A particle is moving along the curve whose equation is (xy^3)/(1+y^2)= 8/5. Assume the x-coordinate is increasing at the rate of 6 units/second when the particle is at the point (1,2). At what rate is the y-coordinate of the point changing at that instant? Is it rising or falling?

Well, let's take a closer look at the situation. We have a particle moving along a curve given by the equation (xy^3)/(1+y^2) = 8/5. We want to find the rate at which the y-coordinate is changing when the x-coordinate is increasing at 6 units/second, and the particle is at the point (1,2).

To solve this problem, we need to use implicit differentiation. Differentiating both sides of the equation with respect to time, we get:

d/dt[(xy^3)/(1+y^2)] = d/dt(8/5)

Now, let's calculate the derivatives. The derivative of the left side of the equation can be calculated using the product rule and the chain rule. For simplicity, let's denote dx/dt as x' and dy/dt as y'. We have:

(x'y^3 + 3xy^2y')/(1+y^2) - (xy^3(2yy'))/(1+y^2)^2 = 0

Next, let's substitute the values of x and y when the particle is at the point (1,2):

(1)(2^3)/(1+2^2) * x' + (3)(1)(2^2)(y')/(1+2^2) - (1)(2^3)(2y')/(1+2^2)^2 = 0

Simplifying this equation, we have:

8/5 * x' + 12y'/(5) - 16y'/(9) = 0

Combining like terms, we get:

(8/5 - 16/9)y' + 12/(5)y' = -8/5 * x'

Simplifying this further, we have:

(40 - 80/9)y' = -8/5 * x'

Multiplying both sides by 9/40, we get:

y' = -9/40 * (8/5) * x'

Simplifying, we have:

y' = -36/200 * x'

Now, let's substitute the given rate of change for x' when the particle is at the point (1,2) and is increasing at 6 units/second:

y' = -36/200 * 6

Calculating this, we find:

y' = -9/25 units/second

So, the y-coordinate is changing at a rate of -9/25 units/second, and since the rate is negative, it implies that the y-coordinate is falling at that instant.

I hope this answers your question, even if it wasn't as funny as a clown juggling bananas!

To find the rate at which the y-coordinate of the point is changing when the particle is at the point (1,2), we can use implicit differentiation.

First, let's differentiate both sides of the equation with respect to time, t, using the chain rule:

d/dt [(xy^3)/(1+y^2)] = d/dt (8/5)

Now, let's differentiate each term separately:

Using the product rule, d(xy^3)/dt = y^3 * dx/dt + 3xy^2 * dy/dt.

Using the chain rule, d(1+y^2)/dt = 2y * dy/dt.

Note that dx/dt is given as 6 units/second.

Substituting these derivatives into the equation, we have:

[y^3 * dx/dt + 3xy^2 * dy/dt] / (1+y^2) - [2y * dy/dt] / (1+y^2) = 0

Now, let's substitute the values of x and y when the particle is at the point (1,2):

[2^3 * 6 + 3(1)(2)^2 * dy/dt] / (1+2^2) - [2(2) * dy/dt] / (1+2^2) = 0

Simplifying further, we have:

[48 + 12 * dy/dt] / 5 - [4 * dy/dt] / 5 = 0

Multiplying through by 5 to eliminate the denominators:

48 + 12 * dy/dt - 4 * dy/dt = 0

Combining like terms:

48 + 8 * dy/dt = 0

Subtracting 48 from both sides:

8 * dy/dt = -48

Dividing by 8:

dy/dt = -6

Therefore, the rate at which the y-coordinate of the point is changing when the particle is at the point (1,2) is -6 units/second. Since the rate is negative, the y-coordinate is falling.

-10/7 units per second

Use implicit differentiation:

xy^3 / (1+y^2) = 8/5

(y^3 + 3xy^2 y')(1+y^2) - xy^3 (2yy') = 0

It's all over (1+y^2)^2, but that can be ignored, since it's never 0.

y'(3xy^2 + 3xy^4 - 2xy^4) = -y^3(1 + y^2)

y' = -y^3 (1+y^2)/(3xy^2 + xy^4)

= -y/x * (1+y^2)/(3 + y^2)

Take it from there.