Calc 2: what is the point of

  • Thread starter jaredmt
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  • #1
jaredmt
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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
 

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  • #2
lurflurf
Homework Helper
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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]/

dx/dt is rate of potatoe ingestion [potatoe]/

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
kuahji
394
2
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

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
zpconn
243
2
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,

[tex]\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} [/tex]

[tex]\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/tex]

This formula let's 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.
 

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