Step 1: Understanding the Concept:
Calculus allows us to determine the geometric properties of a curve at any specific point using derivatives.
The first derivative of a function, denoted as \( \frac{dy}{dx} \) or \( f'(x) \), represents the slope of the tangent line touching the curve at that point.
A "Tangent" is a line that just grazes the curve at a point, reflecting the curve's direction at that instant.
A "Normal" line, on the other hand, is defined as the line that is perpendicular to the tangent at the point of tangency.
In coordinate geometry, the slopes of two perpendicular lines (\( m_1 \) and \( m_2 \)) have a specific relationship: their product is always \( -1 \).
This means that if we know the slope of the tangent, the slope of the normal is its negative reciprocal.
Step 2: Key Formula or Approach:
1. Find the general derivative \( \frac{dy}{dx} \) of the given curve.
2. Evaluate this derivative at the point \( x = 0 \) to find the tangent slope, \( m_t \).
3. Calculate the normal slope using the formula:
\[ m_n = -\frac{1}{m_t} \]
Step 3: Detailed Explanation:
The given equation of the curve is \( y = 2x^2 + 3\sin x \).
We apply the sum rule and power rule of differentiation. The derivative of \( x^2 \) is \( 2x \), and the derivative of \( \sin x \) is \( \cos x \).
\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3\sin x) \]
\[ \frac{dy}{dx} = 4x + 3\cos x \]
We need to find the slope of the tangent at the specific point where \( x = 0 \).
Substituting \( x = 0 \) into the derivative expression:
\[ m_t = 4(0) + 3\cos(0) \]
Recalling trigonometric values, we know that \( \cos(0) = 1 \).
\[ m_t = 0 + 3(1) = 3 \]
So, the slope of the tangent line at the origin (or where \( x = 0 \)) is 3.
Since the question asks for the slope of the normal, we must take the negative reciprocal of the tangent's slope.
\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} \]
\[ m_{\text{normal}} = -\frac{1}{3} \]
This value represents the steepness of the line perpendicular to the curve at \( x = 0 \).
Step 4: Final Answer:
The slope of the normal line at the given point is \( -\frac{1}{3} \).