Question

A particle moves on the hyperbola xy=18 for time t≥0 seconds. At a certain instant, y=6 and dy/dt=8. Which of the following is true about x at this instant? a. x is decreasing by 4 units per second. b. x is increasing by 4 units per second. c. x is decreasing by 1 unit per second. d. x is increasing by 1 unit per second.

Answer 1

When y=6 and dy/dt=8 for a particle moving on the hyperbola xy=18, the value of x is decreasing by 4 units per second. (Option a)

Step-by-step explanation:

The student's question asks for the rate of change of x given that a particle is moving along the hyperbola defined by xy=18, at the instant when y=6 and dy/dt=8. Since xy=18, when y=6, x=3. We can find dx/dt by implicitly differentiating the equation xy=18.

x(dy/dt) + y(dx/dt) = 0

Here, substitute y=6 and dy/dt=8 into the differentiated equation to solve for dx/dt:

3(8) + 6(dx/dt) = 0

dx/dt = -(3(8))/6 = -4

Thus, x is decreasing by 4 units per second.

Answer 2

The direction and rate of change of x at a certain instant on the hyperbola xy = 18, we use the derivative. In this case, x is decreasing by 4 units per second.

This problem, we'll use the given information about the particle's motion on the hyperbola xy = 18. First, let's differentiate the given equation with respect to time t:

xy = 18

Taking the derivative with respect to time t using the product rule:

y * dx/dt + x * dy/dt = 0

Now, we're given that at a certain instant, y = 6 and dy/dt = 8. We need to find dx/dt at this instant. Substitute the given values into the equation:

6 * dx/dt + x * 8 = 0

Simplify the expression:

dx/dt = -8x/6

Now, we know dx/dt = -8x/6 at the given instant. To determine the sign (whether x is increasing or decreasing), we can use the fact that dx/dt = -8x/6. Since dx/dt and x have opposite signs, we can conclude that x is decreasing. The rate of decrease is 4/3 times the value of x (in absolute terms). Therefore, option (a) 'x is decreasing by 4 units per second' is the correct answer.