Finding exact point at this tanget line of a graph

In summary, the homework statement asks for the slope of the tangent line at x=0. Using implicit differentiation, the student found that the slope is positive and therefore safe to say. However, the slope may not be correct as it was calculated using arctan 1, which is not the correct function to use.
  • #1
Flappy
25
0

Homework Statement


Given the equation: x - cos^2(x^2 - y^2) = xy - sin^2(x^2 - y^2)

Here's a graph of this equation: http://img240.imageshack.us/my.php?image=graphtq8.png

What is the exact point on the tangent line?
And what is the slope of the tangent line?

The Attempt at a Solution



I know since x=0 at the point, I would need to find y when x=0. But this is confusing me a bit since when I plug in 0 I'm left with -cos(-y^2) = -sin^2(-y^2). And I don't really know when to go from there. I also figure I am going to need to derive the equation by implicit differentiation to find the slope I assume. I'm just looking for some suggestions.
 
Physics news on Phys.org
  • #2
well, now you need to find the y-intercept right.
you are left with sth like this, after u plug in 0 for x.
[tex]cos^{2}(-y^{2})=sin^{2}(-y^{2})[/tex] since sine is squared we can get rid of the minus sign, for cosine we know it is even so it does not matter

[tex]cos^{2}(y^{2})=sin^{2}(y^{2})[/tex] divide by [tex]cos^{2}(y^{2})[/tex], but we lose solutions here for y^2=kpi , so
[tex]tan^{2}(y^{2})=1[/tex] from here we get [tex]tan(y^{2})=+-1[/tex], but since y^2 can never be negative we get [tex]tan(y^{2})=1[/tex], now
[tex]y^{2}=arctan{1}[/tex]
so du u see now?
 
  • #3
How is the question stated? You said there is a tangent line, but it's equation is not given, only that it appears to be a tangent line at x=0.

You can find the exact point on the graph at x=0 by simply plugging x=0 into the equation of graph. You'll get a trigo equation, which you must then solve for a general solution in terms of y. The exact point on the tangent line would then be the first instance where y is negative according to the general solution.

You have to use implicit differentiation here. Just treat y as a function of x. Once you get [tex]-cos^{2}(-y^{2}) = -sin^2(-y^2)[/tex], which you already did, then note that sine is an odd function which means you can simplify RHS and solve for y to get the general solution.
 
  • #4
I do not think he will need to implicitly differentiate at all.-That is what i first thought though. because he just needs the y value for which x=0, so basically that will be
[tex]y=-sqrt(pi)/2[/tex]
 
  • #5
He'll need to differentiate it implicitly because he's also asking for the slope of the tangent line.
 
  • #6
I did give the equation on the first line. The equation of the tangent line isn't given, they just want to know the coordinates at that point on the graph and also the slope of the tangent line.

Anyways, the sqrt of the arctan(1) gave me about -.8662269255, which seems reasonable by looking at the graph.
 
  • #7
You shouldn't use the calculated value. Arctan 1 = pi/4 Square root that for the exact expression.
 
  • #8
Defennnder said:
He'll need to differentiate it implicitly because he's also asking for the slope of the tangent line.

Sorryyyyy! My bad, i did not look at the second question at all, i just missed it, that one about the slope. So you are right, he defenitely will need to differentiate implicitly.
 
  • #9
I see, thanks for the correction.

Anyways, after using implicit on the equation I ended up with: y` = [tex]\frac{1}{xy - 2x + 2y}[/tex]

So is it safe to say the slope is [tex]\frac{1}{2(-\sqrt{pi/4})}[/tex] ?
 
  • #10
For the op: after you differentiate it implicityly, you need to bring on one side dy/dx, while on the other side the rest. After that you can easily find the slope of the tangent line, just plug in x=0 and y=-sqrt(pi)/2
 
  • #11
Yep, if you did the calculations correctly when you differentiated implicitly, then i guess that is safe to say so!
 
  • #12
Flappy said:
I see, thanks for the correction.

Anyways, after using implicit on the equation I ended up with: y` = [tex]\frac{1}{xy - 2x + 2y}[/tex]

So is it safe to say the slope is [tex]\frac{1}{2(-\sqrt{pi/4})}[/tex] ?
I have this feeling that you have done some mistakes somewhere, because by just looking at the graph you provided, one could tell that the slope of the tangent line at that point is positive. So, look somewhere if you haven't missed some minus sign somewhere.
 
