Question 1

I understand that the derivative at a sharp point of a function doesn't exist because the slopes of the points around them don't reach the same value. But something is happening at that point rather than nothing, right? I mean, if we throw a ball up in the air, it makes a curve (Parabola) and comes back to the ground and the derivative of the highest point on the curve of the ball is 0. Now, if we throw a ball at a wall at a sharp angle, the ball reflects back making a sharp angle right at the moment it hit the wall w.r.t to the ground, right and so according to the derivative definition we have, the    slope at the point ball hits the ball doesn't exist, but if we want to make sense of that, how do we move further ?

Accepted Answer

If we take data points very, very finely, we can actually trace the ball's velocity function through the moment that it hits the wall. It reality, its change in velocity was not instantaneous. The ball touched the wall, compressed, thereby slowing to a stop, then decompressed and began moving in the other direction. This translates to an extremely high acceleration for that moment, and a standard acceleration due to gravity elsewhere.

Question 2

i get the point of the "sharp" point or turn, since the slopes are not equal.
but why would it matter if the slope is equal or not to the Differentiability?

Accepted Answer

Each point in the derivative of a function represents the slope of the function at that point. The slope of a point in the graph that is "sharp" is undefined: we could view it as the slope as we approach it from the left side, or as we approach it from the right side. In case of a sharp point, the slopes differ from both sides.

Or, more mathetical: if you look at how we find the derivative, it's about finding the limit of the change in y over the change in x, as the delta approaches zero: 
lim h->0 (f(x+h) - f(x)) / h

In the case of a sharp point, the limit from the positive side differs from the limit from the negative side, so there is no limit. The derivative at that point does not exist.

Question 3

At 2:40 in the video, how does one tell if a turn is too sharp?

Accepted Answer

Formally, if taking the limit of the derivative up to a certain value from both the right and left side results in different values, then the turn is too sharp. The turn not being too sharp simply means that the rate of change from both sides of a certain point should converge at the same value, i.e. for some input value a:

  lim      f(h)-f(a)        lim      f(h)-f(a)
h -> a^-  -----------  =  h -> a^+  -----------
            h - a                      h - a

```
  lim      f(h)-f(a)        lim      f(h)-f(a)
h -> a^-  -----------  =  h -> a^+  -----------
            h - a                      h - a
```

Question 4

at 4:23 why is f not differentiable on 1 ?

Accepted Answer

Starting  at 4:06 Sal said the situation where it is not differentiable.
- Vertical tangent (which isn't present in this example)
- Not continuous (discontinuity) which happens at x=-3, and x=1
- Sharp point, which happens at x=3

So because at x=1, it is not continuous, it's not differentiable.

Question 5

Hey,
4:12 , isn't the vertical asymptote at -3 a vertical tangent too? I thought that it is a vertical tangent because it is approaching infinity, which would mean it is not differentiable!
Thanks

Accepted Answer

No, they are not the same. Since f(-3) doesn't exist or undefined, there are no tangents at x=-3. But because it is undefined, created a discontinuity which made it not differentiable.

Question 6

I might be nit picking, but for clarification, if there is a vertical tangent at x=3 wouldn't that make the said function f technically not a function?

Accepted Answer

It can be a function as long as you set the domain to x not equal to 3

Question 7

In a parabola the slopes also don't reach the same value right? I'm not understanding why sharp turns and parabolas are different when it comes to this.

Accepted Answer

Well, the derivative of a function at a point, as you know, is nothing but the slope of the function at that point. In a parabola or other functions having gentle turns, the slope changes gradually. So, it does not matter whether we approach a point on a parabola from the left or the right, the slope we find will be equal ,or in other words, the left hand derivative equals the right hand  derivative. On the other hand, imagine a sharp turn . If you approach the point from the left the slope will seem something, and if you approach it from the right the slope seems something else. That is why LHD won't equal the RHD. The point will have 2 slopes at the point of the sharp turn ,which is absurd. Hence, it is non-differentiable at that point. I hope that helps.

Question 8

Why can't we find the derivative if the function is not continuous?

Accepted Answer

Because if there's a discontinuity, we can fit multiple tangent lines to that point.

Question 9

1:10 I don't understand Sal's explanation of the vertical tangent as having infinite dy. I know that a vertical line has an undefined slope because there's no change in x so dy is being divided by zero, but what does infinity have to do with that?  
Also, how does a small change in x give an undefined slope when it's still a defined number that returns another real number when divided by a defined change in y?

Accepted Answer

But we don't have a "defined change in y" for a vertical line. The change could be anything. If we have a vertical line, then y is not a function of x.
Would you be happier if we said dy was undefined for a vertical line, rather than "infinite"?
To be honest I'm not sure if you think a vertical line _is_ differentiable with respect to x or something else is troubling you.

Question 10

why sharp edges matter for example in parabola the point in the very top has a positive slope from the right and negative slope from the left and i don't understand why one of them is differentiable while the other one is not ?

Accepted Answer

f isn't even defined at x=-3, so it can't be continuous there. And the function makes a jump at x=1, i.e. it has a jump discontnuity.

A parabola is differentiable at its vertex because, while it has negative slope to the left and positive slope to the right, the slope from both directions shrinks to 0 as you approach the vertex. But in, say, the absolute value function, the slopes are -1 to the left and 1 to the right, constantly. There's no smooth way to jump from -1 to 1 in a single point. It's that sudden change in slope that we see as a sharp corner.

Calculus, all content (2017 edition)

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

Differentiability at a point: graphical

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