Find the distance of the point on `y=x^4+3x^2+2x` which is nearest to the line `y=2x-1`

Answer

To find the distance of the point on the curve \( y = x^4 + 3x^2 + 2x \) that is nearest to the line \( y = 2x - 1 \), we will follow these steps:

### Step 1: Identify the slope of the line
The given line is \( y = 2x - 1 \). The slope (m) of this line is 2.

### Step 2: Differentiate the curve
We need to find the derivative of the curve \( y = x^4 + 3x^2 + 2x \) to find the slope of the tangent at any point on the curve.

\[ \frac{dy}{dx} = 4x^3 + 6x + 2 \]

### Step 3: Set the derivative equal to the slope of the line
To find the points on the curve where the tangent is parallel to the line, we set the derivative equal to the slope of the line:

\[ 4x^3 + 6x + 2 = 2 \]

### Step 4: Simplify the equation
Rearranging the equation gives:

\[ 4x^3 + 6x + 2 - 2 = 0 \]
\[ 4x^3 + 6x = 0 \]

### Step 5: Factor the equation
Factoring out \( 2x \):

\[ 2x(2x^2 + 3) = 0 \]

### Step 6: Solve for x
Setting each factor to zero gives:

1. \( 2x = 0 \) → \( x = 0 \)
2. \( 2x^2 + 3 = 0 \) → This does not yield real solutions since \( 2x^2 + 3 > 0 \) for all real \( x \).

Thus, the only solution is \( x = 0 \).

### Step 7: Find the corresponding y-coordinate
Substituting \( x = 0 \) back into the curve equation to find \( y \):

\[ y = 0^4 + 3(0^2) + 2(0) = 0 \]

So, the point on the curve is \( (0, 0) \).

### Step 8: Find the distance from the point to the line
To find the distance from the point \( (0, 0) \) to the line \( y = 2x - 1 \), we can use the formula for the distance \( D \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \):

\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Rewriting the line \( y = 2x - 1 \) in standard form gives:

\[ 2x - y - 1 = 0 \quad \Rightarrow \quad A = 2, B = -1, C = -1 \]

Substituting \( (x_1, y_1) = (0, 0) \):

\[ D = \frac{|2(0) - 1(0) - 1|}{\sqrt{2^2 + (-1)^2}} = \frac{|-1|}{\sqrt{4 + 1}} = \frac{1}{\sqrt{5}} \]

### Final Answer
The distance of the point on the curve that is nearest to the line is:

\[ \frac{1}{\sqrt{5}} \]