# Word problem and second derivatives

• Dec 17th 2008, 03:40 AM
struck
Word problem and second derivatives
Colin sets off for school, which is 800m from home. His speed is proportional to the distance he still has to go.

Let x meters be the distance that he has gone, and y meters be the distance he still has to go.

a) Sketch the graph of x against t and y against t.

I need help with writing an expression for this word problem in both formats so that I can sketch the graph. I am bad with word problems (Worried)
• Dec 17th 2008, 04:01 AM
Moo
Hello,
Quote:

Originally Posted by struck
Colin sets off for school, which is 800m from home. His speed is proportional to the distance he still has to go.

Let x meters be the distance that he has gone, and y meters be the distance he still has to go.

a) Sketch the graph of x against t and y against t.

I need help with writing an expression for this word problem in both formats so that I can sketch the graph. I am bad with word problems (Worried)

First of all, translate all the information given in the text. Then you may have useless information, but you cannot know at first sight.
(No, the fact that the boy's name is Colin is not an important information (Rofl))

Note that y=800-x. Why ?

Let s be Colin's speed.
We know that the speed is proportional to y. This means that there exists a constant k such that $s=k*y$.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let's deal with "x against the time"

Remember that the speed corresponds to the derivative of the distance with respect to the time.
Thus $s=\frac{dx}{dt}$

$s=k*y=k*(800-x)=k'-kx$, where k'=800k.

So you're now with the following equation :
$k'-kx=\frac{dx}{dt}$

I hope you know how to solve a differential equation (Surprised)
• Dec 17th 2008, 04:26 AM
struck
Quote:

Note that y=800-x. Why ?
Actually, that didn't occur to me. Thanks :D
• Dec 18th 2008, 12:54 AM
struck
On another thought, I still don't get it (Doh) ...

Is this the derivative dx / dt?
$

k'-kx=\frac{dx}{dt}
$

P.s. I am not sure why is the k' notation used. Also, I need the $\frac{d^2x}{dt^2}$

• Dec 19th 2008, 12:17 AM
struck
bump
• Dec 19th 2008, 12:26 AM
Moo
Quote:

Originally Posted by struck
On another thought, I still don't get it (Doh) ...

Is this the derivative dx / dt?
$

k'-kx=\frac{dx}{dt}
$

P.s. I am not sure why is the k' notation used. Also, I need the $\frac{d^2x}{dt^2}$

k' is just a constant. I wrote it above : k'=800k

$\frac{d^2x}{dt^2}=-k \frac{dx}{dt}=-kk'+k^2 x$ (differentiate implicitly and then substitute dx/dt)