equation for a tangent line?

Question

what is an equation for a tangent to the graph of y = Arctan(x/3) at the origin?

Answer 1

y = Arctan(x/3)

First, take the derivative.

dy/dx = { 1/[ (x/3)^2 + 1 ] } {1/3}

This tells us the slope at a given point x. Since we want to find the tangent line at the origin (0, 0), make x = 0. This will be our slope m.

m = { 1/[ (0/3)^2 + 1] } {1/3}
m = { 1/[0 + 1] } {1/3}
m = (1/1)(1/3) = 1/3

Now, this becomes a high school problem, where we want to find the equation of the tangent line with slope 1/3 and through
(0, 0). Using the slope formula,

(y2 - y1)/(x2 - x1) = m

Let (x1, y1) = (0, 0) and (x2 ,y2) = (x, y). Plugging these values in, along with m = 1/3, will give us the equation of the tangent line.

(y - 0) / (x - 0) = (1/3)

y/x = 1/3
y = (1/3)x

Answer 2

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 is equal 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 + b eq. 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 the curve 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 take the 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 simply take 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 + 1 eq. 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 the tangent 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 that we 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 eq. 4.5-3

Step 2: Evaluate the derivative at the point given. The problem says do it at x = 2. The derivative at that x is y'(2) = 4. That is your m. Write it down. m = 4.

Step 3: Solve for b. Remember, y = mx + b. We know what x is, what y is, and what m is. If you plug them all in, you get
5 = 4×2 + b eq. 4.5-4a
And it's trivial algebra to go from there to
b = -3 eq. 4.5-4b

Step 4: Write the equation. You know m and b now. Simply substitute them into y = mx + b. You get
y = 4x - 3 eq. 4.5-5
That's it. We're done

Answer 3

Differentiate.
y' = 1/3(1+(x/3)^2)
At (0,0), y' = 1/3
So,
y = x/3