HELP!! Intersection question!! calculus derivatives.

Question

Wendy and Kayla drive towards an intersection. Wendy is 200 meters north of the intersection and moving towards the intersection at 20 m/s, while Kayla is 190 meters west of the intersection and moving away from the intersection at 18 m/s. How rapidly are they approaching each other 2 seconds later?

also are the cars getting closer or farther apart?

Answer 1

this is a "related rates" problem. Usually a subtopic in your calculus book under the major topic of "derivatives" or somewhere online in endless websites

some of the "related rates" problems can get surprisingly complicated.

use the Pythagorean Theorem

d^2 = w^2 + k^2 where d= distance between them, w = wendy's distance to the intersection and k = kayla's distance to the intersection

d(0)= distance at time t=0

= sqr(200^2 + 190^2)

d(2) = distance after 2 seconds

= sqr((200-2(20))^2 + (190+2(18))^2

= sqr(160^2 + 226^2)

= sqr76676

= about 276.9 meters

take the derivative with respect to time, t, measured in seconds

2dd' = 2ww' + 2kk'

divide by 2

dd' = ww' + kk'

plug in the values

solve for d' = dd/dt = rate of change in distance between Wendy & Kayla

276.9 d' = 160(20) + 216(18)

d' = (3200+ 3888)/276.9

= 7088/276.9

= about 25.6 meters per second

it helps to draw a diagram, a graph of the positions & intersection

IF you do, you might feel uncomfortable with the above solution and it may contain

an arithmetic error somewhere, occasionally mistakes are made,

but it's the basic method for a solution

as Wendy gets closer to the intersection, the rate of change of distance should be closer to 18 m/s

the farther away from the intersection Wendy is, the rate of change is closer to 20 m/s

so why would d' be greater than 18 or 20? still it's not that much greater. It's in the general

range of an "expected" solution

So, rework the problem a few times and see where you end up. 25.6 looks about right, rounded off.

More complicated, more interesting question is where does the rate of change of the distance between them reach a maximum and what is that maximum. my initial guess would be when they are equidistant from the intersection, which is just about where they are now, with about 26.8 m/s

Answer 2

Let W be moving in the -y direction at -20 m/s and K moving in the -x direction at -18 m/s and set the origin at the intersection.

D = sqrt((200 - 20t)² + (-190 - 18t)²)

You want to know dD/dt (t = 2),

Using chain rule:

dD/dt = 1/2((200-20t)² + (-190 - 18t²))^-1/2 (2(200-20t)(-20) +2(-190 - 18t)(-18))

Factor out the 2 and multiply out

dD/dt = (1/D^1/2)(-4000+400t + 3420 + 324t) = (-580 + 724t)/D^1/2

Plug in t=2 and the sign of dD/dt tells you whether D is increasing or decreasing (It depends on t and initial conditions)

Please check the math as there are a lot of places to mess up.