Cracking The Ap Calculus ab Exam 2018

EQUATIONS OF TANGENT LINES AND NORMAL LINES

Finding the equation of a line tangent to a certain curve at a certain point is a standard calculus problem.
This is because, among other things, the derivative is the slope of a tangent line to a curve at a particular
point. Thus, we can find the equation of the tangent line to a curve if we have the equation of the curve and
the point at which we want to find the tangent line. Then all we have to do is take the derivative of the
equation, plug in the x-coordinate of the point to find the slope, then use the point and the slope to find the
equation of the line. Let’s take this one step at a time.

Suppose we have a point (x 1 , y 1 ) and a slope m. Then the equation of the line through that point with that

slope is

(y  −   y 1 )   =   m(x −   x 1 )

You should remember this formula from algebra. If not, memorize it!

Next, suppose that we have an equation y = f(x), where (x 1 , y 1 ) satisfies that equation. Then f ́(x 1 ) = m,

and we can plug all of our values into the equation for a line and get the equation of the tangent line. This
is much easier to explain with a simple example.

Example 1: Find the equation of the tangent line to the curve y = 5x^2 at the point (3, 45).

First of all, notice that the point satisfies the equation: when x = 3, y = 45. Now, take the derivative of the
equation.

    =   10x

Now, if you plug in x = 3, you’ll get the slope of the curve at that point.

= 10(3) =30

Thus, we have the slope and the point, and the equation is

(y  −   45) =   30(x    −   3)

By  the way,    the notation    for plugging    in  a   point   is   .  Learn   to

recognize   it!

It’s customary to simplify the equation if it’s not too onerous.

y   =   30x −   45