Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x^2+y^2=(2x^2+4y^2−x)^2
(0, 0.25)
(cardioid)
To find the equation of the tangent line to a curve at a given point using implicit differentiation, follow these steps:
1. Take the derivative of both sides of the equation with respect to x. Treat y as a function of x and differentiate using the chain rule when necessary.
In this case, we have the equation x^2 + y^2 = (2x^2 + 4y^2 - x)^2.
Differentiating both sides with respect to x, we get:
2x + 2yy' = 2(2x^2 + 4y^2 - x)(4x - 1).
2. Now, substitute the x-coordinate and y-coordinate of the given point into the derivative equation to find the value of y'.
In this case, the given point is (0, 0.25). Substituting these values, we have:
2(0) + 2(0.25)y' = 2(2(0)^2 + 4(0.25)^2 - 0)(4(0) - 1).
Simplifying, we get:
0.5y' = 2(2(0.25)^2 - 0)(-1).
Further simplification gives:
0.5y' = 2(2(0.0625))(-1).
Calculating, we find:
0.5y' = -0.25.
Dividing both sides by 0.5 gives:
y' = -0.5.
3. Finally, use the found derivative value, y', and the given point (0, 0.25) to find the equation of the tangent line using the point-slope form:
The equation of a line in point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
Plugging in the slope y' = -0.5 and the point (0, 0.25), we have:
y - 0.25 = -0.5(x - 0).
Simplifying, we get:
y - 0.25 = -0.5x.
Rearranging, the equation of the tangent line is:
y = -0.5x + 0.25.
Therefore, the equation of the tangent line to the curve x^2 + y^2 = (2x^2 + 4y^2 - x)^2 at the point (0, 0.25) is y = -0.5x + 0.25.
2 x dx + 2 y dy = 2 (2 x^2+4 y^2 -x)(4 x dx + 8 y dy -dx)
at (0, 1/4)
0 +(1/2)dy = 2(0+4(1/16)-0)(8(1/4)dy -dx)
(1/2) dy = 2 (1/4)(2 dy-dx) = dy - (1/2) dx
dy = 2 dy - dx
dy = dx
dy/dx = slope = 1
so y = x + b
put in your point to find b