Find dy/dx 4x^2+y^2-8x+4y+4=0, find points on graph of the equation where there is a vertical or horizontal tangent line
This can be solved without taking derivatives.
4x^2 +y^2 -8x +4y +4 = 0
4(x^2 -2x +1) +(y^2 +4y +4) = 8
4(x-1)^2 + (y+2)^2 = 8
That is the equation of an ellipse with center at (1,-2)
There are vertical tangents at y = -2, where x-1 = + or - sqrt2
x = 1 +/-sqrt2
There are horizontal tangents at x =1, where y+2 = +/1 sqrt8
y = 2 +/-sqrt8
If you want an equation for dy/dx, try implicit differentiation
8x + 2y dy/dx -8 +4 dy/dx = 0
dy/dx (2y +4) = (-8x +8)
dy/dx = (-8x +8)/(2y +4)
This clearly shows that the derivative is zero at x=1 (horizontal slope) and infinity at y= -2 (vertical slope), in agreement with the other method. The y(x) equation can be used to solve for the other coordinates, when x=1 or y = -2.
To find dy/dx, we will differentiate the given equation implicitly.
Differentiating each term with respect to x, we have:
8x + 2yy' - 8 + 4y + 4y' = 0
Rearranging the equation, we get:
8x - 8 + 4y = -2yy' - 4y'
Combining like terms, we have:
8x - 8 + 4y = -4y' - 2yy'
Rearranging further, we get:
4y' + 2yy' = 8x - 8 + 4y
Factoring out the common term on the left side, we have:
y'(4 + 2y) = 8x - 8 + 4y
Dividing both sides by (4 + 2y), we get:
y' = (8x - 8 + 4y) / (4 + 2y)
This is the expression for dy/dx.
For a horizontal tangent line, dy/dx = 0. To find the points on the graph where there is a horizontal tangent line, we need to solve the equation:
(8x - 8 + 4y) / (4 + 2y) = 0
Cross-multiplying, we have:
8x - 8 + 4y = 0
Rearranging, we get:
8x + 4y = 8
Dividing both sides by 4, we have:
2x + y = 2
This is the equation of the line where the graph has a horizontal tangent line.
For a vertical tangent line, the derivative is undefined. In this case, the denominator of dy/dx becomes zero. Therefore, we need to find the points where (4 + 2y) = 0.
Solving for y, we have:
2y = -4
y = -2
At y = -2, the denominator becomes zero, indicating a vertical tangent line.
Therefore, the points on the graph where there is a vertical or horizontal tangent line are the solutions to the equation:
2x + y = 2
and
y = -2.
To find dy/dx for the given equation, we need to differentiate the equation implicitly with respect to x.
Let's differentiate each term one by one:
Differentiating 4x^2 with respect to x gives us 8x.
Differentiating y^2 with respect to x, we need to apply the chain rule. Since y is a function of x, we have dy/dx * 2y.
Differentiating -8x with respect to x gives -8.
Differentiating 4y with respect to x gives us 4 * dy/dx.
Differentiating 4 with respect to x gives us 0.
Now, let's substitute these derivatives back into the equation:
8x + 2y * dy/dx - 8 + 4 * dy/dx = 0
Now, we need to isolate dy/dx by moving all the terms with dy/dx to one side of the equation:
2y * dy/dx + 4 * dy/dx = -8x + 8
Combining the terms with dy/dx:
(2y + 4) * dy/dx = -8x + 8
Finally, we can solve for dy/dx:
dy/dx = (-8x + 8) / (2y + 4)
To find the points on the graph where there is a vertical tangent line, we need to find the points where dy/dx approaches infinity. This happens when the denominator (2y + 4) is equal to 0.
Setting 2y + 4 = 0, we can solve for y:
2y = -4
y = -2
So, the points on the graph where there is a vertical tangent line are the solutions where y = -2.
To find the points on the graph where there is a horizontal tangent line, we need to find the points where dy/dx equals 0.
Setting (-8x + 8) / (2y + 4) = 0, we can solve for x:
-8x + 8 = 0
-8x = -8
x = 1
So, the points on the graph where there is a horizontal tangent line are the solutions where x = 1.