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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 40: Higher-order derivatives (parametric & vector-valued functions)

# Second derivatives (vector-valued functions)

Sal finds the first and the second derivatives of the vector-valued function h(t)=(-t⁵-6,4t⁴+2t+1).

## Want to join the conversation?

• Why would we want to find the second derivative of a vector-valued function?
• well, as sal pointed out, higher order derivatives give different things, an example being, in physics, derivatives of position with respect to time.

p(t) = position, p'(t) = velocity, p''(t) = acceleration, p'''(t) = jolt or jerk, p''''(t) = jounce or snap etc.

while jolt, snap and pop or whatever the higher ones are might seem kind of abstract, you can probably see the use in finding the acceleration of something.
• "H prime prime of t" is that really how it is called?
• Yes, or sometimes 'h double prime of t'.
• So it pretty much seems like vector valued functions are the same exact thing as parametric curves (I don't care about the "well, technically..." explanation, thank you).
So is the derivative of a parametric curve WITH RESPECT TO T, found in the same manner as this? Because in the section on parametric curves, we only went over the derivatives of y with respect to x.
Also, conceptually, what is the connection, if any, between the derivatives of the curve with respect to t vs. the derivatives of the y-component with respect to the x-component?
• Yes, the derivative of the parametric curve with respect to the parameter is found in the same manner.

If you have a vector-valued function r(t)=<x(t), y(t)> the graph of this curve will be some curve in the plane (y will not necessarily be a function of x, i.e. it may not pass the vertical line test.) Each point on the curve will have some value of t associated with it as well. You can think of r as the path traced out by a particle as it moves over time.

In this case, dy/dx is a scalar function of t, and it gives the slope of the tangent line to r at the point associated with t.

dr/dt is another vector-valued function, which we can interpret as the velocity of the particle tracing r.
• In the previous video (Second derivatives of parametric functions) we found (d^2y)/(dx^2).
In this video, we found (d^2x)/(dt^2) and (d^2y)/(dt^2). If we divide the latter by the former, we get (d^2y)/(d^2x).
What is the difference between that and (d^2y)/(dx^2)?
• You see ( d²y)/(dx²) is the same thing as second derivative of y w.r.t x . Look it as d/dx [dy/dx]. However (d²y/d²x) is more like (dy/dx)²
• I thought i understand it but i can't. What is the difference between the second derivative of a vector ( acceleration w.r.t position) and the second derivative of a paremtric ecuation.
As far as i've investigated they are both the same, since any vector can be expressed as an x coordinate and y coordinate with respect to another parameter.
Therefore why for the parametric ecuations we have d(dy/dx)/dx while in the vectors we have (d^2y/dt^2)/(d^2x/dt^2)).
If the parametric ecuation gives the slope of the tangent line to the point at t and the vector gives the same but they are calculated differently what is the difference between them?