How do you decide what is dy/dt, or dx/dt

[AllanW]

Homework Statement

if y=2t+3 and x=t^2, find dy/dx

Homework Equations

dy=dy/dt*dt/dx

The Attempt at a Solution

dy/dt=2
dx/dt=2t
therefore dy/dx=1/t
what I don't understand is how the dy/dt etc. is found when attempting this problem[/B]

 

[SteamKing]

Homework Statement

if y=2t+3 and x=t^2, find dy/dx

Homework Equations

dy=dy/dt*dt/dx

The Attempt at a Solution

dy/dt=2
dx/dt=2t
therefore dy/dx=1/t
what I don't understand is how the dy/dt etc. is found when attempting this problem[/B]

Well, look at each equation in turn

The first equation is y = 2t + 3. What is dy/dt?

If this equation were written y = 2x + 3, would you have any problem finding dy/dx?

The second equation is x = t²

What if you wanted to find dy/dx for y = x² ?

The rules of differentiation that you are supposed to learn are written around finding the change in the dependent variable with respect to the change in the independent variable. Since y is usually the dependent variable and x is the independent variable, the rules all state dy/dx = whatever.

For example, if y = xⁿ, then dy/dx = n ⋅ x^n-1

 

[andrewkirk]

what I don't understand is how the dy/dt etc. is found when attempting this problem

Do you know how to differentiate the formula 2t+3 with respect to t?

 

[AllanW]

i understand differentiation in that you could find the derivative by going (2*1)t^(1-1) =2 I'm not clear on how it relates to dy/dt and combining it with dx/dt

 

[AllanW]

i suppose what I am asking would be; what does dy/dt mean? I can't wrap my head around the explanation in my textbook and I am looking for a dumbed down version

 

[andrewkirk]

i suppose what I am asking would be; what does dy/dt mean

It is the rate of increase of y as t increases. Loosely put, it is the size of the tiny increase in y that would arise from making a tiny increase to t and putting that in the formula for y. If you do a line graph of y on the vertical axis against t on the horizontal axis it is the slope (gradient) of the line.