Use implicit differentiation to find an equation of the tangent line to the cardioid at the point (0,0.5)
x^2+y^2=(2x^2+2y^2-x)^2
Y=
2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)
at the point (0, 0.5)
0+y'=2*2*0.25*(4*0.5*y'-1)
y'=2y'-1
y'=1
Y=0.5+1*(X-0)
Y=X+0.5
To find the equation of the tangent line to the cardioid at the point (0,0.5), we need to use implicit differentiation to find the derivative of y with respect to x.
Given the equation:
x^2 + y^2 = (2x^2 + 2y^2 - x)^2
Let's differentiate both sides of this equation implicitly with respect to x.
Differentiating x^2 + y^2 with respect to x gives:
2x + 2y * dy/dx = 2(2x^2 + 2y^2 - x)(4x - 1)
Now, let's solve for dy/dx, which represents the derivative of y with respect to x.
2x + 2y * dy/dx = 2(2x^2 + 2y^2 - x)(4x - 1)
Rearranging this equation, we have:
2y * dy/dx = 2(2x^2 + 2y^2 - x)(4x - 1) - 2x
Simplifying further,
2y * dy/dx = 4(2x^2 + 2y^2 - x)(4x - 1) - 2x
Now, let's substitute the coordinates of the point (0,0.5) into the equation to find dy/dx at that point.
When x = 0 and y = 0.5,
2(0.5) * dy/dx = 4(2(0)^2 + 2(0.5)^2 - 0)(4(0) - 1) - 2(0)
Simplifying this equation, we get:
dy/dx = 1
So, the derivative of y with respect to x at the point (0,0.5) is equal to 1, which represents the slope of the tangent line.
Now, we have the point (0,0.5) and the slope 1. We can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
where (x1, y1) is the point on the line and m is the slope.
Using the point (0,0.5) and slope 1, we have:
y - 0.5 = 1(x - 0)
y - 0.5 = x
Therefore, the equation of the tangent line to the cardioid at the point (0,0.5) is y = x.
To find the equation of the tangent line to the cardioid at the point (0, 0.5) using implicit differentiation, follow these steps:
Step 1: Differentiate both sides of the equation with respect to x, treating y as a function of x.
For the left side:
d/dx [x^2 + y^2] = d/dx [x^2] + d/dx [y^2]
Using the power rule for differentiation, we get:
2x + 2yy' = 2x + 2y * dy/dx
For the right side:
d/dx [(2x^2 + 2y^2 - x)^2] = 2(2x^2 + 2y^2 - x)(4x + 4yy' - 1)
Step 2: Substitute the coordinates of the given point (0, 0.5) into the resulting equation from Step 1.
Using the point (0, 0.5):
2(0) + 2(0.5)y' = 2(0) + 2(0.5) * dy/dx
0 + y' = 0 + 1 * dy/dx
y' = dy/dx
Step 3: Solve for dy/dx given the point (0, 0.5).
From Step 2, we have:
y' = dy/dx
Step 4: Substitute this value of dy/dx into the derived equation from Step 1.
For the left side:
2x + 2yy' = 2(0) + 2(0.5) * dy/dx
0 + y' = 1 * dy/dx
y' = dy/dx
For the right side:
2(2x^2 + 2y^2 - x)(4x + 4yy' - 1) = 2(2(0)^2 + 2(0.5)^2 - 0)(4(0) + 4(0.5) * dy/dx - 1)
2(0.5)(4(0.5) * dy/dx - 1) = 0
Step 5: Simplify the resulting equation.
2(0.5)(2dy/dx - 1) = 0
(2dy/dx - 1) = 0
2dy/dx = 1
dy/dx = 1/2
Step 6: Write the equation of the tangent line using the slope-intercept form (y = mx + b) with the slope dy/dx and the point (0, 0.5).
Using the point-slope form:
y - y1 = m(x - x1)
y - 0.5 = (1/2)(x - 0)
y - 0.5 = (1/2)x
y = (1/2)x + 0.5
Therefore, the equation of the tangent line to the cardioid at the point (0, 0.5) is y = (1/2)x + 0.5.