Use implicit differentiation to find an equation
of the tangent line to the curve
y^2 = x^3 (26 − x)
at the point (1, 5).
2 y dy = x^3(-dx)+3x^2dx(26-x)
2 y dy = -x^3 dx + 78 x^2 dx - 3 x^3 dx
dy/dx = (-4 x^3 + 78 x^2)/ (2 y)
dy/dx = m = (39 x^2 - 2 x^3)/y
at (1,5)
m = (39-2)/5 = 37/5
so y = (37/5) x + b
5 = 37/5 + b
b = 25/5 - 37/5 = -12/5
so
y = (37/5) x - 12/5
5 y = 37 x - 12
To find the equation of the tangent line to the curve using implicit differentiation, follow these steps:
1. Differentiate both sides of the equation with respect to x.
d/dx [ y^2 ] = d/dx [ x^3 (26 - x) ]
2. Apply the chain rule on the left side of the equation.
2y * dy/dx = 3x^2 * (26 - x) + x^3 * (-1)
3. Simplify the expression.
2y * dy/dx = 78x^2 - 3x^3 - x^3
2y * dy/dx = 78x^2 - 4x^3
4. Substitute the coordinates of the given point (1, 5) to find dy/dx at that point.
2(5) * dy/dx = 78(1)^2 - 4(1)^3
10 * dy/dx = 78 - 4
10 * dy/dx = 74
dy/dx = 74/10
dy/dx = 7.4
5. The slope of the tangent line at the point (1, 5) is found to be 7.4.
6. Substitute the slope and the coordinates of the given point in the point-slope form to find the equation of the tangent line.
y - y1 = m(x - x1), where (x1, y1) = (1, 5) and m = 7.4
y - 5 = 7.4(x - 1)
7. Simplify the equation.
y - 5 = 7.4x - 7.4
y = 7.4x - 2.4
Therefore, the equation of the tangent line to the curve y^2 = x^3 (26 − x) at the point (1, 5) is y = 7.4x - 2.4.