Find the equation of tangent at (1,2) to the curve y = 2x3. | Mathematics I 2078

Mathematics I20785 marksNumerical: tangent line equationTangents and velocity

Find the equation of tangent at (1,2) to the curve $y = 2x^3$ . [5]

Find the equation of tangent at (1,2) to the curve $y = 2x^3$. [5]

Curve: $y = 2x^3$ - Point of tangency: $(x1, y1) = (1, 2)$ --- Substitute $x = 1$ : $y = 2(1)^3 = 2 \checkmark$ The point $(1, 2)$ lies on the curve. --- $\frac{dy}{dx} = 6x^2$ --- $m = 6(1)^2 = 6$ --- Using $y - y1 = m(x - x1)$ : $y - 2 = 6(x - 1)$ $y - 2 = 6x - 6$ $$\boxed{y = 6x - 4}...

Curve: $y = 2x^3$ - Point of tangency: $(x1, y1) = (1, 2)$ --- Substitute $x = 1$: $$y = 2(1)^3 = 2 \checkmark$$ The point $(1, 2)$ lies on the curve. --- $$\frac{dy}{dx} = 6x^2$$ --- $$m = 6(1)^2 = 6$$ --- Using $y - y1 = m(x - x1)$: $$y - 2 = 6(x - 1)$$ $$y - 2 = 6x - 6$$ $$\boxed{y = 6x - 4}...