Expert: Paul Klarreich - 2/28/2006

Question
My name is Jake and i am studying rates of change in calculus.  The question i am having troubles with is as follows "Let P be any point in the first quadrant on the function f(x)=1/2x.  The tangent at P meets the x-axis at A and the y-axis at B.  If 0 is the origin, show that the area of triangle AOB is 1."

I first attempted this question by using formula 2
[f(a+h)-f(a)/h] to get the slop of the tangent at P. I did not know how to solve this part or go any further.

Answer
Hi, Jake,

You wrote:
Subject:  Rate of change
Question:  My name is Jake and i am studying rates of change in calculus. The question i am having troubles with is as follows "Let P be any point in the first quadrant on the function f(x)=1/2x. The tangent at P meets the x-axis at A and the y-axis at B. If 0 is the origin, show that the area of triangle AOB is 1."

I first attempted this question by using formula 2
[f(a+h)-f(a)/h] to get the slop of the tangent at P. I did not know how to solve this part or go any further.
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WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


I will assume your function, written 1/2x, is:
       1
f(x) = ---
      2x

You are interested in the tangent line.  You will have plenty of these problems and the 'recipe' is:

To find the equation of the tangent line to  y = f(x) at x = x0.

1. Find the derivative of the function, f'(x).

2. At the given point, (Don't worry that you don't have one, yet.) substitute  x0 into f'(x) to get the slope, m = f'(x0).

3. Find the value of  y0 = f(x0).

4. Combine m,x0,y0 into the  point-slope form of the equation of a straight line:  y - y0 = m(x - x0)

4.1 Simplify as appropriate. (Season to taste?)

To simplify the typing, I will write  x0 = c.  Then y0 = 1/2cYour line has a slope equal to the derivative at the point (c, 1/2c).

So first find  f'(x) = -1/(2x^2).

At the point (c,1/2c), the slope is:

f'(c) = -1/(2c^2)

So the equation is:

y - y0 = m(x - x0)
    1      -1
y - ---  = ----(x - c)
   2c     2c^2

That will do, for now.  About the problem: show that the area of triangle AOB is 1, where A is the x-intercept and B is the y-intercept.

What is A?  The x-intercept is the value of x when y = 0:

    1      -1
0 - ---  = ----(A - c)
   2c     2c^2

    1      1
  ---  = ----(A - c)
   2c     2c^2

        1
  1  = ----(A - c)
        c
A - c = c

A = 2c



What is B? The y-intercept is the value of y when x = 0:

    1      -1
B - ---  = ----(0 - c)
   2c     2c^2

    1      c
B - ---  = ----
   2c     2c^2

    1      1
B - ---  = ---
   2c     2c

    1     1     2     1
B = --- + --- = --- = ---
   2c    2c    2c     c


Now what is the area of the triangle?  The two legs are OA and OB, with length A and B, so the area is:
                 1      1
Area = (1/2)AB = ---(2c)--- = 1
                 2      c