The radius of a right circular cylinder is given by sqr( t + 6) and its height is 1/6 sqr(t)

, where t is time in seconds and the dimensions are in inches. Find the rate of change of the volume with respect to time.

v = 1/3 pi r^2 h

= pi/3 (t+6)(1/6 √t)
= (pi/18)t^(3/2) + (pi/3)√t

dv/dt = pi/12 √t + pi/(6√t)

Or, using the chain rule,

v = pi/3 r^2 h
dv/dt = 2pi/3 rh dr/dt + pi/3 r^2 dh/dt
= (2pi/3)√(t+6)(1/6 √t)(1/(2√(t+6)) + (pi/3)(t+6)(1/(12√t))
= 2pi/3 √t/6 + pi/36 √t + pi/(6√t)
= (pi/12)√t + pi/(6√t)

or (pi/12√t)(t+2)

To find the rate of change of the volume with respect to time, we first need to determine the formula for the volume of the right circular cylinder. The formula for the volume of a cylinder is given by:

V = πr^2h

where V is the volume, r is the radius, and h is the height.

Given that the radius of the cylinder is given by sqrt(t + 6) and the height is 1/6 sqrt(t), we can substitute these values into the formula:

V = π(sqrt(t + 6))^2(1/6 sqrt(t))

Simplifying, we have:

V = π(t + 6)(1/36) sqrt(t)

To find the rate of change of the volume with respect to time (dV/dt), we can differentiate the volume formula with respect to t:

dV/dt = π(1/36) (t + 6) (1/2) (1/sqrt(t)) + π(1/36) sqrt(t)

Simplifying further, we have:

dV/dt = π(1/72) [(t + 6)/sqrt(t) + 6√t]

Therefore, the rate of change of the volume with respect to time is π(1/72) [(t + 6)/sqrt(t) + 6√t] inches^3/second.

To find the rate of change of the volume of the cylinder with respect to time, we first need to express the volume of the cylinder in terms of time.

The volume of a right circular cylinder is given by the formula V = πr^2h, where r is the radius and h is the height.

In this case, the radius is given by sqrt(t + 6) and the height is (1/6)sqrt(t).

Substituting these values into the formula, we get:

V = π(sqrt(t + 6))^2(1/6)sqrt(t)

Simplifying the equation, we have:

V = π(t + 6)(1/6)sqrt(t)

Next, we can differentiate the equation with respect to time (t) using the chain rule.

dV/dt = π * (1/6) * (sqrt(t))*(d/dt(t + 6)) + π * (t + 6) * (d/dt((1/6)sqrt(t)))

The derivative of (t + 6) with respect to t is simply 1, and the derivative of (1/6)sqrt(t) with respect to t can be found using the power rule.

dV/dt = π * (1/6) * (sqrt(t)) + π * (t + 6) * ((1/6) * (1/2) * (t^(-1/2)))

Simplifying further, we have:

dV/dt = π * (1/6) * (sqrt(t)) + π * (t + 6) * (1/12) * (t^(-1/2))

Now, we can simplify this expression to get the final answer:

dV/dt = π * (1/6) * (sqrt(t)) + (π/12) * ((t + 6) * (1/sqrt(t)))

Please note that the above expression is the rate of change of the volume with respect to time in terms of t. To find the numerical value of dV/dt at a specific point in time, substitute the value of t into the expression.