Finding the rate of change of the distance between 2 cars

[tnutty]

Homework Statement

Car A is traveling north on Highway 16 and car B is traveling west on Highway 83. Each car is approaching the intersection of these highways. At a certain moment, car A is 0.3 km from the intersection and traveling at 90km/h while car B is 0.4km from the intersection and traveling at 80km/h. How fast is the distance between the cars changing at that moment?



Attempt :

D = sqrt(x^2 + y ^ 2)

Let C = sqrt(x^2 + y^2);

d(D) = dD/dx dx + dD/dy dy

= x/C dx + y/C dy

we are given x = 0.3, y = 0.4 , dx = 90 and dy = 80, so

d(D) = dx/6 + 8dy/5;

Now just plug in dx and y ,

so the answer I got is 143 .

Can you check this?

 

[lanedance]

Homework Statement

Car A is traveling north on Highway 16 and car B is traveling west on Highway 83. Each car is approaching the intersection of these highways. At a certain moment, car A is 0.3 km from the intersection and traveling at 90km/h while car B is 0.4km from the intersection and traveling at 80km/h. How fast is the distance between the cars changing at that moment?



Attempt :

D = sqrt(x^2 + y ^ 2)

Let C = sqrt(x^2 + y^2)

I think i see what you are trying to do, but i think its poor notation, be explicit in variables if confused

let the distance between the 2 cars, D, be exressed in terms of their distances form the intersection, x,y
$$D(x,y) = \sqrt{x^2 + y^2}$$
then at the point in time when x = 0.3, y = 0.4, then you let
$$C = D(0.3,0,4)$$
i don't think there is a real need for C and its a little confusing...

d(D) = dD/dx dx + dD/dy dy

= x/C dx + y/C dy

this the small change formula, change in D for a small change in x & y - though you need to show how you got to the next step (the differentiation)

that said it should be in form of the partial derivatives of D wrt x & y, and you might as well relate it directly to the rate of cahnge with time of D, which is what you want at the end of the day ie.

$$\frac{d D(x,y)}{dt} = \frac{\partial D(x,y)}{\partial x}\frac{dx}{dt} +\frac{\partial D(x,y)}{\partial y}\frac{dy}{dt}$$

this is true for all values, not just at C = D(0.3,0.4)

we are given x = 0.3, y = 0.4 , dx = 90 and dy = 80, so

these dx & dys are not small changes - so even if it gives you the corrcet answer (whih I'm not sure of) you should probably format it as given above

ALso as x,y are the sides of a triangle, tey are decreasing in length
$$\frac{d x}{dt} = -90km/hr, \frac{d x}{dt} = -80km/hr,$$

d(D) = dx/6 + 8dy/5;

Now just plug in dx and y ,

so the answer I got is 143 .

Can you check this?

and finally what are your units?