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

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 :)

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

Hello, Errisa!

The problem has a strange answer . . .

Quote:

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 $\displaystyle A$ drives east from $\displaystyle P$ at 20 km/hr.

By 10:20, it has moved to point $\displaystyle Q:\;PQ = \tfrac{20}{3}$ km.

In the next $\displaystyle t$ hours, it has moved $\displaystyle 20t$ km to point $\displaystyle R.$

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

. . $\displaystyle PS\,=\,60t$

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

Pythagorus: .$\displaystyle 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 $\displaystyle x$, set $\displaystyle \tfrac{dx}{dt} = 0$ and solve.

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

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

Negative time?

. . . The minimum distance occured two minutes *before* 10:20.

At 10:18, car $\displaystyle A$ was 6 miles east of point $\displaystyle P.$

At 10:18, car $\displaystyle B$ was 2 miles *north* of point $\displaystyle P.$

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

1 Attachment(s)

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

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

http://mathhelpforum.com/attachment....1&d=1340072325

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

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

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

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

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

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

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

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

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

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

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

:o 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 :D