Growth and Decay with Differential equation (1 Viewer)

[vds700]

A scientist found that the amount Q(t) of a substance present in a mineral at time t >or= 0 satisfies the equation 4d²Q/dt² + 4dQ/dt + Q = 0.

a) Verify that Q(t) = A(1 + t)e^-0.5t satisfies this equation for any constant A > 0.

b) If Q(0) = 10 mg, find the maximum value of Q(t) and the time at which this occurs.

c) Describe what happens to Q(t) as t increases indefinitely.

 

[conics2008]

Im not going to bother with a) but im going to tell you use product rule from

Ae^-0.5t(1+t) and then basically accept that 4x(double dot) + 4x(dot) + Q=0

b) t=1 sub t=1 into Q >>>>>>> Q(1)=5e^-.5(2) >> 10e^-0.5

c) it just approaches 0

 

[vds700]

Im not going to bother with a) but im going to tell you use product rule from

Ae^-0.5t(1+t) and then basically accept that 4x(double dot) + 4x(dot) + Q=0

b) t=1 sub t=1 into Q >>>>>>> Q(1)=5e^-.5(2) >> 10e^-0.5

c) it just approaches 0

i tried a) by differntiating and subbing in but it didn't equal 0. Ill try again tomorrow and see if i can do it when i can think clearer. Thanks for your help anyways

 

[conics2008]

VDS: Ill post my working out for part a)

what about part b and c. are they fine or wrong ?

 

[conics2008]

OK For simplicity purpose lets expand the given.

Q=Ae^-.5(1+t)
Ae^-0.5t+Ate^-0.5t

Now

dQ/dT= -0.5Ae^-0.5t - -0.5Ate^-0.5t + Ae^-0.5t ( Hint * product rule * )

Simplify = 1/2Ae^-0.5t-1/2Ate^-0.5t

d^2Q/dt^2= -1/4Ae^-0.5t + 1/4Ate^-0.5t - 1/2Ae^-0.5t * product rule again *

simplify = -3/4Ae^-0.5t + 1/4Ate^-0.5t

Now since the give was 4d^2Q/dt^2+4dQ/dT+Q

you know what needs to be done right ?

-3Ae^-0.5t + Ate^-0.5t + 2Ae^-0.5t - 2Ate^-0.5t + Ae^-0.5t + Ate^-0.5t = 0

Good night =] sorry but i dont like typing long working out on the net.. hope you understood this...

 

[vds700]

VDS: Ill post my working out for part a)

what about part b and c. are they fine or wrong ?

thanks for that conics2008. I got part a) using your derivatives. Mine must have been incorrect cooz it wasnt making the differential equation equal to 0.

btw your answer to b) is incorrect. A is 10, not 5. So the answer is 20e^-0.5t.

 

[vds700]

VDS: Ill post my working out for part a)

what about part b and c. are they fine or wrong ?

thanks for that conics2008. I got part a) using your derivatives. Mine must have been incorrect cooz it wasnt making the differential equation equal to 0.

btw your answer to b) is incorrect. A is 10, not 5. So the answer is 20e^-0.5t.

 

[conics2008]

VDS yeah i was going to tell you that because a= 10/1+t where t= 0
=]