contestada

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.
y5+ x3 = y2 + 12x, slope at (0, 1)
0-2
02
04

Respuesta :

adioabiola

The slope of the curve described by the equation at the given point (0,1) as in the task content is; 4.

What is the slope of the curve, the line tangent to the curve at the given point; (0, 1)?

According to the task content, it follows that the slope of the curve can be determined by means of implicit differentiation as follows;

y⁵+ x³ = y² + 12x

5y⁴(dy/dx) -2y(dy/dx) = 12 - 3x²

(dy/dx) = (12 -3x²)/(5y⁴-2y)

Hence, since the slope corresponds at the point given; (0, 1); we have;

(dy/dx) = (12 -3(0)²)/(5(1)⁴-2(1))

dy/dx = 12/3 = 4.

Hence, slope, m = 4.

Consequent to the mathematical computation above, it can then be concluded that the slope of the curve, the line tangent to the curve at the given point is; 4.

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