Question

Find the equation of tangent to the curve y = 2x^3 - x^2 + 2 at { 1\2 , 2 }

Answer 1

Step-by-step explanation:

y=2x3−x2+3

Differentiate both sides w.r.t. x,

⇒  dxdy=6x2−2x

⇒  Slop of tangent =(dxdy)(1,4)=6(1)2−2(1)=6−2=4

⇒  Slop of normal =Slopoftangent−1=4−1

⇒  (x1,y1)=(1,4)                [ Given ]

Equation of normal is,

y−y1=m(x−x1)

⇒  y−4=4−1(x−1)

⇒  4y−16=−x+1

⇒  x+4y=17

Answer 2

The equation of tangent to the curve is $4y - 2x = 7$.

Step-by-step explanation:

Given: curve $y = 2x^3 - x^2 + 2$ and points $(\frac{1}{2} , 2 )$

To Find:  The equation of tangent to the curve

Solution:

• Finding the equation of tangent to the curve

We have given the curve $y = 2x^3 - x^2 + 2$ such that the slope of the tangent to the curve will be $dy/dx$. Therefore, the slope is,

$\Rightarrow \text{slope, m} = \dfrac{dy}{dx} = \dfrac{d}{dx} (2x^3 - x^2 + 2)$

$\Rightarrow \text{slope, m} = 6x^2 - 2x$

And, the slope at points (1/2, 2) is,

$\Rightarrow \text{slope, m} |_{(1/2, \ 2)} = 6 \Big( \dfrac{1}{2} \Big)^2 - 2 \Big(\dfrac{1}{2}\Big) = \dfrac{1}{2}$

Now, the equation of tangent at (x₁, y₁) is given as,

$\Rightarrow (y - y_1) = m (x - x_1)$

Therefore, the equation of tangent at (1/2, 2) is given as,

$\Rightarrow (y - 2) = \dfrac{1}{2} (x - \dfrac{1}{2})$

$\Rightarrow 4 (y - 2) = (2x - 1)$

$\Rightarrow 4y - 2x = 7$

Hence, the equation of tangent to the curve is $4y - 2x = 7$.