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Expert: Steve Holleran - 3/6/2008


Question

The question and diagr
Hi, I'm having real trouble figuring out what to do for this question, can you please help. Attached is THE question.  

Answer
Hi Jaime,

Okay, let's see what we have here.

Since the graph passes through (0,0) and (pi/3, 2), their coordinates must satisfy the equation:

  (0,0) :  0 = a sec(0) - b -->  0 = a * 1 - b--> a - b = 0

(pi/3, 2):  2 = a sec(pi/3) - b --> 2 = a * 2 - b --> 2a - b = 2

Solving, multiply the first one by -1 and add:

          -a + b = 0

          2a - b = 2
        ---------------

            a = 2   and then you get b = 2.

The equation is y = 2 sec x - 2.

Then y' = 2 sec x tan x

y'(pi/3) = 2 sec(pi/3) tan(pi/3) = 2 * 2 * sqrt3= 4 sqrt(3)=slope of the tangent at (pi/3, 2).

Therefore the slope of the normal at that point is -1/(4sqrt(3))

So an equation for the normal at (pi/3, 2) is

                  y = (-1/4sqrt(3)) * x + b

so                 2 = -1/ 4sqrt(3) * (pi/3) + b

and b = 2 + [pi/(12sqrt(3))]

So, point C has y-coordinate  2 + [pi/(12sqrt(3))] = 2.1511

To find the point on the curve with this y-coordinate,

                    2.1511 = 2 sec x - 2

so                   4.1511 = 2 sec x

and                     sec x = 2.07555

and                     cos x = 1/2.07555 = 0.4818

and                   x = Arccos(.4818) = 1.068

So then AB = 2 * 1.068 = 2.1362

This is my best shot!

Hope its okay.
Steve
    Questioner's Rating
    Rating(1-10)Knowledgeability = 10Clarity of Response = 8Politeness = 10
    CommentThanks so much Steve that question was so hard, i didn't even know where to start. Thanks.


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Steve Holleran

Expertise

I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

Experience

33 years teaching experience in NJ public schools

Education/Credentials
B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

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