Question

Find the equations of the tangent and normal to the curve \( y = 2x^3 - x^2 + 2 \) at the point \( \left( \frac{1}{2}, 2 \right) \).

Hint:

To find the equation of the tangent and normal, first calculate the slope of the tangent using the derivative, then use the point-slope form to write the equations.

Answer 1

Step 1: Find the derivative of the curve.
The equation of the curve is: \[ y = 2x^3 - x^2 + 2 \] The derivative of \( y \), which gives the slope of the tangent, is: \[ \frac{dy}{dx} = 6x^2 - 2x \]

Step 2: Find the slope of the tangent at \( x = \frac{1}{2} \).
Substitute \( x = \frac{1}{2} \) into the derivative to find the slope of the tangent: \[ \frac{dy}{dx} \Big|_{x=\frac{1}{2}} = 6 \left( \frac{1}{2} \right)^2 - 2 \left( \frac{1}{2} \right) = 6 \times \frac{1}{4} - 1 = \frac{3}{2} - 1 = \frac{1}{2} \] So, the slope of the tangent at \( \left( \frac{1}{2}, 2 \right) \) is \( \frac{1}{2} \).

Step 3: Find the equation of the tangent.
The equation of the tangent line is given by the point-slope form: \[ y - y_1 = m(x - x_1) \] Where \( m = \frac{1}{2} \), \( x_1 = \frac{1}{2} \), and \( y_1 = 2 \). Substituting the values: \[ y - 2 = \frac{1}{2} \left( x - \frac{1}{2} \right) \] Simplifying: \[ y - 2 = \frac{1}{2}x - \frac{1}{4} \] \[ y = \frac{1}{2}x + \frac{7}{4} \] So, the equation of the tangent is: \[ \boxed{y = \frac{1}{2}x + \frac{7}{4}} \]

Step 4: Find the slope of the normal.
The slope of the normal is the negative reciprocal of the slope of the tangent: \[ m_{\text{normal}} = -\frac{1}{\frac{1}{2}} = -2 \]

Step 5: Find the equation of the normal.
Using the point-slope form for the normal: \[ y - 2 = -2 \left( x - \frac{1}{2} \right) \] Simplifying: \[ y - 2 = -2x + 1 \] \[ y = -2x + 3 \] So, the equation of the normal is: \[ \boxed{y = -2x + 3} \]

Final Answer: - Equation of the tangent: \( \boxed{y = \frac{1}{2}x + \frac{7}{4}} \) - Equation of the normal: \( \boxed{y = -2x + 3} \)