Find the points on the lemniscate, given below, where the tangent is horizontal
2(x^2 + y^2)^2 = 49(x^2 − y^2)
find where dy/dx = 0
4 (x^2 + y^2)(2x + 2 y dy/dx) = 49 (2 x - 2 y dy/dx)
when dy/dx = 0
4(x^2+y^2)(2x) = 49(2x)
x^2+y^2 = 49/4 = r^2 it is a circle
center at (0,0) , r = 7/2
i need four points how to find?
thanks
Find the points on the lemniscate where the tangent is horizontal. (Order your answers from smallest to largest x, then from smallest to largest y.)
2(x2 + y2)2 = 49(x2 − y2)
Why did the lemniscate get dizzy? Because its tangent went for a wild spin!
To find the points on the lemniscate where the tangent is horizontal, let's differentiate the equation with respect to x:
d/dx [2(x^2 + y^2)^2] = d/dx [49(x^2 − y^2)]
Applying the chain rule and simplifying, we get:
4(x^2 + y^2)(2x + 2yy') = 49(2x - 2yy')
Since we want the tangent to be horizontal, the slope of the tangent (dy/dx) should be zero. Thus, we can set the equation equal to zero:
4(x^2 + y^2)(2x) = 49(2x)
8x(x^2 + y^2) = 49(2x)
Simplifying further, we can divide both sides by 2x (assuming x ≠ 0):
4(x^2 + y^2) = 49
Rearranging the equation, we get:
x^2 + y^2 = 49/4
This shows that all the points on the lemniscate where the tangent is horizontal lie on the circle with radius r = sqrt(49/4) = 7/2. The center of this circle is at the origin (0, 0).
So, the points on the lemniscate where the tangent is horizontal are located on the circle with a radius of 7/2 centered at the origin (0, 0).
To find the points on the lemniscate where the tangent is horizontal, we need to determine the values of x and y that satisfy two conditions:
1. The equation of the lemniscate: 2(x^2 + y^2)^2 = 49(x^2 − y^2)
2. The slope of the tangent line is 0 (since we want a horizontal tangent)
Let's break down the steps to find these points:
Step 1: Find the derivative of the lemniscate equation with respect to x.
The derivative will allow us to find the slope of the tangent line at any point on the lemniscate.
Differentiating both sides of the equation with respect to x, we get:
d/dx [2(x^2 + y^2)^2] = d/dx [49(x^2 − y^2)]
Expanding the derivatives using the chain rule:
4(x^2 + y^2)(2x + 2y(dy/dx)) = 49(2x - 2y(dy/dx))
Simplifying, we get:
4(x^2 + y^2)(2x + 2y(dy/dx)) = 49(2x - 2y(dy/dx))
Step 2: Set the derivative equal to 0 and solve for dy/dx.
Since we are looking for the points where the tangent is horizontal (slope is 0),
we can set the derivative equal to 0 and solve for dy/dx.
Setting the derivative equal to 0, we have:
4(x^2 + y^2)(2x + 2y(dy/dx)) = 49(2x - 2y(dy/dx))
Simplifying further:
4(x^2 + y^2)(2x) = 49(2x)
Canceling out the common factors:
2(x^2 + y^2) = 49
Step 3: Substitute the previous result into the original equation of the lemniscate.
Now, we substitute the value we obtained from the previous step (2(x^2 + y^2) = 49) into the original equation of the lemniscate:
2(x^2 + y^2)^2 = 49(x^2 − y^2)
With the substitution:
2(49)^2 = 49(x^2 − y^2)
Simplifying:
98 = x^2 − y^2
Step 4: Solve the resulting equation for the values of x and y.
Now, we solve the resulting equation for the values of x and y to obtain the coordinates of the points on the lemniscate where the tangent is horizontal.
x^2 − y^2 = 98
This equation represents a hyperbola. By finding the points on this hyperbola, we'll get the coordinates of the points on the lemniscate where the tangent is horizontal.
To graphically visualize these points or find exact coordinates, you can plot the graph of the lemniscate equation and the hyperbola equation using a graphing software or calculator. The intersection points of these two curves will give you the points where the tangent is horizontal on the lemniscate.
well, x^2 +y^2 = 49/4
put that back in
2 (49/4)^2 = 49 (x^2-y^2)
49/8 = x^2 - y^2
49/4 = x^2 + y^2
2 x^2 = 3*49/8
well that is two points x values :)
you have 2 y values for each x so 4