Interesting Parabola: Learn About t Slope and dx/dt Properties

[srfriggen]

On a test in multivariable calculus I came across, what I thought, was an interesting parabola. Figured I'd ask the forum because my professor is pretty unavailable.


In parametric form it is,

x=t+4 , y = (1/2)t^2 + 2


The reason I found it interesting is because t is always equal to the slope of the curve...

recall; dy/dx = dy/dt / dx/dt = t

I've seen a similar property in f(x)= e^x, where x is always equal to dy/dx, which I always found pretty amazing.


Is there a name for such "behavior"? Or are some of you looking at this saying, "that's not so special at all" lol

Just curious.

Thanks

 

[issacnewton]

I've seen a similar property in f(x)= e^x, where x is always equal to dy/dx

that's not true.

Or are some of you looking at this saying, "that's not so special at all" lol

i don't think anything special here(your first example). just some interesting property..

 

[srfriggen]

that's not true.









i don't think anything special here(your first example). just some interesting property..

true, I typed that incorrectly. f'(e^x) = e^x, is what I meant. still fascinating.
ok thanks. I just thought there may have been a name for a curve that exhibits such property.

 

[Office_Shredder]

Because dx/dt=1, dy/dt=dy/dx. Now we can see that the parametrization is just getting in the way:

If dy/dx=x, then we can integrate and y=x²/2+C So we find precisely the set of parabolas for which we have dy/dx=x, and then any choice of parametrization for which x=t+D for some constant is going to have the same property as your parabola.

 

[srfriggen]

Because dx/dt=1, dy/dt=dy/dx. Now we can see that the parametrization is just getting in the way:

If dy/dx=x, then we can integrate and y=x²/2+C So we find precisely the set of parabolas for which we have dy/dx=x, and then any choice of parametrization for which x=t+D for some constant is going to have the same property as your parabola.

thanks, that makes a lot of sense :)