Question

The slope of the normal to the curve $y = 2x^2 + 3\sin x$ at $x = 0$ is:

• A. $-\frac{1}{3}$
• B. $\frac{1}{3}$
• C. $-3$
• D. $3$

Hint:

Remember: $m_{\text{tangent}} \times m_{\text{normal}} = -1$. Once you find the derivative value as $3$, its negative reciprocal is immediately $-\frac{1}{3}$.

Answer 1

The Correct Option is A

Step 1: Know the tangent and normal link.
The slope of the tangent to a curve at a point is just $\frac{dy}{dx}$ there. The normal is the line at right angles to the tangent, so its slope is the negative reciprocal.
\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} \]

Step 2: Write the curve.
The curve is $y = 2x^2 + 3\sin x$. We need its slope at $x = 0$ first.

Step 3: Differentiate the curve.
The derivative of $2x^2$ is $4x$, and the derivative of $3\sin x$ is $3\cos x$.
\[ \frac{dy}{dx} = 4x + 3\cos x \]

Step 4: Put $x = 0$.
At $x = 0$, $4x$ becomes $0$ and $\cos 0 = 1$.
\[ m_{\text{tangent}} = 4(0) + 3(1) = 3 \]

Step 5: Take the negative reciprocal.
The normal slope is minus one over the tangent slope.
\[ m_{\text{normal}} = -\frac{1}{3} \]

Step 6: State the answer.
So the normal to the curve at $x = 0$ has slope $-\frac{1}{3}$.
\[ \boxed{-\dfrac{1}{3}} \]