mathgeek22
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Let P be a point in the intersection of the curves x^2-y^2=c and xy=d, where c and d are constants. Show that the tangents to the curve at P are perpendicular.
Tangent line: Let P be a point in the intersection of the curves x^2-y^2=c and xy=d.
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jonah2.0
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Beer soaked ramblings follow.
You show me what you've done so far first and maybe I'll show you what I think.
Steven G
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Can you please show us the work you have done so far and tell where you are stuck?
Have you found the point(s)of intersection (in terms of c and d)? Have you found the slope of the tangent lines at point P for both curves? Just do that and say wow, the slopes are perpendicular.
mathgeek22
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Hello, So far,
I calculated the derivative of x^2-y^2=c using implicit diff.
I got 2x-2y(dy/dx)=0
x/y=y '
Next I calculate the derivative of xy=d. I did y=d/x and then used the quotient rule to solve the derivative...
Then I got y'=-d/x^2
Now, I am confused as to where I should proceed next. The two slopes that I got aren't perpendicular. I am also confused as to how to find the slope of Point P.
HallsofIvy
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You don't need to find the point of intersection! All you need is the fact that the x and y values are the same for each curve.
Yes, if \(\displaystyle x^2- y^2= e\) then \(\displaystyle 2x- 2yy'= 0\) so \(\displaystyle y'= \frac{x}{y}\). If \(\displaystyle xy= d\) then instead of solving for y I would do the same as before: \(\displaystyle y+ xy'= 0\) so \(\displaystyle y'= -\frac{y}{x}\). Now the result is obvious.
Yes, if \(\displaystyle x^2- y^2= e\) then \(\displaystyle 2x- 2yy'= 0\) so \(\displaystyle y'= \frac{x}{y}\). If \(\displaystyle xy= d\) then instead of solving for y I would do the same as before: \(\displaystyle y+ xy'= 0\) so \(\displaystyle y'= -\frac{y}{x}\). Now the result is obvious.
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Steven G
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The two slopes that I got aren't perpendicular. Ok, so what is the product of your two slopes? You really do have to answer this before you get any more help.
Where did you get \(\displaystyle y+ xy'= 0\)?
topsquark
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This is an example of implicit derivatives, along with the product rule.
\(\displaystyle x y = d\)
Taking the derivative wrt x:
\(\displaystyle \dfrac{d}{dx} (x y) = \dfrac{d}{dx} (d)\)
Product rule on the LHS:
\(\displaystyle \dfrac{d}{dx} (x) \cdot y + x \cdot \dfrac{d}{dx} (y) = 0\)
What do you do next?
-Dan
Thank you.
\(\displaystyle y+ xy'= 0\)
then
\(\displaystyle y'= -\frac{y}{x}\)
but how can i use this to prove that the tangent lines are perpendicular?
Bob Brown MSEE
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Move to origin
Move point (a,b) to (0,0)
CLICK HERE
then throw out quadratic terms (limit near (0,0))
get {y=(a/b)x, y=(-b/a)x}
NOTE:
Obtain similar result by
Move point (-a,-b) to (0,0)
or simply use symmetry arguments
EXAMPLE:
To see example let (a,b)=(2,1)
CLICK HERE
Let c=a^2-b^2 and d=ab
Move point (a,b) to (0,0)
CLICK HERE
then throw out quadratic terms (limit near (0,0))
get {y=(a/b)x, y=(-b/a)x}
NOTE:
Obtain similar result by
Move point (-a,-b) to (0,0)
or simply use symmetry arguments
EXAMPLE:
To see example let (a,b)=(2,1)
CLICK HERE
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jonah2.0
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Beer soaked ramblings follow.
Yes.
And without using chain rule? I am wondering if there is a visual proof? Can't see anything yet.
Yes, but it is a little less convenient.
I am going to assume that \(\displaystyle d \ne 0.\)
Now we can turn \(\displaystyle x^2 - y^2 = c\) into a function by considering two cases.
