# Thread: velocity and acceleration with more than one variable!

1. ## velocity and acceleration with more than one variable!

a particle moves along a horizontal line so that its coordinate at time t is x=(b2+c2t2)^(1/2), t is greater than or equal to 0, where b and c are positive constants
a.) find the velocity and acceleration functions
b.)show that the particle always moves in the positive direction

2. ## Re: velocity and acceleration with more than one variable!

your title is misleading ... x is defined in terms of a single variable, t

$\displaystyle x = (b^2+c^2t^2)^{\frac{1}{2}}$ , $\displaystyle t \ge 0$

chain rule ...

$\displaystyle \frac{dx}{dt} = \frac{1}{2}(b^2+c^2t^2)^{-\frac{1}{2}} \cdot 2c^2 t$

clean up the algebra and use the quotient/chain rule to find $\displaystyle \frac{d^2x}{dt^2}$

note that $\displaystyle \frac{dx}{dt}$ > 0 for $\displaystyle t > 0$

3. ## Re: velocity and acceleration with more than one variable!

.
Originally Posted by skeeter
your title is misleading ... x is defined in terms of a single variable, t

so i got the second derivative equaling
c^2t(-1/2(b^2+c^2t)^-3/2*2c^2t)+(b^2+c^2t^2)^-1/2*c^2dx/dt
is that right? and how do i show that the particle always moves in a positive direction?

chain rule ...

$\displaystyle \frac{dx}{dt} = \frac{1}{2}(b^2+c^2t^2)^{-\frac{1}{2}} \cdot 2c^2 t$

clean up the algebra and use the quotient/chain rule to find $\displaystyle \frac{d^2x}{dt^2}$

note that $\displaystyle \frac{dx}{dt}$ > 0 for $\displaystyle t > 0$

4. ## Re: velocity and acceleration with more than one variable!

I get ...

$\displaystyle \frac{d^2x}{dt^2} = \frac{b^2c^2}{(b^2+c^2t^2)^{\frac{3}{2}}}$

if you clean up the algebra of the expression for $\displaystyle \frac{dx}{dt}$ , it's fairly simple to see why the velocity > 0 for t > 0