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Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2 y2 = (3x2 2y2 − x)2 (0, 0.5) (cardioid)

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nobillionaireNobley

x^2 + y^2 = (3x^2 + 2y^2 - x)^2
2x + 2y f'(x) = 2(3x^2 + 2y^2 - x)(6x + 4y f'(x) - 1) = 36x^3 + 24x^2yf'(x) + 24xy^2 + 16y^3f'(x) - 4y^2 - 18x^2 - 8xyf'(x) + x
f'(x)(2y - 24x^2y - 16y^3 + 8xy) = 36x^3 + 24xy^2 - 4y^2 - 18x^2 - x
f'(x) = (36x^3 + 24xy^2 - 4y^2 - 18x^2 - x)/(2y - 24x^2y - 16y^3 + 8xy)
f'(0, 0.5) = -4(0.5)^2/(2(0.5) - 16(0.5)^3) = -1/(1 - 2) = -1/-1 = 1

Let the required equation be y = mx + c; where y = 0.5, m = 1, x = 0
0.5 = 1(0) + c = 0 + c
c = 0.5

Therefore, the tangent line at point (0, 0.5) is
y = x + 0.5

isyllus

Answer:

The equation of tangent is y=x+0.5

Step-by-step explanation:

Given: \(x^2+y^2=(3x^2+2y^2-x)^2\)

We need to find the equation of tangent on curve at (0,0.5)

First we find slope of tangent using differentiation at given point.

\(x^2+y^2=(3x^2+2y^2-x)^2\)

Differentiate both side w.r.t x

\(2x+2yy'=2(3x^2+2y^2-x)(6x+4yy'-1)\)

Put x=0 and y=0.5 to solve for y'

\(2(0)+2(0.5)y'=2(3(0)^2+2(0.5)^2-0)(6(0)+4(0.5)y'-1)\)

\(0+y'=2(0+0.5-0)(0+2y'-1)\)

\(y'=2y'-1\)

\(y'=1\)

Derivative at given point is slope of tangent line.

Thus, Slope = 1 and point: (0,0.5)

Using point slope form to find the equation of tangent line.

\(y-y_1=m(x-x_1)\)

\(y-0.5=1(x-0)\)

\(y-0.5=x\)

\(y=x+0.5\)

Hence, The equation of tangent is y=x+0.5

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