  • #13
sutupidmath said:
Yep, if you did the calculations correctly when you differentiated implicitly, then i guess that is safe to say so!


Edit: i guess it is not safe at this case!
 
  • #14
Hmm, this is the work i did for getting the derivative

d/dx [ x - cos^2(x^2 - y^2) ] = d/dx [xy - sin^2(x^2 - y^2)]

1 + sin^2(x^2 - y^2)*(2x-2y)*y` = x * y` * y - cos^2(x^2 - y^2)*(2x-2y)*y`

1 = x * y` * y - cos^2(x^2 - y^2)*(2x-2y)*y` - sin^2(x^2 - y^2)*(2x-2y)*y`

1 = y`[ xy - cos^2(x^2 - y^2)*(2x-2y) - sin^2(x^2 - y^2)*(2x-2y)]

1 = y`[xy - (2x-2y)( cos^2(x^2 - y^2) + sin^2(x^2 - y^2) ) ]

1 = y`[xy - (2x-2y)(1)]
1 = y`[xy - 2x +2y]
y` = (1)/(xy - 2x +2y)

Can you spot my mistake by any chance?
 
  • #15
Flappy said:
Hmm, this is the work i did for getting the derivative

d/dx [ x - cos^2(x^2 - y^2) ] = d/dx [xy - sin^2(x^2 - y^2)]

1 + sin^2(x^2 - y^2)*(2x-2y)*y` = x * y` * y - cos^2(x^2 - y^2)*(2x-2y)*y`



Can you spot my mistake by any chance?
Here it is the first mistake i can spot, there might be others, but i will show you the idea

well think of it this way say u=u(x) that is u is a dependent variable, then

[cos^2(u)]'=[cos(u)cos(u)]'=[cos(u)]'cos(u)+cos(u)[cos(u)]'=-sin(u)u'cos(u)+cos(u)(-sin(u)u' =-2cos(u)sin(u)u'

i hope this helps!
By the way, your mistake is on the second line!
 
Last edited:

Related to Finding exact point at this tanget line of a graph

1. How do you find the exact point at the tangent line of a graph?

To find the exact point at the tangent line, you need to first identify the slope of the tangent line by taking the derivative of the function at the given point. Then, using the point-slope formula, plug in the slope and the coordinates of the given point to find the equation of the tangent line. Finally, solve for the x and y values of the point of intersection between the tangent line and the graph.

2. Can you explain the concept of tangent lines?

A tangent line is a straight line that touches a curve at only one point, without crossing over or intersecting the curve. At this point of contact, the tangent line has the same slope as the curve, allowing us to approximate the behavior of the curve at that specific point.

3. How is the slope of the tangent line related to the derivative?

The slope of the tangent line is equal to the value of the derivative at that point on the curve. The derivative measures the rate of change of the function at a given point, which is also the slope of the tangent line at that point.

4. What is the point-slope formula and how is it used to find the equation of a tangent line?

The point-slope formula is y-y1 = m(x-x1), where (x1,y1) is a point on the line and m is the slope of the line. To find the equation of a tangent line, we first take the derivative of the function at the given point to find the slope (m). Then, we plug in the x and y values of the given point into the formula to find the equation of the tangent line.

5. Are there any other methods for finding the exact point at the tangent line of a graph?

Yes, there are other methods such as using the limit definition of the derivative, using the slope-intercept form of a line, or using calculus techniques such as L'Hopital's rule. However, the point-slope formula is the most commonly used method and is usually the most efficient way to find the exact point at the tangent line of a graph.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
860
  • Calculus and Beyond Homework Help
Replies
1
Views
615
  • Calculus and Beyond Homework Help
Replies
3
Views
399
  • Calculus and Beyond Homework Help
Replies
2
Views
857
Replies
1
Views
583
  • Calculus and Beyond Homework Help
Replies
1
Views
446
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
381
  • Calculus and Beyond Homework Help
Replies
6
Views
741
  • Calculus and Beyond Homework Help
Replies
8
Views
645
Back
Top