Tangent to Curves Problems

Problems that require students to determine the equation of a line tangent to a function in a specific point are frequent in AP Calculus AB tests. We have created a few examples to help exam takers improve their scores. Examples.

1. Some of the problems ask to find the line that is tangent to a function f(x) at point (x₀, y₀).
You have to determine the derivative f '(x) and then the slope of the line, f '(x₀).
The equation ot the line is y - y₀ = f '(x₀)[x - x₀].

Example.

2. A slightly different case is when you are given the equation of a curve in the plane (x, y), f(x, y) = 0, and you are asked to find the line of the tangent in (x₀,y₀).
You can find the slope f '(x₀) by treating y as an implicitly defined function of x and differentiating the equation with respect to x.
The equation ot the line is y - y₀ = f '(x₀)[x - x₀].

Example.

3. Another type of question may ask the coordinates of the points where the tangent to a specific curve are vertical.
By treating y as an implicitly defined function of x and differentiating the equation of the curve with respect to y, you can obtain an expression for dx/dy.
The solution of the equation dx/dy = 0 gives the coordinates of the points where the tangent to the curve are vertical.

Example.

4. In some cases the derivative of the function x is given, f '(x). The problem also specifies that f(x₀) = y₀ and asks for the equation of the line tangent to f(x) in x₁.
As long as we have f '(x), we can easily fing the slope of the line, f '(x₁).
We also need to find the value of the function at x₁:
ap, calculus,

The equation of the line is y - f(x₁) = f '(x₁)[x - x₁]

Examples of problems with lines tangent to a function

1.Write an equation for the line tangent to the curve x² + 4x + y³ = 4 at the point (1, -1).

Solution: Treating y as an implicitly defined function of x and differentiating both sides of the equation with respect to x, we obtain:
2x + 4 + 3y²y' = 0
y' = -(2x + 4)/(3y²)
At point (1,-1), y'(1) = -2
y = -1 - 2·(x-1)

2. dy/dx = (x + 1)²y
Let y = f(x) be the solution of this equation with the initial condition f(1) = 1. What is the equation of the tangent to the graph of f at x = 1?

Solution: f '(1) = (1 + 1)²1 = 4
The tangent equation is y = 1 + 4(x - 1), or
y = 4x - 3.

3. Let f be the function given by f(x) = (2x² + 5x -1)⁷.
Write the equation for the line tangent to f(x) in x = 0.

Solution: Let u(x) = (2x² + 5x -1)
Then, f '(x) = (du⁷/du)d(2x² + 5x -1)/dx = 7u⁶(4x + 5) = 7(4x + 5)(2x² + 5x -1)⁶
f '(0) = 35 and f(0) = -1.
The line of slope 35 that passes through (0,-1) is y = 35x - 1

4. Find the coordinates of the two points on the x² - 2x + 4y² + 16y + 1 = 0 closed curve where the line tangent to the curve is vertical.

Solution: Treating y as an implicitly defined function of x and differentiating both sides of the equation with respect to y, we obtain:

2xx' - 2x' + 8y + 16 = 0
x'(2x - 2) = -8(y + 2)
x' = -4(y + 2)/(x - 1) and x' should be 0 if the line tangent is vertical.
y = -2
At the points with vertical tangents, x² - 2x + 16 - 32 + 1 = 0
x² - 2x - 15 = 0
The solutions are x₁ = 5 and x₂ = -3