Question

find the slope of normal to the curve y= 3x^2 at the point whose x coordinate is 2​

Answer 1

Answer:

Slope of the normal = -1/12

Step-by-step explanation:

Given:

• Equation of the curve = y = 3x²
• X coordinate = 2

To Find:

• Slope of the normal to the curve

Solution:

Here we have to first find the slope of the tangent to the curve.

Differentiating on both sides of the given function with respect to x,

$\sf \dfrac{d}{dx} (y) = \dfrac{d}{dx} (3x^{2})$

$\sf {\dfrac{dy}{dx} =6x}$

Now finding the slope of the tangent at the given point,

$\sf Slope\:of\:tangent=\bigg(\dfrac{dy}{dx} \bigg)_{at\:x=2}$

$\implies 6\times 2$

$\implies 12$

Hence the slope of the tangent to the curve at x =2 is 12.

Now finding the slope of the normal to the curve,

We know that,

$\sf Slope\:of\:the\:normal=\dfrac{-1}{Slope\:of\:the\:tangent}$

Substitute the data,

$\sf Slope\:of\:the\:normal=\dfrac{-1}{12}$

Hence the slope of the normal to the given curve is -1/12.

Answer 2

Answer:

the answer is 2 because it's 2