MHB Solve Distance Formula Problem: Parametric Equations & Slope

AI Thread Summary
Sally and Hannah start 5 km north and 3 km east of each other in a desert, walking east at different speeds. The distance between them can be modeled with parametric equations, leading to the conclusion that they will be 10 km apart after approximately 11.66 hours. To find when the line through their locations becomes perpendicular to their starting line, the calculation shows this occurs at 34/3 hours. The discussion includes clarifications on how the parametric equations for their movements were derived based on their speeds and initial positions. Overall, the thread emphasizes solving distance problems using parametric equations and slope concepts.
MarkFL
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Here is the question:

Distance formula problem?

Sally and Hannah are lost in the desert. Sally is 5 km north and 3 km east of Hannah. At the same time, they both begin walking east. Sally walks at 2 km/hr and Hannah walks at 3 km/hr.

1. When will they be 10 km apart?
2. When will the line through their locations be perpendicular to the line through their starting locations?

I found that answer for #1 is 11.66. #2 answer is 34/3 but I don't know how to solve for #2.

Here is a link to the question:

Distance formula problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Re: eb's question from Yahoo! Answers regarding distance, parametric equations and slope

Hello eb,

I would place Hannah initially at the origin (0,0), and Sally at (3,5).

1.) I would choose to represent the positions of Hannah and Sally parametrically. The unit of distance is km and the unit of time is hr. So, we may state:

$$H(t)=\langle 3t,0 \rangle$$

$$S(t)=\langle 2t+3,5 \rangle$$

and so the distance between them at time $t$ is:

$$d(t)=\sqrt{((2t+3)-3t)^2+(5-0)^2}=\sqrt{t^2-6t+34}$$

Now, letting $d(t)=10$ and solving for $0\le t$, we find:

$$10=\sqrt{t^2-6t+34}$$

Square both sides and write in standard form:

$$t^2-6t-66=0$$

Use of the quadratic formula yields one non-negative root:

$$t=3+5\sqrt{3}\approx11.66$$

2.) Initially the slope of the line through their positions is $$\frac{5}{3}$$, and so we want to equate the slope of the line through their positions at time $t$ to the negative reciprocal of this as follows:

$$\frac{5-0}{2t+3-3t}=-\frac{3}{5}$$

$$\frac{5}{t-3}=\frac{3}{5}$$

Cross-multiply:

$$3t-9=25$$

$$3t=34$$

$$t=\frac{34}{3}$$

To eb and any other guests viewing this topic, I invite and encourage you to post other parametric equation/analytic geometry questions in our http://www.mathhelpboards.com/f21/ forum.

Best Regards,

Mark.
 
Last edited:
Re: eb's question from Yahoo! Answers regarding distance, parametric equations and slope

A new member to MHB, Niall, has asked for clarification:

I'm a bit confused on how you got S(t) = (2t + 3, 5) and H(t) = (3t, 0) could you explain that?

Hello Niall and welcome to MHB! (Rock)

The $y$-coordinate of both young ladies will remain constant as they are walking due east. Since they are walking at a constant rate, we know their respective $x$-coordinates will be linear functions, where the slope of the line is given by their speed, and the intercept is given by their initial position $x_0$, i.e.

$$x(t)=vt+x_0$$

Sarah's speed is 2 kph and her initial position is 3, hence:

$$S(t)=2t+3$$

Hannah's speed is 3 kph and her initial position is 0, hence:

$$H(t)=3t+0=3t$$

Does this explain things clearly? If not, I will be happy to try to explain in another way.
 
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