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**john-1** · March 12th 2010, 03:04 PM

two cars approach an intersection at a certain time, each car is 5km from it. One car travels west at 90km/h and the other south at 80km/h. How fast is the distance between the two cars decreasing after 2min?

I got an answer of -55 km/h but the answer in my text says 119 km/h..... Can anyone verify if my answer is wrong or correct and explain why. Thanks

**jameselmore91** · March 12th 2010, 03:29 PM

Originally Posted by john-1

two cars approach an intersection at a certain time, each car is 5km from it. One car travels west at 90km/h and the other south at 80km/h. How fast is the distance between the two cars decreasing after 2min?

I got an answer of -55 km/h but the answer in my text says 119 km/h..... Can anyone verify if my answer is wrong or correct and explain why. Thanks

So the distance between them is defined as
$\sqrt{x\left( t \right)^{2}\; +\; y\left( t \right)^{2}}$

where x(t) is given by: $x\left( t \right)\; =\; 5km\; -\; \left( \frac{3}{2}t \right)\frac{km}{\min }$
and y(t) similarly is given by: $y\left( t \right)\; =\; 5km\; -\; \left( \frac{4}{3}t \right)\frac{km}{\min }$

You then derive the expression with respect to t and you should get:
$\frac{d\left( dis\tan ce \right)}{dt}\; =\; \frac{x\left( t \right)x'\left( t \right)\; +\; y\left( t \right)y'\left( t \right)}{\sqrt{x\left( t \right)^{2}\; +\; y\left( t \right)^{2}}}$

or if you want it only in terms of t you can insert the function x(t), y(t), x'(t), and y'(t) in their respective places to get:

$\frac{d\left( dis\tan ce \right)}{dt}\; =\; \frac{\frac{145}{36}t\; -\; \frac{25}{2}}{\sqrt{\left( \frac{145}{36} \right)t^{2}\; -\; \frac{85}{3}t\; +\; 50}}$

and plug in t = 2 to get:

$\frac{d\left( dis\tan ce \right)}{dt}\; =\; -\frac{8}{3}\sqrt{\frac{5}{17}}$

I believe that is right, but that answer certainly is not pretty.

**TKHunny** · March 12th 2010, 03:30 PM

It would be even more useful if you would explain what you did to get that answer.

The easterer lives at (5 km - 90km/h * t min, 0) = (f(t),0)

The northerner lives at (0, 5 km - 80km/h * t min) = (0,g(t))

The distance between them is $h(t) = \sqrt{f^{2}+g^{2}}$

Now what?

**john-1** · March 12th 2010, 03:43 PM

Originally Posted by TKHunny

It would be even more useful if you would explain what you did to get that answer.

The easterer lives at (5 km - 90km/h * t min, 0) = (f(t),0)

The northerner lives at (0, 5 km - 80km/h * t min) = (0,g(t))

The distance between them is $h(t) = \sqrt{f^{2}+g^{2}}$

Now what?

Ok that's that formula that I did use. I differentiated and substitued t = 1/30 h into the equation and found that dD/dt = -55 km/h.. so I was wondering if my answer was correct

**jameselmore91** · March 12th 2010, 03:47 PM

So you have the distance in terms of two functions in t and you can plug those functions of t into your distance formula, then you derive it and plug in the value of 2 mins. (keep in mind that your rate is in hours and your t value is in minutes so I converted 90 km/h to 3/2 km/min by dividing by 60 and similarly converting 80km/h to 4/3 km/min.

I guess and easier approach would be to plug in before deriving so you have:
distance = $\sqrt{\left( 5\; -\; \frac{3}{2}t \right)^{2}\; +\; \left( 5\; -\; \frac{4}{3}t \right)^{2}\; }$

then simplifying you get: distance = $\sqrt{\frac{145}{36}t^{2}\; -\; \frac{85}{3}t\; +\; 50\; }$

Derive that with respect to t to get:
D(distance) = $\left( \frac{1}{2} \right)\frac{\left( \frac{145}{18}t\; -\; \frac{85}{3} \right)}{\sqrt{\frac{145}{36}t^{2}\; -\; \frac{85}{3}t\; +\; 50}}$

I derived that using first the power rule $d\left( x^{p} \right)\; =\; px^{p-1}$ and the chain rule: $d\left( f\left( g\left( x \right) \right) \right)\; =\; f'\left( g\left( x \right) \right)g'\left( x \right)\; dx$

Then plug in your value of t = 2 min to get the answer.

**john-1** · March 12th 2010, 04:03 PM

Originally Posted by jameselmore91

So you have the distance in terms of two functions in t and you can plug those functions of t into your distance formula, then you derive it and plug in the value of 2 mins. (keep in mind that your rate is in hours and your t value is in minutes so I converted 90 km/h to 3/2 km/min by dividing by 60 and similarly converting 80km/h to 4/3 km/min.

I guess and easier approach would be to plug in before deriving so you have:
distance = $\sqrt{\left( 5\; -\; \frac{3}{2}t \right)^{2}\; +\; \left( 5\; -\; \frac{4}{3}t \right)^{2}\; }$

then simplifying you get: distance = $\sqrt{\frac{145}{36}t^{2}\; -\; \frac{85}{3}t\; +\; 50\; }$

Derive that with respect to t to get:
D(distance) = $\left( \frac{1}{2} \right)\frac{\left( \frac{145}{18}t\; -\; \frac{85}{3} \right)}{\sqrt{\frac{145}{36}t^{2}\; -\; \frac{85}{3}t\; +\; 50}}$

I derived that using first the power rule $d\left( x^{p} \right)\; =\; px^{p-1}$ and the chain rule: $d\left( f\left( g\left( x \right) \right) \right)\; =\; f'\left( g\left( x \right) \right)g'\left( x \right)\; dx$

Then plug in your value of t = 2 min to get the answer.

I did it the way the person who replied to my post did it and I got -55.2 km/h. When I did it using your method, I got -1.8 something.. It's still not correct with the answer in my textbook. Any answers?

**ione** · March 12th 2010, 04:46 PM

Originally Posted by john-1

two cars approach an intersection at a certain time, each car is 5km from it. One car travels west at 90km/h and the other south at 80km/h. How fast is the distance between the two cars decreasing after 2min?

I got an answer of -55 km/h but the answer in my text says 119 km/h..... Can anyone verify if my answer is wrong or correct and explain why. Thanks

Let x be the distance of the westbound car from the intersection, y be the distance of the southbound car from the intersection, and r be the distance between the two cars.

Then $x^2 + y^2 = r^2$

$2x\frac{dx}{dt}+2y \frac{dy}{dt}=2r\frac{dr}{dt}$

You are given $\frac{dx}{dt}=-90$ and $\frac{dy}{dt}=-80$

$x=5-90t$

$y=5-80t$

After 2 minutes, $t=\frac{1}{30},x=2,y=\frac{7}{3},r=\sqrt{\frac{85} {9}}$

$(2)(-90)+(\frac{7}{3})(-80)=(\sqrt{\frac{85}{9}})(\frac{dr}{dt})$

Solve for $\frac{dr}{dt}$

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