A particle moves along the curve `y=x^2 + 2x` . At what point(s) on the curve are x and y coordinates of the particle changing at the same rate?

Answer

To find the point(s) on the curve \( y = x^2 + 2x \) where the x and y coordinates of the particle are changing at the same rate, we can follow these steps:

### Step 1: Understand the relationship between the rates of change
We know that the rates of change of x and y with respect to time t are given by \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). We want to find when these rates are equal:
\[ \frac{dy}{dt} = \frac{dx}{dt} \]

### Step 2: Differentiate the curve equation
Given the curve \( y = x^2 + 2x \), we differentiate both sides with respect to time t:
\[ \frac{dy}{dt} = \frac{d}{dt}(x^2 + 2x) \]
Using the chain rule, we have:
\[ \frac{dy}{dt} = 2x \frac{dx}{dt} + 2 \frac{dx}{dt} \]
This simplifies to:
\[ \frac{dy}{dt} = (2x + 2) \frac{dx}{dt} \]

### Step 3: Set the rates equal
Since we want \( \frac{dy}{dt} = \frac{dx}{dt} \), we can set the two expressions equal:
\[ (2x + 2) \frac{dx}{dt} = \frac{dx}{dt} \]

### Step 4: Simplify the equation
Assuming \( \frac{dx}{dt} \neq 0 \), we can divide both sides by \( \frac{dx}{dt} \):
\[ 2x + 2 = 1 \]

### Step 5: Solve for x
Now, we solve the equation:
\[ 2x + 2 = 1 \implies 2x = 1 - 2 \implies 2x = -1 \implies x = -\frac{1}{2} \]

### Step 6: Find the corresponding y-coordinate
Now that we have \( x = -\frac{1}{2} \), we can find the corresponding y-coordinate using the original equation of the curve:
\[ y = \left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) \]
Calculating this gives:
\[ y = \frac{1}{4} - 1 = -\frac{3}{4} \]

### Conclusion
Thus, the point on the curve where the x and y coordinates of the particle are changing at the same rate is:
\[ \left(-\frac{1}{2}, -\frac{3}{4}\right) \]