A line that is tangent to a curve has two properties:
The line shares a point with the curve in question.
At the shared point, the derivative of the curve isequal to the slope of the line.
If you solve for those two properties, you have completed the problem.
The equation of a line is, as you learned in algebra
y = mx + beq. 4.5-1
So all you have to do is find m and b.
Let's take the second property first. The slope of the line is m. That value has to match the derivative of thecurve at the point they give you. So for finding m you need to know the derivative of the curve. And you need to evaluate it at the point they give in the problem. And that value is m.
Then takethe first property. You need to find the b that makes that first property true. And you now know what m is equal to. If, for example, the point they want you to be tangent to is (2, 3), then simplytake y = mx + b, put in 2 for x, 3 for y, whatever you came up with for m, and solve for b. It's that easy.
Let's run through an example. Let the curve be
y(x) = x2 + 1eq. 4.5-2
Usually it's understood that y is a function of x, so very often this would be written as y = x2 + 1. It means exactly the same thing.
Let's find thetangent at x = 2. So what point is that? We know the x-coordinate of the point. To find the y-coordinate, simply use the equation of the curve (given in 4.5-2). That gives us y = 5, so the point thatwe want our line to be tangent at is (2, 5).
Step 1: Find the derivative of the curve. The equation of the curve is given in 4.5-2. You know how to take its derivative. It's
y'(x) = 2x...