Question

The equation of the normal to the curve y = 2sinx at (0, 0) is:

• A. \(x+\frac{1}{2}y=0\)
• B. x-2y=0
• C. \(x-\frac{1}{2}y=0\)
• D. x+2y=0

Answer 1

The Correct Option is D

To find the equation of the normal to the curve \(y=2\sin x\) at the point \((0,0)\), we need to determine the slope of the tangent at the given point and then find the negative reciprocal to get the slope of the normal.

1. Derive the Curve: The derivative \(\frac{dy}{dx}\) gives the slope of the tangent to the curve. For \(y=2\sin x\), the derivative is:

\(\frac{dy}{dx}=2\cos x\)

2. Evaluate at \(x=0\): Substitute \(x=0\) to find the slope of the tangent at \((0,0)\).

\(\frac{dy}{dx}\big|_{x=0}=2\cos(0)=2\)

3. Slope of the Normal: The slope of the normal is the negative reciprocal of the tangent slope:

\(m_{\text{normal}}=-\frac{1}{2}\)

4. Equation of the Normal: Using the point-slope form of the line equation \(y-y_1=m(x-x_1)\) with point \((0,0)\) and slope \(-\frac{1}{2}\), we have:

\(y-0=-\frac{1}{2}(x-0)\)

Simplifying, the equation is:

\(y=-\frac{1}{2}x\)

Or in standard form:

\(x+2y=0\)

This matches the correct option, which is: x+2y=0.