Calculus - How fast is this distance changing?

[erik05]

Hey guys. Just a quick calculus question for you all.

Two people start from the same point. One walks east at 3 km/h and the other walks northeast at 2 km/h. How fast is the distance between them changing after 15 min? Ans: 2.125 km/h

Here's what I did:

dz/dt= 3 km/h
dx/dt= 2 km/h
dy/dt= ?

2 km/h * 0.25h= 0.5 km = x
3 km/h * 0.25h= 0.75 km= z
y= 0.5590
x^2 + y^2 = z^2

2x (dx/dt) + 2y (dy/dt)= 2z (dz/dt)
dy/dt= (z (dz/dt) - x (dx/dt))/ y

So I put all the numbers in and I get 2.236. ...and that's kind of close to the answer. Anyways, if anyone could point out what I'm doing wrong, it would be much appreciated. Thanks.

 

[dextercioby]

I think those 2 answers (yours & the one given) are wrong...

After 1/4 hr

a=0.75 Km;b=0.5 Km;\alpha=45 \ \mbox{deg}

The distance is

D=\sqrt{a^{2}+b^{2}-2ab\cos\alpha}\simeq 0.53 Km

Then

\frac{dD}{dt}=\frac{\partial D}{\partial a}\frac{da}{dt}+\frac{\partial D}{\partial b}\frac{db}{dt}

\frac{dD}{dt}=\frac{2a-b\sqrt{2}}{2D}\cdot 3+\frac{2b-a\sqrt{2}}{2D}\cdot 2 \simeq 2.36 \frac{\mbox{Km}}{\mbox{hr}}


Daniel.

 

[erik05]

Yeah, I thought the answer was wrong. A question though, why is the formula D=\sqrt{a^{2}+b^{2}-2ab\cos used and not pythagoras?

 

[dextercioby]

Because the triangle is not rectangular.That formula gives u the distance at any time.


Daniel.

 

[HallsofIvy]

Because this is not a right triangle. The Pythagorean theorem works only for right triangles. What dextercioby used was the "cosine law", a generalized form of the Pythagorean theorem: c²= a²+ b²- 2ab cos C where C is the angle opposite side c.

 

[erik05]

Ah, I see. Sorry,I have another question. Would you apply pythagoras to the question if it told you that the person that is walking northeast remains north of the person walking east at all times?

 

[dextercioby]

You apply the cosine law in every possible case,because you'd still have triangle at any moment of time and you'd need to find one of its sides...


Daniel.