IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: yasser37 on March 15, 2011, 02:04:12 pm

Title: Urgent Help Needed
Post by: yasser37 on March 15, 2011, 02:04:12 pm: Hi ppl

I need urgent help.. I have an exam in two days but I can't seem to solve differential equations correctly

can someone please do June 10 paper 32 Question 7?

Thanks
Title: Re: Urgent Help Needed
Post by: elemis on March 15, 2011, 02:34:10 pm: Hang on.
Title: Re: Urgent Help Needed
Post by: elemis on March 15, 2011, 02:42:19 pm: $\frac{\mathrm{dx} }{\mathrm{d} t} = \frac{cos^2x}{e^{2t}}$

Separating the variables

$\int \frac{1}{cos^2x}dx=\int e^{-2t}~dt$

$\int sec^2x~dx=\int e^{-2t}~dt$

$<br />tanx=-\frac{1}{2}e^{-2t} +C$

When x = 0 t= 0. Input these values into the below equation and find the value of C

$tanx=-\frac{1}{2}e^{-2t} +C$

C = 0.5

Hence,

$tanx=-\frac{1}{2}e^{-2t} + \frac{1}{2}<br /><br />x = tan^{-1}(-\frac{1}{2}e^{-2t} + \frac{1}{2})$
Title: Re: Urgent Help Needed
Post by: elemis on March 15, 2011, 02:46:10 pm: Did you understand ?
Title: Re: Urgent Help Needed
Post by: yasser37 on March 15, 2011, 06:40:55 pm: Thanks a lot mate

yes I got it

if possible question 8 in November 08 and November 09 p31 question 10 part 1
Title: Re: Urgent Help Needed
Post by: yasser37 on March 16, 2011, 05:30:46 am: anyone?
:-[

question 8 in November 08 and November 09 p31 question 10 part 1
Title: Re: Urgent Help Needed
Post by: elemis on March 16, 2011, 02:14:25 pm: $volume~=\frac{4}{3}h^3$

Differentiating :

$\frac{\mathrm{dV} }{\mathrm{d} h}=4h^2$

The rate of change of volume with respect to time is equal to the water input minus the water loss :

$\frac{\mathrm{dV} }{\mathrm{d} t}=20-kh^2$

-kh² represents the loss via leakage

$\frac{\mathrm{dV} }{\mathrm{d} t}\times \frac{\mathrm{dh} }{\mathrm{d} t}$

$\frac{\mathrm{dV} }{\mathrm{d} t}\times \frac{\mathrm{dh} }{\mathrm{d} V} = \frac{\mathrm{dh} }{\mathrm{d} t}<br />$

$\frac{\mathrm{dh} }{\mathrm{d} t}=\frac{20-kh^2}{4h^2}$

Simplify to :

$\frac{5}{h^2}-\frac{k}{4}$

$When~h=1,\frac{\mathrm{dh} }{\mathrm{d} t}=4.95$

Input h=1 into above equation and find k. You will see the end solution is the equation given in the paper.