Derivatives and application Pythagorean: A North-South highway intersects....

**Errisa** · June 18th 2012, 05:52 PM

Question : A North-South highway intersects an East-West highway at point P. An
automobile crosses P at 10:00 am, traveling east at a constant speed of 20 km/h.
A car north of P, twenty minutes later passes that same point, traveling south at 60
km/h. Find the approximate the minimum distance between the automobiles.
Answer: 6.3 km

I keep getting 7.8 km as my answer- don't know if I'm setting up the equations wrong or if the answer is wrong. Please help~ and thanks in advance

**Soroban** · June 18th 2012, 06:35 PM

Hello, Errisa!

The problem has a strange answer . . .

A north-south highway intersects an east-west highway at point P.
A car crosses P at 10:00 am, traveling east at a constant speed of 20 km/hr.
A car north of P, twenty minutes later passes that same point, traveling south at 60 km/hr.
Find the approximate the minimum distance between the automobiles.
Answer: 6.3 km

Code:

      |
      | 20/3  Q   20t   R
    P o - - - o - - - - o
      |              *
      |           *
  60t |        * x
      |     *
      |  *
    S o

At 10:00, car $A$ drives east from $P$ at 20 km/hr.
By 10:20, it has moved to point $Q:\;PQ = \tfrac{20}{3}$ km.
In the next $t$ hours, it has moved $20t$ km to point $R.$

In the same $t$ hours, car $B$ moves from $P$ to $S$ at 60 km/hr.
. . $PS\,=\,60t$

Let $x$ = distance between the cars, $RS.$

Pythagorus: . $x \;=\;\sqrt{(20t + \tfrac{20}{3})^2 + (60t)^2} \;=\;\left(4000t^2 + \tfrac{800}{3}t + \tfrac{400}{9}\right)^{\frac{1}{2}}$

To minimize $x$ , set $\tfrac{dx}{dt} = 0$ and solve.

$\frac{dx}{dt} \;=\;\tfrac{1}{2}\left(4000t^2 + \tfrac{800}{3}t + \tfrac{400}{9}\right)^{-\frac{1}{2}}\left(8000t + \tfrac{800}{3}\right) \;=\;0$

We have: . $8000t + \tfrac{800}{3} \:=\:0 \quad\Rightarrow\quad t \:=\:-\tfrac{1}{30}$

Negative time?
. . . The minimum distance occured two minutes before 10:20.

At 10:18, car $A$ was 6 miles east of point $P.$
At 10:18, car $B$ was 2 miles north of point $P.$

Their distance was: . $\sqrt{6^2 + 2^2} \:=\:\sqrt{40} \:=\:6.32455532 \;\approx\;6.3\text{ km.}$

**Reckoner** · June 18th 2012, 06:37 PM

We need to be careful with our units. I'm going to use kilometers for distance and hours for time, with $t=0$ representing 10:00 am. The horizontal position of the first car relative to the intersection at time $t$ is $20t$ and the vertical position of the second car is $60\left(\frac13-t\right)$ (because 20 minutes = 1/3 hours). Draw a picture:

Using the Pythagorean Theorem, the distance between A and B turns out to be

$s = 20\sqrt{10t^2-6t+1}$

To make the differentiation easier, let's minimize the square of the distance.

$s^2 = 400\left(10t^2-6t+1\right)$

$\frac d{dt}\left[s^2\right] = 400(20t-6)$

$\frac d{dt}\left[s^2\right] = 0\Rightarrow t = \frac3{10}$

So the cars are closest at 10:18 am. At this time $\left(t = \frac3{10}\right),$ the distance between the cars is

$s = 20\sqrt{10\left(\frac3{10}\right)^2-6\left(\frac3{10}\right)+1}$

$= 20\sqrt{\frac1{10}} = 2\sqrt{10} \approx 6.325\ \mathrm{km}$

Edit: Beaten by two minutes! Dang it Soroban, nice work.

**Errisa** · June 18th 2012, 08:48 PM

thank you! Both of you~ I see which step I messed up on I didn't add 20/3 to 20t!

Sorry about late reply I was finishing up my other questions~

Thanks once again

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