A point is in motion along a hyperbola `y=(10)/(x)` so that its abscissa x increases uniformly at a rate of l unit per second. Then, the rate of change of its ordinate, when the point passes through (5, 2)

Options

A. increases at the rate of `(1)/(2)` unit per second
B. decreases at the rate of `(1)/(2)` unit per second
C. decreascs at the rate of `(2)/(5)` unit per second
D. increases at the rate of `(2)/(5)` unit per second

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the given information
We have a hyperbola defined by the equation:
\[ y = \frac{10}{x} \]
We know that the abscissa \( x \) is increasing at a rate of \( \frac{dx}{dt} = 1 \) unit per second. We need to find the rate of change of the ordinate \( y \) when the point passes through \( (5, 2) \).

### Step 2: Differentiate the equation
To find the rate of change of \( y \) with respect to time \( t \), we will differentiate both sides of the equation \( y = \frac{10}{x} \) with respect to \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{10}{x}\right) \]
Using the chain rule, we have:
\[ \frac{dy}{dt} = -\frac{10}{x^2} \cdot \frac{dx}{dt} \]

### Step 3: Substitute known values
Now we will substitute the values into the differentiated equation. We know that when the point passes through \( (5, 2) \), \( x = 5 \) and \( \frac{dx}{dt} = 1 \):
\[ \frac{dy}{dt} = -\frac{10}{5^2} \cdot 1 \]

### Step 4: Calculate \( \frac{dy}{dt} \)
Now we can calculate:
\[ \frac{dy}{dt} = -\frac{10}{25} \cdot 1 = -\frac{2}{5} \]

### Step 5: Interpret the result
The negative sign indicates that the ordinate \( y \) is decreasing at the rate of \( \frac{2}{5} \) units per second when the point passes through \( (5, 2) \).

### Final Answer
Thus, the rate of change of the ordinate \( y \) when the point passes through \( (5, 2) \) is:
\[ \frac{dy}{dt} = -\frac{2}{5} \text{ units per second} \]

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