Find an equation of the tangent line to the curve at the given point.

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Homework Help Overview

The discussion revolves around finding the equation of the tangent line to the curve defined by y = tan(x) at the point (π/4, 1). Participants are exploring the necessary steps to determine the slope of the tangent line and the implications of trigonometric functions in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of finding the derivative of tan(x) and its implications for determining the slope at the given point. There are attempts to clarify the calculation of sec^2(π/4) and its relationship to the tangent line's slope. Some participants express confusion regarding the correct coordinates and the values of trigonometric functions at π/4.

Discussion Status

The conversation is ongoing, with participants providing insights and corrections regarding the calculations of trigonometric functions. There is a recognition of miscalculations and misunderstandings about the coordinates associated with the angle, indicating a productive exploration of the topic.

Contextual Notes

There are indications of confusion regarding the relationship between angles and their corresponding coordinates on the unit circle, as well as the proper use of trigonometric functions in calculations. Some participants question the accuracy of their previous assumptions and calculations.

bobsmith76
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Homework Statement



Find an equation of the tangent line to the curve at the given point.

y = tan x at point (pi/4,1)


Homework Equations





The Attempt at a Solution



step 1. find the derivative of tan x, which sec^2 x
step 2. find the slope.
this is where I mess up. I assume that I have to square x where sec is pi/4. Sec^2 of pi/4 I think would be the hypotenuse/x which is 1/2√2. So I square that, which to me would be 1/ (4*2) or 1/8. But the solution manual says it's the square of 2√2. why? Obviously I don't know how to square secant.

step 3. insert the slope into the y - y = m(x - x) formula, which I can do.

In the solution manual below, they don't have negative signs, so there's a few negative signs missing but I know where they are.
Screenshot2012-01-25at35634PM.png
 
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Ok, I got a new idea. The slop at pi/4,1, the coordinates are 2√2, 2√2. So maybe I should just find the slope of that, don't worry about squaring it. Well, the slop of that coordinate is 2 root 2, divided by 2 root 2, which is 1, not 2 like the book says. still lost.

(5 minutes later)

No, that can't be right, the book clearly says that you find the slope by sec^2 pi/4 = the square root of two squared. still don't see why.
 
Last edited:
You miscalculated sec(pi/4).

sec(x) = 1/cos(x).

What is cos(pi/4)?
 
at pi/4, the coordinates are (2√2,2√2), so the cosine would be x/hyp, which is 2√2/1 and the secant would be

1/(2√2/1), if you multiply the numerator by the inverse of the denominator you get:

1 * (1/2√2) and if you square that you get 1/2.

when i put in my calculator cos^-1 3.14/4 i get .67 radians.
 
bobsmith76 said:
at pi/4, the coordinates are (2√2,2√2), so the cosine would be x/hyp, which is 2√2/1

You should be able to see immediately that this is wrong. The cosine (and sine) of any angle must be between -1 and 1.

You have the coordinates wrong.
 
now I got it. i still have troubles getting the concepts of trig right. i was confusing the coordinates with the angle.

cos (45) = .52, sec (45) = 1.92 = square that and you almost get 4, so I guess it's right.

thanks for your help.
 
I'm not sure what you mean by almost 4. That's not the answer I got for the slope. Your method in the previous post looked correct, you just got the coordinates wrong.

At pi/4, the unit circle coordinates are (1/sqrt(2), 1/sqrt(2)).

So what's sec(pi/4)?
 
P.S. I'm going to go out on a limb and guess that you plugged 45 degrees into cos while your calculator was in radians mode. Oops.

This is really easy to do without a calculator, give it a try.
 

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