Find Tangents on Velocity-Time Graph: Discontinuities?

[aspodkfpo]

Homework Statement

Velocity-time graph

Relevant Equations

n/a

If I had a velocity-time graph starting at t=0 and ending at t=8, would I be able to differentiate at these two points for an acceleration value? Or are the start point and end point discontinuities?

 

[Adesh]

Homework Statement:: Velocity-time graph
Relevant Equations:: n/a

If I had a velocity-time graph starting at t=0 and ending at t=8, would I be able to differentiate at these two points for an acceleration value? Or are the start point and end point discontinuities?

You cannot differentiate at end points.

1. You can not differentiate at $t =0$ because by the definition of the derivative we can write $$ v’(0) = \lim_{h\to 0} \frac{ v(h) - v(0) }{h} $$ the RHS of the above expression exists only if $$ \lim_{h\to 0^- } \frac{v(h) - v(0) }{h} = \lim_{h\to 0^+} \frac{v(h) - v(0)}{h} $$ the LHS in the above expression doesn’t exist because $v$ is not defined for values of $t$ less than $0$.

2. You cannot differentiate at $t=8$ because for derivative to exist at $t=8$ we have to have $$ \lim_{h\to 0^-} \frac{v(8+h) - v(8) }{h} = \lim_{h\to 0^+} \frac{v(h+8) - v(8)}{h} $$ the RHS cannot exist because $v(8+h)$ doesn’t exist for any positive $h$ no matter how small.

 

[Orodruin]

You cannot differentiate at end points.

1. You can not differentiate at $t =0$ because by the definition of the derivative we can write $$ v’(0) = \lim_{h\to 0} \frac{ v(h) - v(0) }{h} $$ the RHS of the above expression exists only if $$ \lim_{h\to 0^- } \frac{v(h) - v(0) }{h} = \lim_{h\to 0^+} \frac{v(h) - v(0)}{h} $$ the LHS in the above expression doesn’t exist because $v$ is not defined for values of $t$ less than $0$.

2. You cannot differentiate at $t=8$ because for derivative to exist at $t=8$ we have to have $$ \lim_{h\to 0^-} \frac{v(8+h) - v(8) }{h} = \lim_{h\to 0^+} \frac{v(h+8) - v(8)}{h} $$ the RHS cannot exist because $v(8+h)$ doesn’t exist for any positive $h$ no matter how small.

This is not entirely true. In particular, all differential equations with boundary conditions on the derivatives of a function would be doomed. What is true is that for a limit $\lim_{h\to 0} f(h)$ to exist, there must be a $\delta > 0$ for every $\epsilon > 0$ such that $|f(h)-f(0)| < \epsilon$ if $|h|<\delta$. In the interior of an interval $[a,b]$, you do have the requirement that both left and right derivative exists. However, this is no longer the case at the interval endpoints. If you go back to the definition of the limit above, if at $a$ the right derivative exists, then the derivative exists as it exists for all $h$ such that $f(a+h)$ is defined and $|h|<\delta$.
This is even more apparent if you instead use the topological definition of the limit and the induced topology on $[a,b]$ as a subset of $\mathbb R$. In that case, it is easy to see that the sets $[a,a+\delta)$ are indeed open sets on $[a,b]$.