Question

Find the slope of the normal to the curve \(y = 2x^2 + 3\sin x\) at \(x = 0\).

• A. \(0\)
• B. \(-1/3\)
• C. \(1/3\)
• D. \(3\)

Hint:

Slope of normal is always the negative reciprocal of the slope of the tangent. If tangent slope is \(m\), normal slope is \( -1/m \).

Answer 1

The Correct Option is B

Concept: The slope of a curve at any point is given by its derivative \( \frac{dy}{dx} \). The slope of the normal to the curve is the negative reciprocal of the slope of the tangent. If slope of tangent is \(m\), then slope of normal is \[ m_n = -\frac{1}{m} \]

Step 1: Differentiate the given function. Given: \[ y = 2x^2 + 3\sin x \] Differentiate with respect to \(x\): \[ \frac{dy}{dx} = 4x + 3\cos x \]

Step 2: Find slope of tangent at \(x=0\). Substitute \(x = 0\): \[ \frac{dy}{dx} = 4(0) + 3\cos 0 \] Since \(\cos 0 = 1\): \[ \frac{dy}{dx} = 3 \] Thus, slope of tangent \(m = 3\).

Step 3: Find slope of the normal. \[ m_n = -\frac{1}{m} \] \[ m_n = -\frac{1}{3} \] \[ \boxed{-\frac{1}{3}} \]