A particle moves along the curve `y=(4)/(3)x^(3)+5`. Find the points on the curve at which the y - coordinate changes as fast as the x - coordinate.

Answer

To find the points on the curve \( y = \frac{4}{3}x^3 + 5 \) where the y-coordinate changes as fast as the x-coordinate, we will follow these steps:

### Step 1: Differentiate y with respect to t
We start by differentiating the given equation with respect to \( t \):

\[ y = \frac{4}{3}x^3 + 5 \]

Differentiating both sides with respect to \( t \):

\[ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{4}{3}x^3 + 5\right) \]

Using the chain rule:

\[ \frac{dy}{dt} = \frac{4}{3} \cdot 3x^2 \cdot \frac{dx}{dt} = 4x^2 \cdot \frac{dx}{dt} \]

### Step 2: Set the condition for equal rates of change
We want to find the points where the rate of change of \( y \) with respect to \( t \) is equal to the rate of change of \( x \) with respect to \( t \):

\[ \frac{dy}{dt} = \frac{dx}{dt} \]

Substituting the expression we found for \( \frac{dy}{dt} \):

\[ 4x^2 \cdot \frac{dx}{dt} = \frac{dx}{dt} \]

### Step 3: Simplify the equation
Assuming \( \frac{dx}{dt} \neq 0 \), we can divide both sides by \( \frac{dx}{dt} \):

\[ 4x^2 = 1 \]

### Step 4: Solve for x
Now we solve for \( x \):

\[ x^2 = \frac{1}{4} \]

Taking the square root of both sides:

\[ x = \frac{1}{2} \quad \text{or} \quad x = -\frac{1}{2} \]

### Step 5: Find corresponding y-coordinates
Now we will find the corresponding \( y \)-coordinates for both values of \( x \).

1. For \( x = \frac{1}{2} \):

\[ y = \frac{4}{3}\left(\frac{1}{2}\right)^3 + 5 = \frac{4}{3} \cdot \frac{1}{8} + 5 = \frac{4}{24} + 5 = \frac{1}{6} + 5 = \frac{1}{6} + \frac{30}{6} = \frac{31}{6} \]

2. For \( x = -\frac{1}{2} \):

\[ y = \frac{4}{3}\left(-\frac{1}{2}\right)^3 + 5 = \frac{4}{3} \cdot -\frac{1}{8} + 5 = -\frac{4}{24} + 5 = -\frac{1}{6} + 5 = -\frac{1}{6} + \frac{30}{6} = \frac{29}{6} \]

### Final Points
The points on the curve where the y-coordinate changes as fast as the x-coordinate are:

\[ \left(\frac{1}{2}, \frac{31}{6}\right) \quad \text{and} \quad \left(-\frac{1}{2}, \frac{29}{6}\right) \]