- #1
christian0710
- 409
- 9
Why do math professors always say that dy/dt is NOT a RATIO?
I’ve sat down and really analyzed it and I can only conclude that dy/dt is a ratio between the function delta value y and the corresponding delta t at ANY point on a continuous function.
Here is why: (This is a bit of a mix of topics I’ve posted which now make sense)
Δv = change in v = v(t2) - v(t1)
Δt = change in time = t2-t1
dv = infinitesemally small change in v
dt = infinitesimalle small change in t
If Δv/ Δt = 2, then this is a ratio between a chunk of function value and a corresponding chunk of time interval, It tells us that “At this specific interval some function the ratio between function value and time is 2”,
Now the term.dv/dt = infinitesmally small change in both v and t as deltat approaches zero, which gives us the tangent to some point of the graph of v
Therefore if
dv/dt =1
then the change in dv is equal to the change in dt
dv=dt
But if dv/dt = 2
dv=2*dt
Then at any point of the graph the small interval dv is 2 times as large as the corresponding interval dt. if dv/dt= 6t^2 then it must be true that dv=6t^2*dt, therefore you can interprete it like this: For the function v(t) you can see that an infitesimally small change in v (on the y-axis) is equal to 6*t^2 times the infinitely small change change in t (on the x-axis), so graphically when you divide the function V(t) up into intervals, the delta t (on the t axis) will correspond to a delta v (on the y-axis) whose vale is about 6*t^2 times larger in value than the delta t value? And this is true for ANY value of t, so for the complete function, so as the function progresses (as t increases) the change in t becomes much smaller compared to the change in v. This means the slope of the tangent increases. So Actually dv/dt can somehow be interpreted like a ratio because if dv/dt = 2, then the change in dv is always twice the change in dt at any given interval on the function.
Just to give an example to make sure I understood it 100%:
I’ve sat down and really analyzed it and I can only conclude that dy/dt is a ratio between the function delta value y and the corresponding delta t at ANY point on a continuous function.
Here is why: (This is a bit of a mix of topics I’ve posted which now make sense)
Δv = change in v = v(t2) - v(t1)
Δt = change in time = t2-t1
dv = infinitesemally small change in v
dt = infinitesimalle small change in t
If Δv/ Δt = 2, then this is a ratio between a chunk of function value and a corresponding chunk of time interval, It tells us that “At this specific interval some function the ratio between function value and time is 2”,
Now the term.dv/dt = infinitesmally small change in both v and t as deltat approaches zero, which gives us the tangent to some point of the graph of v
Therefore if
dv/dt =1
then the change in dv is equal to the change in dt
dv=dt
But if dv/dt = 2
dv=2*dt
Then at any point of the graph the small interval dv is 2 times as large as the corresponding interval dt. if dv/dt= 6t^2 then it must be true that dv=6t^2*dt, therefore you can interprete it like this: For the function v(t) you can see that an infitesimally small change in v (on the y-axis) is equal to 6*t^2 times the infinitely small change change in t (on the x-axis), so graphically when you divide the function V(t) up into intervals, the delta t (on the t axis) will correspond to a delta v (on the y-axis) whose vale is about 6*t^2 times larger in value than the delta t value? And this is true for ANY value of t, so for the complete function, so as the function progresses (as t increases) the change in t becomes much smaller compared to the change in v. This means the slope of the tangent increases. So Actually dv/dt can somehow be interpreted like a ratio because if dv/dt = 2, then the change in dv is always twice the change in dt at any given interval on the function.
Just to give an example to make sure I understood it 100%: