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## Calculus, all content (2017 edition)

### Unit 2: Lesson 3

Derivative as instantaneous rate of change

# Tangent slope as instantaneous rate of change

Sal finds the average rate of change of a curve over several intervals, and uses one of them to approximate the slope of a line tangent to the curve. Created by Sal Khan.

## Want to join the conversation?

• • Have anyone notice that the secant line in the interval [6.5,7.5] is equal to the sum of the secant line in the intervals [6.5,7] and [7,7.5] divided by two?
Is this a property of derivatives or slopes? • •   Most people can "get" the problem more easily by observing what the curve or line or triangle looks like. They can then label angles and foci and xnought, vertex and whatever, and from that basis, they can more easily set up the necessary equations.

But I know that SOME people can figure out characteristics of functions like x³-6x²+5x -2 in their heads and can describe the transformations, and announce where the vertex will be if the directrix of x²+4x -3 is increased by 2, or what the instantaneous rate of change will be when x = -2. If you can do that, awesome! You don't need to draw graphs.

Or you don't need to draw graphs until you try to explain and justify your method of solution to someone without your gifts. Also, teachers and professors usually require it so that you can prove that you know the significance of the numbers and formulas.

I don't know of a profession where you don't need to show results to someone else in your company, or to potential clients or peers in academia, and often the most interesting thing is increase over time, maximum profit, rate of decline or decay, etc. --exactly this stuff that Sal is teaching us--unless you are planning to stack firewood or something.
• At Sal did an incomplete equation. Isn't the equation in the form of y = mx + c? • The equation of a line can be written several different ways. The form you mention is called "slope-intercept form," because we can easily read off the slope (m) and the y-intercept (c in your version, although it's more usual to use b for the y-intercept).

This problem requests the equation in a different form called "point-slope form." This form allows us to read off the slope (1.9) and the coordinates of a point through which the line passes (7, 109.45). This equation is not incomplete. With a little algebraic manipulation it can be converted into slope-intercept form, but an equation in that form was not requested for this problem.
• Why do we call it a secant line ? Why not cosine line or sine line ? • That makes sense with one definition of a secant. But in fact there are two definitions. The one you refer to is the secant trig identity. The other is a line that intersects the perimeter of a circle at exactly two points but is not the diameter. The second definition is the one referred to in the name secant line. When a line intersects two points on a curve it creates a kind of secant line.
• At , isn't point-slope form y=m(x-h)+k? • Why does he always say y=f(x) on the y axis? is this true? I thought f(x) is the function of x, why can we say the y-axis is equal to this function? • If I say y=x², and then I ask, "if x=2, what is y and what point on the x-y graph should I plot?", what would be your reply?
Hopefully your reply would be "y=4 and the point to plot would be where x=2 and y=4, or, (2, 4)."

Now, if I say f(x)=x², and then I ask, "if x=2, what is f(x) and what point of the x-y graph should I plot?", hopefully you would say "f(x)=4 and the point to plot would be where x=2 and y=4, or, (2, 4)."

In both cases the value of y and of f(x) are dependent on what x is. In the f(x) =x² case, the function is given a name, f. In the y=x² case, it is still a function, but it has no name. So if you have some function of x, lets call it g, and you say, "y=g(x)" all you are doing is associating the output of g(x) to a value on the y axis so that the graph of the function can be plotted.
• • I think you're referring to the y's in the point-slope form of the equation seen at . The general form of an equation in point-slope form is y - y1 = m(x - x1) where m is the slope and (x1,y1) is the point. Our point is (7,109.45) and the slope is the average slope between [6.5,7.5] which is 1.9. Plug these into the equation and you get an approximation of the equation of a tangent line at (7,109.45).
• does this video means we are using triangles to solve the problems? • I do understand that the gradient of a line is always taken as (rise/run), (delta y/delta x) and so on. However, even after knowing this formula for many years, I have never understood as to why DELTA Y is the numerator and not the denominator. Any help? Thanks
(1 vote) • That is because y is the dependent variable and you need to do something to x to convert it to y.
Suppose that x represents time in seconds and y represents distance in meters. Then, a linear relationship would look have units like this:
y in meters = [(∆y in meters) / (∆x in sec ] ∙ (x in sec) + (y intercept in meters)
Notice that the unit "seconds" cancels out. But if we had the gradient with the x in the numerator and the y in the denominator, then the units wouldn't work out. So, you couldn't use that.

Also, when there is 0 slope or gradient, then if you had the ∆y in the denominator, you would get division by 0 and not be able to solve problems involving horizontal lines.