# Calc 2: what is the point of

1. Apr 26, 2008

### jaredmt

ok we're learning the part of calc 2 where you have y = 3 + 4t; x = 4 + 3t

then u multiply dy/dt * dy/dx / dx/dt

but then u just plug in the value of t into dy/dt.... so what was the point of finding the whole second part? lol it seems random and pointless to me. my friend says when u do this t "sometimes" cancells out. seems like a lot of work for nothing

2. Apr 27, 2008

### lurflurf

what are you trying to do?
dy/dt * dy/dx / dx/dt=(dy/dx)^2

maybe you intended
dy/dx=(dy/dt)/(dx/dt)?
in which you see how y changes with x
since you know how x and y change with t and maybe that is not interesting
maybe y is how happy you are at time t
and x is how many potatoes you have eaten at time t
when what you would really like to know is how happy does eating potatoes make you
so
dy/dt is increase in happiness [happy]/[hr]
dx/dt is rate of potatoe ingestion [potatoe]/[hr]
dy/dx is the happyness given by each potatoe [happy]/[potatoe]

so maybe knowing you get 4 happier each hour and eat 3 potatoes is not as interesting as knowing each potatoe makes you 4/3 happier.

3. Apr 27, 2008

### kuahji

The textbook our school uses talks about those problems dealing with the motion of particles. Overall, like many things, its just preparing you for what lies beyond... :)

4. Apr 28, 2008

### zpconn

Are you trying to find the derivative of y with respect to x where y and x are defined parametrically? In this case, the "point" is to avoid having to solve for y as a function of x and instead to use the provided parametric formulas.

Notice that by the chain rule,

$$\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx}$$

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

This formula lets you find the derivative of y with respect to x directly from the parametric equations. The formula you gave was incorrect if this is the goal you wanted to accomplish.