A right circular cone has a constant volume. The height h and the base radius r can both vary. Find the rate at which h is changing with respect to r at the instant when r and h are equal.

v = pi/3 r^2 h

since v is a constant, dv = 0

0 = 2rh dr + r^2 dh

When r=h,

2r^2 dr + r^2 dh = 0
2dr + dh = 0

dh = -2dr

dh/dr = -2

To find the rate at which the height (h) is changing with respect to the base radius (r) at the instant when they are equal, we need to apply the concept of related rates.

Let's assume the volume of the right circular cone is V, and we know that it is constant. The formula for the volume of a cone is given by V = (1/3)πr^2h, where π is Pi, r is the base radius, and h is the height.

Now, we need to find the relationship between r and h when they are equal. So, at the instant when r and h are equal, we have r = h.

Next, we differentiate both sides of the equation with respect to time (t) using the chain rule:

d/dt(r) = d/dt(h)
d/dt(r) = dh/dt

Now, we will differentiate the volume formula implicitly with respect to time (t):

d/dt(V) = d/dt((1/3)πr^2h)

Since the volume is constant, its derivative is zero:

0 = d/dt((1/3)πr^2h)

Using the product rule of differentiation, we can expand this expression:

0 = (1/3)π(2r)(dr/dt)(h) + (1/3)π(r^2)(dh/dt)

Simplifying further, we get:

0 = (2/3)πr(dr/dt)(h) + (1/3)π(r^2)(dh/dt)

Substituting r = h, we have:

0 = (2/3)πh(dr/dt)(h) + (1/3)π(h^2)(dh/dt)

0 = (2/3)πh^2(dr/dt) + (1/3)πh^2(dh/dt)

Since we are interested in finding dh/dt when r = h, substitute h for r:

0 = (2/3)πh^2(dr/dt) + (1/3)πh^2(dh/dt)

Now, we can solve for dh/dt when r = h:

0 = (2/3)πh^2(dr/dt) + (1/3)πh^2(dh/dt)

Rearranging the equation:

(1/3)πh^2(dh/dt) = -(2/3)πh^2(dr/dt)

Finally, divide both sides by (1/3)πh^2:

dh/dt = -(2/3)(dr/dt)

Therefore, at the instant when r and h are equal, the rate at which h is changing with respect to r is -(2/3) times the rate at which r is changing with respect to t.