
#1
Dec906, 02:15 PM

P: 84

Here's a problem that I just found in my book and to my dismay, I couldn't figure out how differentiation can be used to solve a particular problem (seeing how we've just finished this unit at school). So here's the problem:
Felicity and Jane start alking at the same time towards an intersection of two roads that meet at right angles. Felicity starts at 9km from the intersection while Jane starts at 13km from the intersection. Their speeds are 4 km/h and 3 km/h respectively. What is the closest that Felicity and Jane will get? I cannot figure out how to relate the two into one equation. Obviously, we need an equation for the distance between them and find the minimum for it (i.e. f'(x) = 0). Anyway, I thought I got differentiation down pat but ... guess not 



#2
Dec906, 02:27 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Well, denote Felicity's position VECTOR as a function of time:
[tex]\vec{r}_{F}(t)=(x_{F}(t),y_{F}(t))[/tex] and similarly, for Jane: [tex]\vec{r}_{J}(t)=(x_{J}(t),y_{J}(t))[/tex] now, choose an intelligent origin, specify the component functions, and find an expression for the distance between them, as a function of time. 



#3
Dec906, 03:12 PM

P: 84

I haven't learned anything about vectors but I think I've got it. x = Felicity's distance to the intersection while y = Jane's distance to the intersection.
So the closest they can get is 5 km Phew, haven't lost my touch yet. Thanks anyway arildno! Hmm I wonder why the images won't actually show up ... 


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