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p3 rates of change question
hi. June 2004 P3 question 7:
a drop of oil is modelled as a circle of radius r. at time t,
r = 4 [1 - e^(-t x lamda)], t > 0
where lamda is a positive constant
a: show that the area A of the circle satisfies
dA/dt = 32pie x lamda [e^( -t x lamda) - e^ - (2 t x lamda)]
really don't know where to go with this one adn i was never really good at reates of change. ay help is appreciated. but please try and explain what you're doing each time.
a drop of oil is modelled as a circle of radius r. at time t,
r = 4 [1 - e^(-t x lamda)], t > 0
where lamda is a positive constant
a: show that the area A of the circle satisfies
dA/dt = 32pie x lamda [e^( -t x lamda) - e^ - (2 t x lamda)]
really don't know where to go with this one adn i was never really good at reates of change. ay help is appreciated. but please try and explain what you're doing each time.
rudeboymcc
hi. June 2004 P3 question 7:
a drop of oil is modelled as a circle of radius r. at time t,
r = 4 [1 - e^(-t x lamda)], t > 0
where lamda is a positive constant
a: show that the area A of the circle satisfies
dA/dt = 32pie x lamda [e^( -t x lamda) - e^ - (2 t x lamda)]
really don't know where to go with this one adn i was never really good at reates of change. ay help is appreciated. but please try and explain what you're doing each time.
a drop of oil is modelled as a circle of radius r. at time t,
r = 4 [1 - e^(-t x lamda)], t > 0
where lamda is a positive constant
a: show that the area A of the circle satisfies
dA/dt = 32pie x lamda [e^( -t x lamda) - e^ - (2 t x lamda)]
really don't know where to go with this one adn i was never really good at reates of change. ay help is appreciated. but please try and explain what you're doing each time.
Okay Area of circle = Pi r²
in this case A= Pi(4 [1 - e^(-t x lamda)])²
A= 16Pi[1- 2e ^-(t x lamda) + e^-(2t x lamda)]
A= 16Pi[1- 2e ^-(t x lamda) + e^-(2t x lamda)]
So dA/dt = 16Pi[2lamda e^-(t x lamda) -2lamda e^-(2t x lamda)]
=32Pi lamda[e^-(t x lamda) - e^-(2t x lamda)]
Hope this helps dude.
You need to look at da/dt = da/dr . dr/dt
I'll just scan the working I did in for you and post a link.
EDIT: a slightly different method, but it's all good - comes out the same and all that. I'll still post up my working, I often find it hard to follow mathematical symbols as typed, so I'll give you the hand written stuff.
I'll just scan the working I did in for you and post a link.
EDIT: a slightly different method, but it's all good - comes out the same and all that. I'll still post up my working, I often find it hard to follow mathematical symbols as typed, so I'll give you the hand written stuff.
This is quite a tough one. I'll be as thorough as I can!
OK, just had a shot at typing the solution in - it's completely incomprehensible.
I'll write it and scan...could be a few mins.
OK, just had a shot at typing the solution in - it's completely incomprehensible.
I'll write it and scan...could be a few mins.
Unparseable latex formula:
\Large[br] r \ = \ 4(1-e^{-\lambda t}) \\ \, \\[br]dr/dt \ = \ 4\lambda e^{-\lambda t} \\ \, \\[br][br]A \ = \ \pi r^2 \\ \, \\[br]dA/dr \ = \ 2 \pi r \\ \, \\[br][br]dA/dt \ = \ \frac{dA}{dr} \frac{dr}{dt} \\ \, \\[br]dA/dt \ = \ 2 \pi r . 4\lambda e^{-\lambda t} \\ \, \\[br]substitute \ r: \ r \ = \ 4-4e^{-\lambda t} \\ \, \\[br][br][br]dA/dt \ = \ 8 \pi \lambda (4 - 4e^{-\lambda t})e^{-\lambda t} \\ \, \\[br]dA/dt \ = \ 32 \pi \lambda (1 - e^{-\lambda t})e^{-\lambda t} \\ \, \\[br]dA/dt \ = \ 32 \pi \lambda (e^{-\lambda t} - e^{-\lambda 2t})\\ \, \\[br]
ah! i get it now! i just missed the link between area and radius i was trying to find dA/dr. thanks!
Lol, you beat me to it! My scanner's to slow.
wow that was fast, you gave a whole other solution whilst i was replying.
k well it gets harder.
even after that there's a second part:
in an alternative model of the drop of oil its area A at time t satisfies
dA/dt = A^(3/2) / t^2, t>0
given that the area of the drop is 1 at t = 1
b:find an expression for A in terms of t for this alternative model
now i'm guessing you just integrate dA/dt but the A and t are on the same side. do i have to make it:
1/A^(3/2) dA = 1/t^2 dt
and then integrate both sides?
i get :
- 2/squareroot A = - 1/3t
A = (3t/2)^2
someone tell me what i'm doing wrong.
k well it gets harder.
even after that there's a second part:
in an alternative model of the drop of oil its area A at time t satisfies
dA/dt = A^(3/2) / t^2, t>0
given that the area of the drop is 1 at t = 1
b:find an expression for A in terms of t for this alternative model
now i'm guessing you just integrate dA/dt but the A and t are on the same side. do i have to make it:
1/A^(3/2) dA = 1/t^2 dt
and then integrate both sides?
i get :
- 2/squareroot A = - 1/3t
A = (3t/2)^2
someone tell me what i'm doing wrong.
rudeboymcc
now i'm guessing you just integrate dA/dt but the A and t are on the same side. do i have to make it:
1/A^(3/2) dA = 1/t^2 dt
and then integrate both sides?
i get :
- 2/squareroot A = - 1/3t
A = (3t/2)^2
someone tell me what i'm doing wrong.
now i'm guessing you just integrate dA/dt but the A and t are on the same side. do i have to make it:
1/A^(3/2) dA = 1/t^2 dt
and then integrate both sides?
i get :
- 2/squareroot A = - 1/3t
A = (3t/2)^2
someone tell me what i'm doing wrong.
Unparseable latex formula:
is correct, but your integration on the right hand side isn't.\Bigint \frac{1}{A^{3/2}} dA \ = \ \Bigint \frac{1}{t^2} dt
It should be: -t^-1 + c
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can someone please explain what principle domain is and why the answer is a not c?Maths
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