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Finding angle b/w line equation and tangent.

  • Thread starter SolCon
  • Start date
  • #1
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Greetings to all. :)
This is my first time posting here, so if I do anything wrong, just tell me. :)
On to the question.

Alright, this question came in 3 parts. I've done the first 2 parts but have no clue on doing the third part. There are two equations (below) and a constant k.

Relevant equations and data:

.Curve: xy=12
.Line 'l': 2x+y=k
.k=10
.One of the points of intersection is P(2,6).

The task is to find the angle (in degrees) between 'l' and the tangent to the curve at P.

I really don't have any idea on how this is to be done. Maybe there's some formula that must be followed but I don't know what it could be.

Thanks for any help. :)
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Greetings SolCon! Welcome to PF! :smile:

Hint: the tangent of the tangent (ie the tangent of the angle between the the tangent and the x axis) is dy/dx. :wink:
This is my first time posting here, so if I do anything wrong, just tell me. :)
:biggrin: Use more smilies! o:)
 
  • #3
HallsofIvy
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As tiny-tim says, the tangent of the angle between a curve (tangent line to the curve) and the x-axis is the derivative, dy/dx, evaluated at that point. The tangent of the angle between a line and the x-axis is, of course, the slope.

To find the tangent of the angle between two lines (the tantgent line of the curve and the given line, say) use the fact that
[tex]tan(\theta_1- \theta_2)= \frac{tan(\theta_1)- tan(\theta_2)}{1+ tan(\theta_1)tan(\theta_2)}[/tex]
(I think that's right- better check whether it is "1+ " or "1- " in the denominator.)

You can also use the dot product. The vector <1, m> lies in the same direction as line y= mx+ b or the tangent line to a curve with derivative dy/dx= m and the dot product of two vectors, u and v, is given by [itex]|u||v|cos(\theta)[/itex] where [itex]\theta[/itex] is, of course, the angle between them.
 
  • #4
33
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Thanks for the replies. :)

Okay, I used the dy/dx method to get the 'm' values but they seem to be incorrect.

For eq. xy=12, I got the point (-1,12) and from eq. 2x+y=k (10 here), I got (6,-2).
So I apply them in the slope formula: y2-y1/x2-x1 and got -2 as m1 and 1/2 as m2. Plugged them in the equation provided by HallsofIvy but did not get the right answer. What am I doing wrong?

Thanks for the help. :) (This is the only smiley I'm used to typing with keys, :) )
 
  • #5
tiny-tim
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Hi SolCon! :smile:
For eq. xy=12, I got the point (-1,12) and from eq. 2x+y=k (10 here), I got (6,-2).
What point? :confused:

(I'm missing the point! :redface:)
Thanks for the help. :) (This is the only smiley I'm used to typing with keys, :) )
PF has a button for smilies on the "QUOTE" page …

alternatively, just type [NOPARSE]:smile:[/NOPARSE] :smile:
 
  • #6
HallsofIvy
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If 2x+ y= 10, then y= 10- 2x so xy= 12 becomes [itex]x(10- 2x)= 10x- 2x^2= 12[/itex] or [itex]x^2- 5x+ 6= (x- 2)(x- 3)= 0[/itex] so the two points of intersection are (2, 6) and (3, 4). You gave (2, 6) in your original post so I have no idea why you are looking at (-1, 12) (Did you mean (1, 12)? (-1, 12) doesn't even satisfy xy= 12.). To find the angle between the line and the curve, you have to look at a point of intersection, so there is an angle between them. The "slope" calculated between a point on the line and a point on the curve is meaningless.

From xy= 12, y= 12/x so [itex]y'= -12/x^2[/itex]. At x= 2, y= 6, that is y'= -3 so the tangent line there is given by y= -3(x- 2)+ 6 or y= -3x+ 11. At x= 3, y= 4, y'= -4/3 so the tangent line is y= (-3/4)(x- 3)+ 4 or y= (-3/4)x+ 25/4.
You want to find the angle between the lines y= -3x+ 11 and y= -2x+ 10 at x= 2 or between y= (-3/4)x+ 25/4 and y= -2x+ 10 at x= 3.
 
  • #7
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Thanks for the help you two. :smile:

HallsofIvy, I had some trouble understanding the last paragraph of your last post, but the rest was quite helpful. The problem has been solved.:smile:
 

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