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eb's question at Yahoo! Answers regarding distance, parametric equations and slope

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MarkFL

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Feb 24, 2012
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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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MarkFL

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Feb 24, 2012
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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:

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

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

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

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

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

\(\displaystyle 10=\sqrt{t^2-6t+34}\)

Square both sides and write in standard form:

\(\displaystyle t^2-6t-66=0\)

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

\(\displaystyle t=3+5\sqrt{3}\approx11.66\)

2.) Initially the slope of the line through their positions is \(\displaystyle \frac{5}{3}\), 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:

\(\displaystyle \frac{5-0}{2t+3-3t}=-\frac{3}{5}\)

\(\displaystyle \frac{5}{t-3}=\frac{3}{5}\)

Cross-multiply:

\(\displaystyle 3t-9=25\)

\(\displaystyle 3t=34\)

\(\displaystyle t=\frac{34}{3}\)

To eb and any other guests viewing this topic, I invite and encourage you to post other parametric equation/analytic geometry questions in our Pre-Calculus forum.

Best Regards,

Mark.
 
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MarkFL

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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.

\(\displaystyle x(t)=vt+x_0\)

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

\(\displaystyle S(t)=2t+3\)

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

\(\displaystyle H(t)=3t+0=3t\)

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