# Math Help - Misc Calculus questions

1. ## Misc Calculus questions

Hi, im having trouble with some questions, here's one, i will post more as i run into more problems.

The acceleration of an object, $a m/s^2$ moving in a line at t seconds is given by $a = sin(2t)$ if the object has an initial velocity of 4 m/s then v is equal to?

Now you'd think i would just have to integrate a and find the c constant, so i got: $-\frac{1}{2}\cos{2t}+\frac{9}{2}$
$\int_{0}^{t} sin(2x)dx + 4$

can someone explain why this is correct?

2. another similar question. I basically am having a problem with the form they give the answer in. I can find the exact answer result, but they always have it in this annoying form.

Given $\frac{dy}{dx}=\frac{2-y}{4}$, x=3 when y=1, then when y=1/2, x is equal to?

i got $x= 4ln(\frac{3}{2})+3$
they wanted: $\int_{1}^{\frac{1}{2}} \frac{4}{2-t}dt + 3$

im confused...

3. this one is seemingly easier, i probably made a silly mistake somewhere, but I'd appreciate some help anyway:

$\frac{dy}{dx}=\frac{1}{5}(y-1)^2$ and y=0 when x=0, then y is equal to?

apparently the answer is $\frac{x}{x+5}$

why?

ps: i think i was overcomplicating something...

4. Originally Posted by scorpion007
Hi, im having trouble with some questions, here's one, i will post more as i run into more problems.

The acceleration of an object, $a m/s^2$ moving in a line at t seconds is given by $a = sin(2t)$ if the object has an initial velocity of 4 m/s then v is equal to?

Now you'd think i would just have to integrate a and find the c constant, so i got: $-\frac{1}{2}\cos{2t}+\frac{9}{2}$
$\int_{0}^{t} sin(2x)dx + 4$

can someone explain why this is correct?
They are the same thing.

$
\int_0^t \sin(2x)\ dx=\left[\frac{-\cos(2x)}{2}\right]_0^t=\frac{-\cos(2t)}{2}+\frac{1}{2}
$

RonL

5. Originally Posted by scorpion007
another similar question. I basically am having a problem with the form they give the answer in. I can find the exact answer result, but they always have it in this annoying form.

Given $\frac{dy}{dx}=\frac{2-y}{4}$, x=3 when y=1, then when y=1/2, x is equal to?

i got $x= 4ln(\frac{3}{2})+3$
they wanted: $\int_{1}^{\frac{1}{2}} \frac{4}{2-t}dt + 3$

im confused...

The result they are using is:

if: $\frac{dx}{dt}=f(x)$, then:

$x(t)=\int_0^t\ f(\xi)\ d\xi\ + \ x(0)$

Which is just the fundamental theorem of calculus in disguise.

RonL

6. Originally Posted by scorpion007
this one is seemingly easier, i probably made a silly mistake somewhere, but I'd appreciate some help anyway:

$\frac{dy}{dx}=\frac{1}{5}(y-1)^2$ and y=0 when x=0, then y is equal to?

apparently the answer is $\frac{x}{x+5}$

why?

ps: i think i was overcomplicating something...
Its a variables seperable differential equation, and has solution:

$
\int\ \frac{5}{(y-1)^2}\ dy=\int\ dx +c
$

and the initial conditions will determin c.

RonL