# Math Help - help cal hw

1. ## help cal hw

1. 1) the population of a country doubles every 20 years. if the population (in millions) of the country was 100 in 2002, what will the population be in 2014?

2) a particle moves along the x axis with a velocity given by v(t)= 2+sint. when t=0 the particle is at x= -2 . where is the particle when t= π (pie)
a)2π b)π c)π-1 d)π-2 e)π+1

3) consider the diff.eq dy/dx=(1-2x)y.if y=10 when x=, find an equation for y.

a) y=e^x-x^2 b)y=10+e^x-x^2 c)y=10e^x-x^2 d)y=x-x^2+10

please show me how to do these..i want to learn

2. Question 1 is fairly straightforward. Since only 12 years have passed the population will only increase by $\frac{12}{20}$. So the population will increase by:
$\frac{12}{20} * 100 = 60 \mbox{ people}$

Moving on to question 2. First of all to get your position you need to integrate your formula for velocity and take into account the constant:

$x(t) = \int v(t) dt$
$= \int (2 + \sin t) dt$
$x(t) = 2t - \cos t + c$
To find the value for the constant use the information given when t= 0, x = -2:
$-2 = 2*0 - \cos 0 + c$
$-2 = -1 + c$
$c = -1$

Formula for $x(t) = 2t - \cos t -1$
To find where the particle is you just need to put each of the values asked for into the equation in place of t.

Finally for 3:
Rearrange the equation so x terms are all on the same side and the same with the y terms
$\frac{dy}{dx} = (1 - 2x)y$
$\frac{1}{y} dy = (1 - 2x) dx$
Then integrate to get:
$\ln y = x -x^2 + c$
Rearrange to make find an equation for y:
$y = e^{x - x^2 + c}$

3. for question 1 i got aproximately 151.572

i did this
p(t)=100(2^t/20)
p(12)=100(2^12/20)
=151.572

for nmber 2 is it b

3) it is c right

just to make sure

thank u so much for your help

4. 1) the population of a country doubles every 20 years. if the population (in millions) of the country was 100 in 2002, what will the population be in 2014?
let 2002 be t = 0

$P(t) = 100 \cdot 2^{\frac{t}{20}}$

evaluate $P(12)$