Writing equation of tangent line AP Calculus AB free response 2011 3a
About this video
Sal solves 2011 AP Calculus AB free response 3a: writing the equation of the line tangent to F(X) = 8X^3 at X = 1/2. He finds the point (1/2, 1), uses F'(X) = 24X^2 to get a slope of 6, and solves for the intercept to get Y = 6X - 2.
You'll be able to
- find the point of tangency by evaluating the function.
- use the derivative to find the slope of a tangent line.
- write the equation of a tangent line in slope-intercept form.
Key words
- Tangent line
- A line that touches the curve at a point and has the same slope as the curve there.
- Slope-intercept form
- The equation of a line written as Y = MX + B, where M is the slope and B is the Y-intercept.
- Point-slope form
- Another way to write a line using a known point and the slope.
Transcript
- [Instructor] Let R be the region in the first quadrant enclosed by the graphs of F of X is equal to 8X to the third, and G of X is equal to sine of Pi X as shown in the figure above. And they drew the figure right over here. Part A, write the equation for the line tangent to the graph of F at X is equal to one-half.
So let me just redraw it here just so that I like drawing on the black background, I guess is the main reason why I'm redrawing it. So the function F of X is equal to 8X to the third looks like this. It looks like that. This is our F of X axis and this is our X axis.
And we want the equation for the line tangent at X is equal to one-half. So this is X is equal to one-half. If you go up here, if you evaluate F of one-half, you get F, you get eight times one-half to the third, which is eight times one-eighth, which is one. And they actually gave us that already on this point.
This is the point, one-half comma one. And we need to find the equation for the tangent line. So the tangent line will look something like that. And to figure out its equation, we just literally figure out its slope and then we know a point that it's on and we could either, you could use point slope or you could just use your kind of standard slope intercept form to give an equation for that line.
So the first part, let's figure out its slope. And the slope of the tangent line is going to be the same slope as the slope of our function at that point. Or another way to think about it, it is going to be F prime, F prime of one-half or the derivative evaluated at one-half.
The derivative gives us the slope of that line at any point. So what is F prime of X? F prime of X is just the derivative of this. So three times eight is 24 times X squared. 24X squared, F prime of one-half is equal to 24 times one-half squared, which is equal to 24 times one-fourth, which is equal to six.
So the slope of this line is equal to six. I'll use M for slope, that's what we, the convention that we use when we first learned it in algebra. So the slope is going to be six. So the general equation for this line is Y is equal to MX plus B. This is a slope, this is the Y intercept.
We already know that the slope is six and then we can use the fact that the line goes through the point one-half, one to figure out what B is. So when Y is one, one is equal to our slope times X, X is one-half, or another way to say it, when X is one-half, Y is one and plus some Y intercept.
Or if I take X is one-half, I multiply it times the slope plus the Y intercept, I should get one. And so I get one is equal to three plus B. I can subtract three from both sides and I get negative two is equal to B. So the equation of the line is going to be Y is equal to 6X minus two.
That is the equation of the tangent line.
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