Tangents to Curves: Intersection in First Quadrant

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In summary, there are two equations with two unknowns: xy = 1 and x^2 - y^2 = 1. The derivatives of these functions are dy/dx = -y/x and dy/dx = x/y. The product of these derivatives for a given (x,y) is -1. In the first quadrant, the tangents to these curves intersect at two points, but the original curves do not intersect. To find the points of intersection, you can solve the system of equations xy = 1 and x^2 - y^2 = 1.
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mstudent123
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Homework Statement



Is there anything special about the tangents to the curves xy=1 and (x^2) - (y^2) =1 at their point of intersection in the first quadrant.

The Attempt at a Solution



I know what the derivatives of both functions are and what they look like when graphed. But, I'm not sure what to do with it beyond that. For the first function I have that dy/dx=-1/x^2 and for the second I have dy/dx = x/sqrt(-1+x^2). What is different about these functions and what happens in the first quadrant?
 
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  • #2


mstudent123 said:

Homework Statement



Is there anything special about the tangents to the curves xy=1 and (x^2) - (y^2) =1 at their point of intersection in the first quadrant.

The Attempt at a Solution



I know what the derivatives of both functions are and what they look like when graphed. But, I'm not sure what to do with it beyond that. For the first function I have that dy/dx=-1/x^2 and for the second I have dy/dx = x/sqrt(-1+x^2). What is different about these functions and what happens in the first quadrant?

Well, according to the problem statement, the functions intersect there. Seems to me that it would be useful to find the point of intersection.
 
  • #3


I have been trying to do that. However, I have not been able to find the intersection of the tangents and there is not intersection of the original functions.
 
  • #4


mstudent123 said:
I have been trying to do that. However, I have not been able to find the intersection of the tangents and there is not intersection of the original functions.

There are in fact two points of intersection. Why do you think there are none?

You need to find (x,y) that satisfies both xy = 1 and x^2 - y^2 = 1. How do you solve two equations with two unknowns?
 
  • #5


You don't need to find where they intersect.

With xy= 1, xy'+ y= 0 so y'= -y/x. With [itex]x^2- y^2= 1[/itex], [itex]2x- 2yy'= 0[/itex] so y'= x/y. Without worrying about what the "points of intersection" are, what is the product of thos two derivatives for a given (x, y)?
 
  • #6


Thank you!
 

1. What is a tangent to a curve?

A tangent to a curve is a straight line that touches the curve at only one point, called the point of tangency. It represents the instantaneous rate of change or slope of the curve at that particular point.

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

To find the equation of a tangent to a curve at a given point, you can use the slope-point form of a line. First, find the slope of the curve at the given point by taking the derivative of the curve. Then, substitute the coordinates of the point and the slope into the equation y-y1=m(x-x1), where (x1, y1) is the point of tangency and m is the slope.

3. What is the difference between a tangent and a secant?

A tangent is a line that touches the curve at one point, while a secant is a line that intersects the curve at two points. A tangent represents the instantaneous rate of change, while a secant represents the average rate of change between two points.

4. Can a curve have more than one tangent at a point?

Yes, a curve can have more than one tangent at a point if the curve has a sharp turn or breaks in continuity at that point. In these cases, the curve has two or more tangents with different slopes at the same point.

5. How are tangents to a curve related to the derivative?

The derivative of a curve at a given point is equal to the slope of the tangent to the curve at that point. This means that if you know the derivative of a curve at a point, you can find the equation of the tangent at that point using the slope-point form of a line.

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