A particle is constrained to move along the curve `y= sqrtx` starting at the origin at time t=0. The point on the curve where the abscissa and the ordinate are changing at the same rate is

Options

A. `((1)/(2), (1)/(sqrt2))`
B. `((1)/(8), (1)/(2 sqrt2))`
C. `((1)/(4), (1)/(2))`
D. (1,1)

Answer

To solve the problem step by step, we need to find the point on the curve \( y = \sqrt{x} \) where the rates of change of the abscissa (x-coordinate) and the ordinate (y-coordinate) are equal.

### Step 1: Understand the relationship between x and y
The curve is given by:
\[ y = \sqrt{x} \]

### Step 2: Differentiate y with respect to time
To find the rates of change, we differentiate both sides of the equation with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(\sqrt{x}) \]
Using the chain rule, we have:
\[ \frac{dy}{dt} = \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt} \]

### Step 3: Set the rates equal
According to the problem, the rates of change of x and y are equal:
\[ \frac{dx}{dt} = \frac{dy}{dt} \]
Substituting the expression for \( \frac{dy}{dt} \) from Step 2:
\[ \frac{dx}{dt} = \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt} \]

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

### Step 5: Solve for x
Now, we can solve for \( \sqrt{x} \):
\[ 2\sqrt{x} = 1 \]
\[ \sqrt{x} = \frac{1}{2} \]
Squaring both sides gives us:
\[ x = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 6: Find the corresponding y value
Now, we can find the corresponding y value using the original equation \( y = \sqrt{x} \):
\[ y = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Conclusion
The point on the curve where the abscissa and the ordinate are changing at the same rate is:
\[ \left(\frac{1}{4}, \frac{1}{2}\right) \]