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

In summary, the Homework Equations are y = tan x at point (pi/4,1), y = tan x at point (1/sqrt(2), 1/sqrt(2)), and sec(pi/4) = 1*(1/2√2)
  • #1
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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  • #2
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:
  • #3
You miscalculated sec(pi/4).

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

What is cos(pi/4)?
 
  • #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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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)?
 
  • #8
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.
 

1. What is a tangent line?

A tangent line is a line that touches a curve at a single point and has the same slope as the curve at that point.

2. How do you find the equation of a tangent line?

To find the equation of a tangent line, you need to find the slope of the curve at the given point and then use that slope and the given point to write the equation in point-slope form.

3. What information do you need to find the equation of a tangent line?

You need to know the coordinates of the point at which the tangent line touches the curve and the slope of the curve at that point.

4. Can you use the derivative to find the equation of a tangent line?

Yes, you can use the derivative to find the slope of the curve at a given point, which is necessary for finding the equation of the tangent line.

5. Why is the equation of a tangent line important in calculus?

The equation of a tangent line is important in calculus because it allows us to approximate the behavior of a curve at a specific point. This is useful in many applications, including optimization problems and curve sketching.

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