Determining a smooth motion when given a function

[vande060]

Show from Eq. 1.1 that the below function is smooth at t = 1 and at t = 2. Is it smooth at any 1 < t < 2?

x(t) =

1.0 + 2.0 t 0 ≤ t ≤ 1
3 + 4(t − 1) 1 ≤ t ≤ 2
7 + 3(t − 2) 2 ≤ t



for equation 1.1 my book gives me:

lim
dt→0 [x(t + dt) − x(t) ]/dt= 0



This problem asks me to show that the motion is smooth, but to me it seems that it would not be smooth at points 1 and 2. the way i understand it, for a motion to be smooth all of the derivatives of x(t) must exist, but if i take the derivatives:

2 0 ≤ t ≤ 1
4 1 ≤ t ≤ 2
3 2 ≤ t

the derivatives at points 1 and 2 conflict, so i don't understand how this meets the requirements of a smooth function. i must be missing something obvious.

Also i was confused by the formula 1.1 in the book, it tell me to use this formula to show the motion is smooth, but then next to the formula in the book it states the formula as being one used to show a function is continuous.

Thank you in advance to anyone who responds.

 

[Delphi51]

Curious question; probably should be over on one of the math forums.
What exactly is the definition of smooth? I graphed the function from 0 to 3 and it is continuous, but the first derivative does not exist at 1 and at 2 because the value of eqn 1.1 is different when you approach from the left or the right. We say the whole limit does not exist if the left limit is not equal to the right limit.

 

[vande060]

Curious question; probably should be over on one of the math forums.
What exactly is the definition of smooth? I graphed the function from 0 to 3 and it is continuous, but the first derivative does not exist at 1 and at 2 because the value of eqn 1.1 is different when you approach from the left or the right. We say the whole limit does not exist if the left limit is not equal to the right limit.

thanks for the reply, the definition of a smooth motion, as given in my book, is : for a motion to be smooth all of the derivatives of x(t) must exist, but if i take the derivatives.

my intuition was that this function is not smooth, because points one and two have trouble with derivatives, any ideas? Again, my book asks me to show the motion as smooth, but i don't understand how it could be by the definition.

 

[Delphi51]

Definitely not smooth at 1 and 2. You can see that on the graph - the slope changes suddenly at those points, so no value can be assigned to the slope or first derivative at t = 1 or t = 2. Looks like the question got mixed up.

 

[rwisz]

As a scientist, my understanding of smooth motion is that it refers to a motion that is continuous and has no sudden changes or discontinuities. In this case, the function x(t) is defined in three different parts for different intervals of t, but it is still continuous at t = 1 and t = 2 because the values of x(t) from both sides of these points match. This means that the function is continuous at these points and therefore, it can be considered smooth at t = 1 and t = 2.

However, for any 1 < t < 2, the function is not smooth because the values of the function from the two different intervals (t = 0 to t = 1 and t = 1 to t = 2) do not match at this point. This leads to a sudden change or discontinuity in the function and therefore, it is not considered smooth at any point between t = 1 and t = 2.

In terms of the formula 1.1, it is used to show the continuity of a function, but in this case, it can also be used to show the smoothness of the function. This is because for a function to be smooth, it needs to be continuous and have all its derivatives exist. The formula 1.1 essentially shows that the function is continuous by showing that the limit of the difference between the function values at two points divided by the difference in those points approaches 0 as the difference tends to 0. This means that the function is continuous at that point and therefore, it is also smooth.

In conclusion, the function x(t) is smooth at t = 1 and t = 2, but not at any point between t = 1 and t = 2 as it has a sudden change or discontinuity at these points.