Related Rates QN (1 Viewer)

[SBHS4EVER]

This question is in Fitzpatrick 3U.

A straight railwaytrack and a straight road intersect at righ angles. At a given instant a motor car at 40km/h and a train at 50km/h are moving away from the intersection and are 40km and 30km respectively from the intersection. At what rate is the distance between them changing one hour later? At what rate would the distance between them be changing at that instant if they both traveling towards the intersection?

This btw is bloody hard as is graffitied all over the question.

 

[underthesun]

here's the solution::

let [x] be the distance of motor from intersection
[dx/dt] = 40 km/h

at the beginning, [x=40], but one hour later, [x=80]

let [y] be the distance of train from intersection
[dy/dt] = 50 km/h

at the beginning, [y=30], but one hour later, [y=80]

now

let D be the distance between them

using pythagoras,

[D^2 = x^2 + y^2]

then, differentiate both with respect to time (t)
(implicit differentiation)

[2D (dD/dt) = 2x(dx/dt) + 2y(dy/dt)]

[dD/dt = ( x(dx/dt) + y(dy/dt) ) / sqrt( x^2 + y^2) ]

(where sqrt is the square root of)

then, you just sub in the values of dx/dt, dy/dt, x and y for the time 1 hour later, and you have the value for dD/dt after 1 hour.

calculator is in my bag, so do the working, and tell me if it's right

 

[SBHS4EVER]

Thanks that helps the only thing is

because theyre moving away from the spot- y isnt dx/dt or dy/dt negative

i dont get that

 

[underthesun]

the distance from the spot is increasing, therefore dy/dx is an increase, in other words positive, i guess.

Unless you want to take that the distance from destination is decreasing which means dy/dx is negative.

but it's always easier to use the first part. who likes negative gearing?

whats negative gearing?

 

[pigs_can_fly]

if you dont like doing implicit, there's another way, but its slightly longer.
the eqn for x is x=50t+30
and for y, y=40t+40
im sure you can see why
but the distance between them is
L=square root of (x squared + y squared)
work that out, then differentiate L with respect to t and you should get the same answer