Question

A point \( P \) is moving on the curve \( x^3 y^4 = 27 \). The x-coordinate of \( P \) is decreasing at the rate of 8 units per second. When the point \( P \) is at \( (2, 2) \), the y-coordinate of \( P \) is:

• A. increases at the rate of 6 units per second
• B. decreases at the rate of 6 units per second
• C. increases at the rate of 4 units per second
• D. decreases at the rate of 4 units per second \bigskip

Hint:

For problems involving rates of change, differentiate the given equation implicitly with respect to time and substitute the known values to find the required rate.

Answer 1

The Correct Option is A

We are given the equation \( x^3 y^4 = 27 \). We need to find the rate of change of \( y \) when \( P(2,2) \) is on the curve, and the x-coordinate is decreasing at the rate of 8 units per second. 

Step 1: Differentiate the given equation with respect to time \( t \). The equation is: \[ x^3 y^4 = 27. \] Differentiate implicitly with respect to \( t \): \[ \frac{d}{dt}(x^3 y^4) = \frac{d}{dt}(27). \] Using the product rule, we get: \[ 3x^2 \frac{dx}{dt} y^4 + x^3 4y^3 \frac{dy}{dt} = 0. \] 

Step 2: Substitute the known values at the point \( P(2, 2) \). At \( x = 2 \) and \( y = 2 \), we know \( \frac{dx}{dt} = -8 \) (since the x-coordinate is decreasing). Substituting these values into the equation: \[ 3(2)^2 (-8) (2)^4 + (2)^3 4(2)^3 \frac{dy}{dt} = 0. \] Simplifying: \[ 3 \times 4 \times (-8) \times 16 + 8 \times 4 \times 8 \frac{dy}{dt} = 0, \] \[ -1536 + 256 \frac{dy}{dt} = 0. \] Solving for \( \frac{dy}{dt} \): \[ 256 \frac{dy}{dt} = 1536, \] \[ \frac{dy}{dt} = \frac{1536}{256} = 6. \] 

Step 3: Conclusion. Thus, the y-coordinate of \( P \) is increasing at the rate of 6 units per second. \bigskip