# Calculus/parametric problem. I can't figure this out : (

• Jan 19th 2009, 09:32 PM
billohbright
Calculus/parametric problem. I can't figure this out : (
Hey all, here's the problem I've been struggling with:

Cameron is located 2 miles due East of an intersection between two country roads. He decides to walk in a straight line (cross-country) toward a position 3 miles due North of the intersection; it takes Cameron one hour to reach that location.

(a) Find parametric equations for the location of Cameron at time t hours.
(b) Find the rate of change of the distance between Cameron and the intersection at time 15 minutes (t= 1/4). (Hint: First find a function d(t) that calculates the distance between Cameron and the intersection at time t, then use the definition of the derivative to calculate the rate of change at time t= 1/4. You will need to rationalize the expression to compute the limit.)

So that's it. Even with the hint I'm brain dead. This is after a day's worth of homework, so my mind is jelly. Thanks in advance to whoever helps! :)
• Jan 19th 2009, 10:07 PM
earboth
Quote:

Originally Posted by billohbright
...

Cameron is located 2 miles due East of an intersection between two country roads. He decides to walk in a straight line (cross-country) toward a position 3 miles due North of the intersection; it takes Cameron one hour to reach that location.

(a) Find parametric equations for the location of Cameron at time t hours.
(b) Find the rate of change of the distance between Cameron and the intersection at time 15 minutes (t= 1/4). (Hint: First find a function d(t) that calculates the distance between Cameron and the intersection at time t, then use the definition of the derivative to calculate the rate of change at time t= 1/4. You will need to rationalize the expression to compute the limit.)

...

1. Use a coordinate system, place the origin at the intersection. Then the startpoint is S(2,0) and the endpoint is E(0,3).

2. The direction vector of Camerons way is: $(0,3) - (2,0) = (-2,3)$

3. The vector equation, which describes Camerons way, is:

$(x,y)=(2,0)+t\cdot(-2,3)~\implies~\boxed{\begin{array}{l}x=2-2t\\y=3t\end{array}}$

4. The distance between any point of this line and the origin(= intersection) is calculated by:

$d(t) = \sqrt{ (2-2t)^2+(3t)^2} = \sqrt{13t^2-4t+4}$

5. Calculate the first derivative of d:

$d'(t)=\frac12 \cdot (13t^2-4t+4)^{-\frac12} \cdot (26t-4) = \dfrac{13t-2}{\sqrt{13t^2-4t+4}}$

6. Now plug in $t=\frac14$ to calculate the rate of change. I've got $d'\left(\frac14\right)=\frac5{61} \sqrt{61} \approx 0.6402$
• Jan 19th 2009, 10:41 PM
billohbright
Quote:

Originally Posted by earboth
1. Use a coordinate system, place the origin at the intersection. Then the startpoint is S(2,0) and the endpoint is E(0,3).

2. The direction vector of Camerons way is: $(0,3) - (2,0) = (-2,3)$

3. The vector equation, which describes Camerons way, is:

$(x,y)=(2,0)+t\cdot(-2,3)~\implies~\boxed{\begin{array}{l}x=2-2t\\y=3t\end{array}}$

4. The distance between any point of this line and the origin(= intersection) is calculated by:

$d(t) = \sqrt{ (2-2t)^2+(3t)^2} = \sqrt{13t^2-4t+4}$

5. Calculate the first derivative of d:

$d'(t)=\frac12 \cdot (13t^2-4t+4)^{-\frac12} \cdot (26t-4) = \dfrac{13t-2}{\sqrt{13t^2-4t+4}}$

6. Now plug in $t=\frac14$ to calculate the rate of change. I've got $d'\left(\frac14\right)=\frac5{61} \sqrt{61} \approx 0.6402$

Going over your solution, I think you made a mistake in multiplying the (2-2t)^2 out. Also, where did you get the equation for the derivative? Thanks again for the help! I'm still stuck hehe.
• Jan 19th 2009, 10:54 PM
earboth
Quote:

Originally Posted by billohbright
Going over your solution, I think you made a mistake in multiplying the (2-2t)^2 out. You are right, so sorry! Should be 4-8t+4t^2 Also, where did you get the equation for the derivative? Thanks again for the help! I'm still stuck hehe.

You have to use the chain rule to calculate the derivative. Since I've made a mistake, the derivative can't be right.

$
d(t) = \sqrt{ (2-2t)^2+(3t)^2} = \sqrt{13t^2-8t+4} = (13t^2-8t+4)^{\frac12}
$

$
d'(t)=\frac12 \cdot (13t^2-8t+4)^{-\frac12} \cdot (26t-8) = \dfrac{13t-4}{\sqrt{13t^2-8t+4}}
$

EDIT: Now I get a rate of change of $-\dfrac15 \cdot \sqrt{5}$ if $t = \frac14$