Introduction

To determine whether the tangent at each point of the curve y = (2/3)x^3 - 2ax^2 + 2x + 5 makes an acute angle with the positive direction of the x-axis, we need to analyze the slope of the tangent line at different points on the curve. The slope of a line represents the angle it makes with the x-axis, and an acute angle is less than 90 degrees.

Slope of the Tangent Line

The slope of the tangent line at any point on the curve can be found by taking the derivative of the equation y = (2/3)x^3 - 2ax^2 + 2x + 5 with respect to x. Let's calculate the derivative:

dy/dx = d/dx [(2/3)x^3 - 2ax^2 + 2x + 5]
= [(2/3) * 3x^2] - [(2a * 2x)] + 2
= 2x^2 - 4ax + 2

The derivative gives us the slope of the tangent line at any given point on the curve.

Condition for an Acute Angle

For the tangent line to make an acute angle with the positive direction of the x-axis, the slope of the tangent line must be positive. This condition can be represented as dy/dx > 0.

Analysis of the Slope

To analyze the slope 2x^2 - 4ax + 2, we need to determine the critical points where the slope changes sign. These critical points occur when dy/dx = 0.

Setting 2x^2 - 4ax + 2 = 0, we can solve for x using the quadratic formula:

x = (-(-4a) ± √((-4a)^2 - 4(2)(2))) / (2(2))
x = (4a ± √(16a^2 - 16)) / 4
x = (a ± √(a^2 - 1))

Analysis of the Slope (continued)

We have two critical points: x = a + √(a^2 - 1) and x = a - √(a^2 - 1). Let's consider these points along with the intervals between them.

1. When x < a="" -="" √(a^2="" -="" 1):="" />
- Substituting a value less than a - √(a^2 - 1) into the derivative, we get a positive value.
- Therefore, the slope is positive in this interval.

2. When a - √(a^2 - 1) < x="" />< a="" +="" √(a^2="" -="" />
- Substituting a value between a - √(a^2 - 1) and a + √(a^2 - 1) into the derivative, we get a negative value.
- Therefore, the slope is negative in this interval.

3. When x > a + √(a^2 - 1):
- Substituting a value greater than a + √(a^2 - 1) into the derivative, we get a positive value.