\(\displaystyle \text {CASE I: } y \ge 0.\)
\(\displaystyle \therefore x^2 - y^2 = c \implies y^2 = x^2 - c \implies y = \sqrt{x^2 - c} = (x^2 - c)^{1/2} \implies \\
\dfrac{dy}{dx} = \dfrac{1}{2} * (x^2 - c)^{\{(1/2)-1\}} * 2x = \dfrac{x}{\sqrt{x^2 - c}} = \dfrac{x}{y}.\)
\(\displaystyle \text { And } xy = d \ne 0 \implies y = \dfrac{d}{x} \implies \dfrac{dy}{dx} = -\ \dfrac{d}{x^2} = \dfrac{-\ xy}{x^2} = \dfrac{-\ y}{x}.\)
\(\displaystyle \dfrac{x}{y} * \dfrac{-\ y}{x} = -\ 1.\)
Now you do \(\displaystyle \text {CASE II: } y < 0.\)
EDIT: You do have to use the chain rule. How else are you going to deal with \(\displaystyle \sqrt{x^2 - c}\)?
As for visualizing it, I suggest you pick some nice numbers for c and d (try c = 0 and d = 2) , and then graph the two curves and the tangents to each at a point of intersection. As Halls said, because x and y are the same for both curves at an intersection, the result is rather obvious once you calculate the derivatives involved. Why do you need to visualize it?
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Bob Brown MSEE
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without calculus
1) move origin (0,0) to the point of intersection.
2) set non-linear terms to zero, leaving 2 intersecting straight lines.
3) compare slopes of the 2 intersecting straight lines.
My proof (above) uses no differentiation.
1) move origin (0,0) to the point of intersection.
2) set non-linear terms to zero, leaving 2 intersecting straight lines.
3) compare slopes of the 2 intersecting straight lines.
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So do you mean something like this?
If (a,b) is the point of intersection of xy=d, x^2-y^2 = k
We can translate each curve to the origin.
so we get
(x+a)(y+b) = d and (x+a)^2 - (y+b)^2 =k
expanding (x+a)(y+b) = d gives xy+bx+ay+ab-d=0, disregarding higher powers
gives equation of tangent at the origin y= -b/a x + (d-ab)/a , so gradient is -b/a
Repeating for (x+a)^2 - (y+b)^2 =k gives a gradient of a/b
Hence perpendicular?
Would we be happy with this reasoning?
Bob Brown MSEE
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Re
Yes, I believe you have the idea.
1)
The reasoning for disregarding higher powers,
is that they become zero much faster than the linear terms
as we look at the functions as they pass through (0,0).
This is true only for polynomials that pass through (0,0).
This is true only at the point (0,0).
Therefore we needed to shift the intersection (a,b) to (0,0).
2)
Translation of a figure preserves the slope at every point.
Yes, I believe you have the idea.
1)
The reasoning for disregarding higher powers,
is that they become zero much faster than the linear terms
as we look at the functions as they pass through (0,0).
This is true only for polynomials that pass through (0,0).
This is true only at the point (0,0).
Therefore we needed to shift the intersection (a,b) to (0,0).
2)
Translation of a figure preserves the slope at every point.
Thank you! It also appears that the tangents at the point of intersection of y=tan x and y= cosx also are perpendicular to each other. I think i know how to prove this using calculus. But can the approach above be used for this situation? Above you say it only works for polynomials?
Bob Brown MSEE
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Power Series
Since you will be throwing away the terms that are higher order,
it doesn't matter that there are an infinite number of them.
However, you will have to deal with regions of convergence.
Power series expansion in one variable will seldom save you as
much complexity as we do here in this implicit differentiation problem.
In your cos&tan case, the translation of the intersection to (0,0) is
complicated by approximating irrational numbers.
Since you will be throwing away the terms that are higher order,
it doesn't matter that there are an infinite number of them.
However, you will have to deal with regions of convergence.
Power series expansion in one variable will seldom save you as
much complexity as we do here in this implicit differentiation problem.
In your cos&tan case, the translation of the intersection to (0,0) is
complicated by approximating irrational numbers.
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Thanks. Steer clear then!