a curve is such that dy/dx=6/(3x-2)^3 and A(1,-3) lies on the curve. a point is moving along the curve and at A the y coordinate of the point is increasing at 3 units per second.
find the rate of increase at A of the x coordinate of the point

To find the rate of increase of the \( x \)-coordinate at point \( A(1, -3) \), given the differential equation of the curve and the information about the rate of change of the \( y \)-coordinate, we need to use the following approach.

### Step 1: Understand the given information
We are given:
- The slope of the curve is \( \frac{dy}{dx} = \frac{6}{(3x - 2)^3} \).
- At point \( A(1, -3) \), the rate of change of the \( y \)-coordinate, \( \frac{dy}{dt} \), is \( 3 \) units per second.

We need to find \( \frac{dx}{dt} \) at \( A \).

### Step 2: Chain rule relation between \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)
By the chain rule, we know:
\[
\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}
\]
This equation connects the rate of change of \( y \) with respect to time \( t \), to the rate of change of \( x \) with respect to time \( t \). We can rearrange this to solve for \( \frac{dx}{dt} \):
\[
\frac{dx}{dt} = \frac{dy}{dt} \div \frac{dy}{dx}
\]

### Step 3: Substitute known values
At point \( A(1, -3) \), we are given \( \frac{dy}{dt} = 3 \). We need to find \( \frac{dy}{dx} \) at \( x = 1 \). Using the equation for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{6}{(3x - 2)^3}
\]
Substitute \( x = 1 \):
\[
\frac{dy}{dx} = \frac{6}{(3(1) - 2)^3} = \frac{6}{(3 - 2)^3} = \frac{6}{1^3} = 6
\]

### Step 4: Calculate \( \frac{dx}{dt} \)
Now, substitute \( \frac{dy}{dt} = 3 \) and \( \frac{dy}{dx} = 6 \) into the chain rule equation:
\[
\frac{dx}{dt} = \frac{3}{6} = \frac{1}{2}
\]

Thus, the rate of increase of the \( x \)-coordinate at point \( A \) is \( \frac{1}{2} \) units per second.

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Would you like further details or clarification? 

Here are 5 related questions to consider:
1. How would the rate of change \( \frac{dx}{dt} \) vary if \( \frac{dy}{dt} \) were different?
2. Can you find the equation of the curve by integrating the slope \( \frac{dy}{dx} \)?
3. How would you handle the case if \( \frac{dy}{dx} \) were more complex, involving trigonometric functions?
4. What happens to \( \frac{dx}{dt} \) if the slope \( \frac{dy}{dx} \) approaches zero?
5. How can implicit differentiation be applied to problems like this in more complex curves?

Tip: Always apply the chain rule when differentiating a function with respect to time, especially when dealing with rates of change in multiple variables.

A curve is such that dy/dx=6/(3x-2)^3 and A(1,-3) lies on the curve. A point is moving along the curve, and at A, the y-coordinate of the point is increasing at 3 units per second. Find the rate of increase at A of the x-coordinate of the point.

Learn how to calculate the rate of increase of the x-coordinate at point A(1, -3) given the slope dy/dx = 6/(3x-2)^3 and the rate of change of y at 3 units per second. This solution applies the chain rule and involves solving for dx/dt.

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