Solving the Intersecting Roads Problem

[naspek]

Homework Statement

hi there.. can someone just give me an idea to solve this problem?

Two straight roads intersect at right angles. Car A, moving on one of the roads,
approaches the intersection at 60km/h and Car B, moving on the other road,
approaches the intersection at 80km/h. At what rate is the distance between the
cars changing when A is 0.5km from the intersection and B is 0.7km from the
intersection?

Homework Equations

The Attempt at a Solution

i don't have any idea..

 

[Identity]

Sounds like a diagram could be very useful since this is involves geometry (please look at the attached diagram)

Let's see what we're given:

$$\frac{da}{dt} = -60km/h$$

$$\frac{db}{dt} = -80km/h$$

We need to find:

$$\frac{dx}{dt}$$

-Try and find an expression for x in terms of a and b.

-Then sub that into the above expression. Remember that

$$\frac{dx}{dt}$$ is just $$\frac{d}{dt} (x)$$

-Then use the chain rule to differentiate that expression, remembering that 'a' and 'b' are functions of t.

-Sub in the conditions, and voila!

 

[naspek]

-Try and find an expression for x in terms of a and b.

thank u very much for such ideas..
erm.. x is a hypotenuse

where x=$$\sqrt{a^{2} + b^{2}}$$

let say the distance is D

D = sqrt(0.5^2 + 0.7^2)
that should be my x or hypotenuse right?

 

[Identity]

$$\frac{dx}{dt} = \frac{d(\sqrt{a^2+b^2})}{dt}$$

Differentiate that using the chain rule (since 'a' and 'b' are functions of t) and you will get an expression

$$\frac{dx}{dt} = ...$$,

where the right hand side contains 'a', 'b', $$\frac{da}{dt}$$ and $$\frac{db}{dt}$$.

Then you sub in a = 0.5, b = 0.7, $$\frac{da}{dt} = -60$$, $$\frac{db}{dt} = -80$$ and voila

 

[naspek]

$$\frac{dx}{dt} = \frac{d(\sqrt{a^2+b^2})}{dt}$$

Differentiate that using the chain rule (since 'a' and 'b' are functions of t) and you will get an expression

$$\frac{dx}{dt} = ...$$,

where the right hand side contains 'a', 'b', $$\frac{da}{dt}$$ and $$\frac{db}{dt}$$.

Then you sub in a = 0.5, b = 0.7, $$\frac{da}{dt} = -60$$, $$\frac{db}{dt} = -80$$ and voila

got one question here..
why is the speed in negative?

 

[naspek]

X dX/dt = A da/dt + B db/dt

am i right?
so.. i should find dX/dt right?
i plug in all the value..

i got 100 km.. am i answered it correctly?

 

[Identity]

Yep I think that's correct (100km/h)

 

[naspek]

Yep I think that's correct (100km/h)

got one question here..
why is the speed in negative?

 

[Identity]

da/dt and db/dt aren't speeds, they are the rate at which the distances a and b decrease. So actually dx/dt should be -100km/h (oops), because it is decreasing as the cars get closer