Two cars, car X and car Y , start moving from the same point P on a cross intersection. Car X is travelling east and car Y is travelling north. Some time later car X is 60 km east of point P and travelling in an easterly direction at 80 km/h and car Y is 80 km north of point P and travelling in a northerly direction at 100 km/h. How fast is the distance between car X and car Y changing?

Question

Two cars, car X and car Y , start moving from the same point P on a cross intersection. Car X is travelling east and car Y is travelling north. Some time later car X is 60 km east of point P and travelling in an easterly direction at 80 km/h and car Y is 80 km north of point P and travelling in a northerly direction at 100 km/h. How fast is the distance between car X and car Y changing?

Answer 1

right triangle 3,4,5 or 60,80,100

x= 60
y = 80
z = hypotenuse = 100
dx/dt = 80
dy/dt = 100

z^2 = x^2+y^2
2 z dz = 2 x dx + 2 y dy
z dz/dt = xdx/dt +y dy/dt
100 dz/dt =60*80+80*100
dz/dt = 6*8+80
dz/dt = 128 km/hr

Answer 2

let the distance traveled by car X be x km

let the distance traveled by car Y by y km
their paths form a right-angled triangle.
Let the distance between them be D km
D^2 = x^2 + y^2
2D dD/dt = 2x dx/dt + 2y dy/dt
dD/dt = (x dx/dt + y dy/dt)/D

at the given case:
x = 60, y = 80 , dx/dt = 80, dy/dt = 100
D^2 = 60^2 + 80^2 = 10000
D = 100

dD/dt = (60(80) + 80(100))/100
= 128

The distance is changing at 128 km/h

check my arithmetic

Answer 3

Yep, it's right :)

Answer 4

Thank you so much guys!!! You guys are awesome! I do have a few more other math problems, should I post them online or ask you directly? Thanks again!

Answer 5

You never know who will be online. All you can do is post the question. There are a lot of math teachers who check in here.

Answer 6

To find the rate at which the distance between car X and car Y is changing, we need to use the concept of derivatives.

Let's denote the distance between car X and car Y as D(t), where t represents time in hours. We want to find dD/dt, which is the rate at which D is changing with respect to time.

First, let's denote the position of car X as X(t), where X(t) is the distance travelled by car X at time t. From the given information, we have X(t) = 60 + 80t (since car X is 60 km east of point P and travelling east at a speed of 80 km/h).

Similarly, let's denote the position of car Y as Y(t), where Y(t) is the distance travelled by car Y at time t. From the given information, we have Y(t) = 80 + 100t (since car Y is 80 km north of point P and travelling north at a speed of 100 km/h).

Now, the distance between car X and car Y at time t is given by the Euclidean distance formula:

D(t) = sqrt((X(t))^2 + (Y(t))^2)

Let's take the derivative of both sides of the equation with respect to time t:

dD/dt = (1/2)*(X(t)^2 + Y(t)^2)^(-1/2) * (2*X(t)*X'(t) + 2*Y(t)*Y'(t))

Plugging in the values we have:

dD/dt = (1/2)*(60 + 80t)^2 + (80 + 100t)^2)^(-1/2) * (2*(60 + 80t)*80 + 2*(80 + 100t)*100)

Now, we can simplify this expression to get the rate at which the distance between car X and car Y is changing, dD/dt, in terms of time t.