- Slope of a line secant to a curve
- Secant line with arbitrary difference
- Secant line with arbitrary point
- Secant lines & average rate of change with arbitrary points
- Secant line with arbitrary difference (with simplification)
- Secant line with arbitrary point (with simplification)
- Secant lines & average rate of change with arbitrary points (with simplification)
- Secant lines: challenging problem 1
- Secant lines: challenging problem 2
Sal finds the slope of the secant line on the graph of ln(x) between the points (e,1) and (x,lnx). Created by Sal Khan.
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- Could someone tell me the difference of tangent lines and secant lines?(23 votes)
- A tangent line has a slope equal to the instantaneous rate of change of the function at one point. It touches the graph at that point.
A secant line has a slope equal to the average rate of change of the function between two points. It touches the graph at those two points.(57 votes)
- 0:55I don't get it, what am I missing. How is e^1 = 0 :S?
ln(0) is undefined, it is impossible?(16 votes)
- At2:00what does e represent, is it a number, if so what is the reason or use of e, and did I miss a video introducing e.(7 votes)
- e = lim h→0 (1+h)^(1/h)
This can also be written as the equivalent:
e = lim h→∞ [1 + (1/h) ]^(h)
e is a number, a constant, and is irrational and transcendental. The first few digits of e are 2.718281828459....
e is used extensively in calculus as well as applied, real-world mathematics. It comes up even more frequently than π. It is used as the base of the natural logarithm as well as the base of most exponential functions. The reason for choosing e will be become very clear when you advance in math, but for now just know that it is the easiest number to work with for log bases and for exponential functions.(16 votes)
- This is my understanding so far of what we've learned in differential calculus: The secant line, being the average slope between two points on the curve, is a rough approximation of the true slope, while the tangent line is the exact value of the slope of the curve at that point. Finding the limit of the secant line should give us the closest approximation of the tangent line. Is this correct?(11 votes)
- Almost correct. The secant line is not the average slope between two points, it is a line drawn between two points, and it has a slope equal to the average slope of the graph between those points.(6 votes)
- So... no matter what the function is, we just need to take
Δ = (y2 - y1) / (x2 - x1)to get the slope of secant line? What if it is a periodic functions such as
f(x) = cos(x)or
f(x) = sin(x) + 2, or a complex function which is only possible to graph?(5 votes)
- The slope of a line between any two points (x1, y1) and (x2, y2) is always m = (y2 - y1) / (x2 - x1). So if you're finding the slope of a secant line, it does not matter what kind of function it is, so long as you have coordinates to work with. f(x) = cos(x) and g(x) = sin(x) + 2 work just fine as well!(7 votes)
I'm from Tunisia in my school we have learned derivatives but we haven't learn "logs" , what i should do ?(2 votes)
- You can learn about Logarithms in this section of Khan Academy-
Logs are pretty simple, once you get their core ideas.(5 votes)
- The question states "Write an expression in x". What does that mean?(2 votes)
- The first one I saw was the following: "A curve has an equation y=cosx and passes through the points P=(0,1) and Q=(x,cosx). Write an expression in x that gives the slope of the secant joining P and Q."
With the secant line being the line joining these two points, (0,1), and (x, cosx), you simply write a slope equation, change in Y/Change in X. In this case, it would be (cosx - 1) / (x - 0). Make sure that you use parentheses so that the exercise knows which operations to perform first. Hope that helps answer your question. Furthermore, I had quite a bit of trouble wrapping my head around another problem which is as follows:
Sometimes this also requires an additional step, like the similar question about the slope of the secant joining P=(π, 0) and Q=(h, Sin(h) / h ). In this case you have the slope equation as follows: (Sin(h) / h - 0) as the change in Y.. all of that over change in x, which is h - π. So your equation would be (Sin(h) / h - 0) / (h - π). But since you have a fraction in the numerator, it would be (Sin(h) / h) * (1 / (h - π)), which would give you (Sin(h)) / (h (h - π)). Hope that helps as well.(7 votes)
- is the concept of instaneous slope physically existent apparent and consistent? or was it invented by humans in order to get a direct answer of a scale for a specific point because it was a high demand?(1 vote)
- Like all of math, instantaneous slope is an abstraction; but, it is an abstraction designed to be very useful in describing reality.(3 votes)
- could you make it (1-ln)/(e-x) and still get the correct answer for this problem?(2 votes)
- While Sid is correct that ln needs input I am assuming that you meant to put ln(x).
If you did mean this then the answer is yes you could do that, because all you did was multiply (ln(x)-1)/(x-e) by -1/-1 which is a valid operation as you are essentially multiplying by 1 which has no effect on the equation except showing it in a new form.
Hope this helped!(2 votes)
A curve has the equation y equals the natural log of x, and passes through the points P equals e comma 1. And Q is equal to x natural log of x. Write an expression in x that gives the slope of the secant joining P and Q. So I think I'm going to need my little scratch pad for this one right over here. So this is the same question over again. Now let's just try to visualize this curve right over here. So let me draw my axes. So let's say this is my y-axis. This is my y-axis, and then this is my x-axis. This is my x-axis right over here. That's my x-axis. And natural log of x-- so let's think about it a little bit. The natural log of 0, e to what power is equal to 0? Well, that's going to be 1. So you're going to have the point 1, 0 on this. So that's 1 right over there. And then the natural log of smaller and smaller numbers as we approach 0 is going to get more and more negative, all the way going down to negative infinity. So this curve is going to look something like this. It's going to look something like that. And we know it also has the point e comma 1 on it. So it has a point, the natural log of e. So if that's 1, that'll be about 2. That'll be about 3. So that is e right over here, roughly e, right over here. And that's the point e comma 1. So that's the point e comma 1 right over there. And I'll just mark that as P there to give us a label for it. So that is P. And we want to find the secant line between P and an arbitrary Q for any given x. The y value is going to be natural log of x. So let's say that this is our Q right over here. So let's say this right over here is our-- actually, I want to make it clear that it is a, I'll do it our here. I'm being a little indecisive. So let's say that is our Q right over there. And that's the point x natural log, natural log of x. And we want to find the slope of the secant line joining these two points. So this, the slope of this line, I want to try to make it so it doesn't look tangent, so it's secant. So you see it cuts through the curve. So that's the secant line right over there. It intersects the curve at P and at Q. So I want to find the slope of that secant line. Well, to find the slope of the secant line, I just need to find the change in y and the change in x between these two points. So what's the change in-- so let's be clear here. This is the point, this is when x is equal to-- well, it's just a kind of arbitrary x. And this right over here is the point lnx. So what is our change in x? Our change in x-- this value right over here, our change in x-- is just going to be x minus e is equal to x minus e. And what is our change in y? Our change in y is going to be lnx is going to be the natural log of x minus 1, minus 1. That's this distance right over here. So the slope of this line, the line that contains both of these points, the slope-- I could write m for slope-- is going to be our change in y over our change in x, which is equal to lnx minus 1 over x minus e. And now we can just input that and make sure that we got that right. So let me try to remember it-- lnx minus 1 over x minus e. So let me input that. So we have the natural log of x minus 1 over-- and it interprets it nicely for us right below, so we make sure that we're doing over x minus e. And now let us check our answer. And we got it right.