# Thread: related rates

1. ## related rates

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

2. 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.

3. 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?

4. 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

5. 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.

6. 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?

7. 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 $\frac{dr}{dt}$