The slope of the normal to the curve $y=2x^{2}+3~sin~x$ at $x=0$ is

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

To find the slope of the normal to the curve $ y = 2x^2 + 3 \sin x $ at $ x = 0 $, we first need to determine the slope of the tangent to the curve at that point. The slope of the tangent is given by the derivative of the function $ y $ with respect to $ x $. 1. **Differentiate the function**: \[ y = 2x^2 + 3 \sin x \] The derivative $ \frac{dy}{dx} $ is: \[ \frac{dy}{dx} = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3 \sin x) \] \[ \frac{dy}{dx} = 4x + 3 \cos x \] 2. **Evaluate the derivative at $ x = 0 $**: \[ \frac{dy}{dx} \bigg|_{x=0} = 4(0) + 3 \cos(0) = 0 + 3 = 3 \] The slope of the tangent to the curve at $ x = 0 $ is 3. 3. **Find the slope of the normal**: The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal is: \[ -\frac{1}{3} \] Thus, the slope of the normal to the curve at $ x = 0 $ is $-\frac{1}{3}$, which corresponds to Option 4. **Explanation of why other options are incorrect**: - **Option 1 (3)**: This is the slope of the tangent, not the normal. The normal is perpendicular to the tangent, so its slope must be the negative reciprocal of the tangent's slope. - **Option 2 ($\frac{1}{3}$)**: This is not the correct negative reciprocal of the tangent's slope. The correct negative reciprocal of 3 is $-\frac{1}{3}$. - **Option 3 (-3)**: This is not the negative reciprocal of the tangent's slope. The negative reciprocal of 3 is $-\frac{1}{3}$, not -3. Therefore, the correct answer is Option 4, $-\frac{1}{3}$, as it accurately represents the slope of the normal to the curve at $ x = 0 $.

Question

Accepted Answer

To find the slope of the normal to the curve $ y = 2x^2 + 3 \sin x $ at $ x = 0 $, we first need to determine the slope of the tangent to the curve at that point. The slope of the tangent is given by the derivative of the function $ y $ with respect to $ x $.

1. **Differentiate the function**:
   $$
   y = 2x^2 + 3 \sin x
   $$
   The derivative $ \frac{dy}{dx} $ is:
   $$
   \frac{dy}{dx} = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3 \sin x)
   $$
   $$
   \frac{dy}{dx} = 4x + 3 \cos x
   $$

2. **Evaluate the derivative at $ x = 0 $**:
   $$
   \frac{dy}{dx} \bigg|_{x=0} = 4(0) + 3 \cos(0) = 0 + 3 = 3
   $$

The slope of the tangent to the curve at $ x = 0 $ is 3.

3. **Find the slope of the normal**:
   The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal is:
   $$
   -\frac{1}{3}
   $$

Thus, the slope of the normal to the curve at $ x = 0 $ is $-\frac{1}{3}$, which corresponds to Option 4.

**Explanation of why other options are incorrect**:

- **Option 1 (3)**: This is the slope of the tangent, not the normal. The normal is perpendicular to the tangent, so its slope must be the negative reciprocal of the tangent's slope.
  
- **Option 2 ($\frac{1}{3}$)**: This is not the correct negative reciprocal of the tangent's slope. The correct negative reciprocal of 3 is $-\frac{1}{3}$.

- **Option 3 (-3)**: This is not the negative reciprocal of the tangent's slope. The negative reciprocal of 3 is $-\frac{1}{3}$, not -3.

Therefore, the correct answer is Option 4, $-\frac{1}{3}$, as it accurately represents the slope of the normal to the curve at $ x = 0 $.