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Calculas1?
At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve, as requested.
y 6 + x 3 = y 2 + 9x, slope at (0, 1)
1 Answer
- 10 months ago
y^6 + x^3 = y^2 + 9x
6y^5 * dy + 3x^2 * dx = 2y * dy + 9 * dx
x = 0 , y = 1
6 * dy + 0 * dx = 2 * dy + 9 * dx
4 * dy = 9 * dx
dy/dx = 9/4
Tangent slope: 9/4
Normal slope: -4/9
y - 1 = m * (x - 0)
y - 1 = mx
y = mx + 1
Tangent line: y = (9/4) * x + 1
Normal line: y = (-4/9) * x + 1
