.

 

[SoulSearcher]

Yes, in the extension 1 course it does, however I do not remember them being in the 2 unit course.

 

[elseany]

hey riviet, the answer you got for Q5 is the same answer tristambrown got for Q5, which he said was wrong.

tristam are you sure the answer in your book for q5 has the units of mm^3/a ?

 

[Riviet]

Sometimes the book's answers can be wrong, unless you can see something wrong in my working.

 

[acmilan]

If you convert to standard units and do the same thing you'll find the answer is 0.02 m³/s which is the answer both Riviet and tristambrown got when you convert to mm³/s

 

[tristambrown]

Re: 2 or 3 unit - this is only in the 3u course (at least at tafe anyway)

Re: elseany "tristam are you sure the answer in your book for q5 has the units of mm^3/a ?"
Yes, but i am going to assume they mean cm^3 as everyone i have asked has come up with this same answer.


Riviet, one last question re your working for q6:

What was different about question 6 that led you to use 1/dx/dv rather than just dx/dv ?

ie
=1/(30 + 3x<SUP>2</SUP>) x 25
rather than (30 + 3x<SUP>2</SUP>) x 25

(as opposed to q5 where you used dv/dt=(dv/dr) x (dr/dt) = 4pi.r<SUP>2</SUP> x 20
rather than dv/dt=1/(dv/dr) x (dr/dt) = 1/(4pi.r<SUP>2</SUP>) x 20 )

Also, thank you all for your input, it is greatly appreciated

 

[Riviet]

I inverted the dv/dx because it allows the dv's to cancel out when applying the chain rule:

dx/dt = (dx/dv) x (dv/dt) => the dv's cancel out.

 

[tristambrown]

AHhhhhhhhhhhhhhhhhhh
*gong sounds as realisation sets in*

Thanks so much !

 

[ianc]

Yes, in the extension 1 course it does, however I do not remember them being in the 2 unit course.

it's in the 2 unit syllabus, but is usually in the extension 1 exam

 

[joesmith1975]

rate of change with 1 varible is in the 2 unit course but with 2 or more is the 3 units course.