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 the tangent (\(m_t\)) at a point is the derivative of the function evaluated at that point. The normal is the line perpendicular to the tangent at the point of contact.
Step 1: Understanding the Question:
We need to find the slope of the normal to the given curve at the specific point where \(x = 0\).
Step 2: Key Formula or Approach:
1. Find the slope of the tangent: \(m_t = \frac{dy}{dx}\).
2. Use the perpendicularity condition: \(m_t \times m_n = -1\).
3. Therefore, slope of the normal \(m_n = -\frac{1}{m_t}\).
Step 3: Detailed Solution:
1. Given: \(y = 2x^2 + 3\sin x\).
2. Differentiating w.r.t. \(x\):
\[ \frac{dy}{dx} = 4x + 3\cos x \]
3. Evaluate \(m_t\) at \(x = 0\):
\[ m_t = 4(0) + 3\cos(0) = 0 + 3(1) = 3 \]
4. Calculate \(m_n\):
\[ m_n = -\frac{1}{m_t} = -\frac{1}{3} \]
Step 4: Final Answer:
The slope of the normal is \(-1/3\).