If 3x^2-2y^2=4xy + 20, find the point(s) at which the graph from Problem 28 has vertical tangents. I found that dy/dx= (3x-2y)/(2x+2y), but now I don't know what to do. Do I set dy/dx to 0?
For a vertical line, the slope would be undefined, that is,
(3x-2y)/(2x+2y) has to be undefined.
When does that happen?
When the denominator is zero, so
2x + 2y = 0
y = -x
sub into the origianal
3x^2 - 2(-x)^2 = 4x(-x) + 20
5x^2 = 20
x^2=4
x = ±2
if x=2, y = -2,
if x = -2, y = 2
points where tangent is vertical are (2,-2) and (-2,2)
thanks! that was really helpful ;)
To find the point(s) at which the graph has vertical tangents, you need to find where the derivative of y with respect to x, dy/dx, is undefined. This is because a vertical tangent occurs when the slope (or derivative) is infinite.
You correctly found the derivative of y with respect to x to be dy/dx = (3x - 2y)/(2x + 2y). To determine where dy/dx is undefined, you need to find the values of x and y that make the denominator equal to zero (since dividing by zero is undefined).
Set the denominator, 2x + 2y, equal to zero and solve for x and y:
2x + 2y = 0
Divide both sides by 2 to isolate y:
y = -x
Now substitute this value of y in the original equation of the graph:
3x^2 - 2(-x)^2 = 4x(-x) + 20
Simplify:
3x^2 - 2x^2 = -4x^2 + 20
Combine like terms:
x^2 = 20
Take the square root of both sides:
x = ± √20
Simplify further:
x = ± 2√5
Now substitute the values of x back into y = -x to find the corresponding y-values:
For x = -2√5:
y = -(-2√5) = 2√5
So, one point at which the graph has a vertical tangent is (-2√5, 2√5).
For x = 2√5:
y = -(2√5) = -2√5
So, another point at which the graph has a vertical tangent is (2√5, -2√5).
Therefore, the graph from Problem 28 has vertical tangents at the points (-2√5, 2√5) and (2√5, -2√5).
To find the points at which the graph has vertical tangents, we need to determine the values of x and y that make the derivative dy/dx equal to zero. In this case, you correctly found that dy/dx = (3x - 2y) / (2x + 2y).
To find the points where dy/dx is equal to zero, you need to set (3x - 2y) / (2x + 2y) equal to zero and solve for x and y. So, the equation becomes:
(3x - 2y) / (2x + 2y) = 0
Now, there are two possible scenarios when a fraction is equal to zero.
Scenario 1: Numerator is equal to zero.
In this case, we have 3x - 2y = 0, which can be rewritten as 3x = 2y.
Scenario 2: Denominator is equal to zero.
In this case, we have 2x + 2y = 0, which can be rewritten as 2x = -2y.
Now, let's solve each scenario to find the corresponding values of x and y.
Scenario 1: 3x = 2y
Divide both sides by 2: (3/2)x = y
Scenario 2: 2x = -2y
Divide both sides by -2: x = -y
Now we have two equations derived from setting the numerator and denominator of dy/dx equal to zero. We need to substitute these equations back into the original equation of the graph, which is 3x^2 - 2y^2 = 4xy + 20.
Using Scenario 1: Substitute y with (3/2)x in the equation.
3x^2 - 2(3/2x)^2 = 4x(3/2x) + 20
Simplify and solve for x.
Using Scenario 2: Substitute y with -x in the equation.
3x^2 - 2(-x)^2 = 4x(-x) + 20
Simplify and solve for x.
Once you find the values of x, you can substitute them back into either of the derived equations to find the corresponding values of y. These (x, y) coordinates will be the points at which the graph has vertical tangents.