1. The problem statement, all variables and given/known data
Car A is going north, car B is going west, each are approaching an intersection on their respective highways. At an instant, car A is .3km from its intersection while car B is .4 km from it's intersection. Car A travels at 90km/h while car B travels 80km/h. Find the rate at which the distance between them is changing at that moment.
3. The attempt at a solution
z=sqrt(x^2+y^2) so by the chain rule (differentiatiing with respect to x*90+differentiating with respect to y*80) I get
90y/sqrt(x^2+y^2)+80x/sqrt(x^2+y^2) and setting y=.3km and x=.4lkm I get 118km/h
but the book used negative values and got -118km/h. It used -80 and -90km/h for its dz/dx's and dz/dy's. Why did it do this?
HallsofIvy
Aug24-08, 11:45 AM
1. The problem statement, all variables and given/known data
Car A is going north, car B is going west, each are approaching an intersection on their respective highways. At an instant, car A is .3km from its intersection while car B is .4 km from it's intersection. Car A travels at 90km/h while car B travels 80km/h. Find the rate at which the distance between them is changing at that moment.
2. Relevant equations
Pythagorean theorem x^2+y^2=z^2
chain rule of partial derivatives.
Why partial derivatives? The only independent variable here is t.
3. The attempt at a solution
z=sqrt(x^2+y^2) so by the chain rule (differentiatiing with respect to x*90+differentiating with respect to y*80) I get
90y/sqrt(x^2+y^2)+80x/sqrt(x^2+y^2)
It would be much simpler to use z2= x2+ y2 so that
2z\frac{dz}{dt}= 2x\frac{dx}{dt}+ 2y\frac{dy}{dt}
I have no idea what you mean by "differentiatiing with respect to x*90+differentiating with respect to y*80". What do "x*90" and "y*80" represent physically in this problem? If x and y are distance as they appear to be then "x*90" and "x*80" would have units of km2 per hour and I don't reconize that as any physical quantity.
and setting y=.3km and x=.4lkm I get 118km/h
but the book used negative values and got -118km/h. It used -80 and -90km/h for its dz/dx's and dz/dy's. Why did it do this?
You never did say what "x", "y", and "z" represent. I guess that x is the distance from the intersection to car B since you take x= 0.4 km at the given instant. If that is correct, then since car B is getting closer to the intersection as time progresses, x is a decreasing function of t and its derivative is negative: -80 km/hr. Similarly, dx/dt= -90 km/hr.
also, your text did NOT use "-80 and -90km/h for its dz/dx's and dz/dy's" because the 80 and 90 here are speeds in km/hr while dz/dx and dz/dy are "km/km" and so dimensionless. Your text is using -80 and -90 for dx/dt and dy/dt.
evilpostingmong
Aug24-08, 12:03 PM
Oh sorry lol I meant to say dy/dt and dx/dt for the speeds. Yeah even though
partial derivatives are not needed, I used them anyway because I'm doing problems from
the chain rule for partial derivatives chapter. I came up with the same answer, the only problem was the signs. Yes x y and z represent the distance from the intersections, you're right.