Let V be the volume of a right circular cone having height h and radius r and assume that h and r vary with time.
a. Express the time rate of change of the cylinder in terms of h, r and their rates of change.
b. At a certain instant, the height is 10 in and decreasing at a rate of 1.5 in/sec, while the
radius is 3 in and increasing at a rate of 2 in/sec. How fast is the volume changing at
that instant and state whether the volume is decreasing or increasing.
V is a function of r(t) and h(t)
so use the product rule and chain rule:
V(t)=(1/3)πr(t)²h(t)
V'(t)=(1/3)π[2r*r'(t)]h(t)+(1/3)πr(t)²h'(t)
Can you take it from here?
To answer this question, we need to relate the volume of the cone to its height and radius, and then differentiate the volume with respect to time to find the rate of change.
a. Expressing the volume of the cone in terms of its height and radius:
The volume of a right circular cone is given by the formula V = (1/3)πr^2h, where π is a constant (approximately 3.14159).
b. To find the time rate of change of the volume, we need to differentiate the volume formula with respect to time (t) using the chain rule. Let's denote the rates of change of h and r with respect to time as dh/dt and dr/dt, respectively.
Taking the derivative of V with respect to t, we get:
dV/dt = (1/3)π(2rh(dr/dt) + r^2(dh/dt))
Now, let's substitute the given values:
h = 10 in (height)
dh/dt = -1.5 in/sec (decreasing rate of height)
r = 3 in (radius)
dr/dt = 2 in/sec (increasing rate of radius)
Substituting these values, we can find the value of dV/dt.
dV/dt = (1/3)π(2(3)(2) + (3^2)(-1.5))
= (1/3)π(12 - 13.5)
= (1/3)π(-1.5)
So, the rate of change of the volume at that instant can be expressed as -(1.5/3)π.
To determine whether the volume is decreasing or increasing, we observe that the rate of change is negative, indicating that the volume is decreasing.
Therefore, at that instant, the volume is changing at a rate of -(1.5/3)π in^3/sec, and it is decreasing.