Understanding Rate of Change for Hemisphere Surface Area

The question asks for the rate at which the total surface area of a hemisphere changes as its radius changes. This is a calculus problem requiring us to find the derivative of the total surface area formula with respect to the radius and evaluate it at a specific radius value.

Total Surface Area of a Hemisphere

A solid hemisphere consists of two parts:

• The curved surface area.
• The circular base area.

The formula for the curved surface area of a hemisphere with radius \(r\) is \(2\pi r^2\).

The formula for the area of the circular base with radius \(r\) is \(\pi r^2\).

Therefore, the total surface area (\(A\)) of a solid hemisphere is the sum of these two areas:

\(A = \text{Curved Surface Area} + \text{Base Area}\)

\(A = 2\pi r^2 + \pi r^2\)

\(A = 3\pi r^2\)

Calculating the Rate of Change with Respect to Radius

The rate of change of the total surface area with respect to the radius \(r\) is given by the derivative of the total surface area formula (\(A\)) with respect to \(r\). We denote this as \(\frac{dA}{dr}\).

We need to find the derivative of \(A = 3\pi r^2\).

Using the power rule for differentiation, \(\frac{d}{dr}(r^n) = nr^{n-1}\):

\(\frac{dA}{dr} = \frac{d}{dr}(3\pi r^2)\)

\(\frac{dA}{dr} = 3\pi \frac{d}{dr}(r^2)\) (Since \(3\pi\) is a constant)

\(\frac{dA}{dr} = 3\pi (2r^{2-1})\)

\(\frac{dA}{dr} = 3\pi (2r)\)

\(\frac{dA}{dr} = 6\pi r\)

This formula \(6\pi r\) gives the rate of change of the total surface area of the hemisphere with respect to its radius \(r\).

Evaluating the Rate of Change at a Specific Radius

The question asks for the rate of change at a specific radius \(r = \sqrt[3]{1.331}\) cm.

First, let's simplify the value of \(r\):

\(1.331 = 1.1 \times 1.1 \times 1.1 = (1.1)^3\)

So, \(r = \sqrt[3]{(1.1)^3} = 1.1\) cm.

Now, substitute this value of \(r\) into the derivative formula \(\frac{dA}{dr} = 6\pi r\):

\(\frac{dA}{dr} \Big|_{r=1.1} = 6\pi (1.1)\)

\(\frac{dA}{dr} \Big|_{r=1.1} = 6.6\pi\)

The rate of change of the total surface area of the hemisphere with respect to the radius at \(r = \sqrt[3]{1.331}\) cm is \(6.6\pi\).

The units of this rate of change are cm²/cm, which simplifies to cm. The question mentions cm²/s, which might imply a context involving change over time, but our calculation is strictly the rate of change with respect to the radius itself.

Comparing with Options

Let's compare our result \(6.6\pi\) with the given options:

1. \(66\pi\)
2. \(6.6\pi\)
3. \(3.3\pi\)
4. \(4.4\pi\)

Our calculated value \(6.6\pi\) matches option 2.

Therefore, the rate of change of the total surface area of a hemisphere with respect to radius r at r = ³√1.331 cm is \(6.6\pi\) cm²/cm.

Revision Table: Hemisphere Surface Area Rate of Change

Additional Information: Calculus and Geometric Shapes

Calculus is a powerful tool for understanding how quantities change. The derivative helps us find instantaneous rates of change. In geometry, this allows us to analyze how areas, volumes, or lengths of shapes change as their dimensions change.

• Rates with respect to Radius: When we find \(\frac{dA}{dr}\), we are calculating how fast the area changes for a small change in radius. This is useful in various applications, like manufacturing or material science, where properties depend on the dimensions of a spherical object.
• Rates with respect to Time: Sometimes, the radius of a shape might be changing over time (e.g., a balloon inflating). In such cases, we might be interested in \(\frac{dA}{dt}\) (rate of change of area with respect to time) or \(\frac{dV}{dt}\) (rate of change of volume with respect to time). This involves the chain rule: \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\). The question here specifically asked for the rate with respect to radius, not time.
• Units: Pay close attention to units. Area is measured in units², radius in units. The rate of change of area with respect to radius (\(\frac{dA}{dr}\)) will have units of units²/unit, which simplifies to units. The cm²/s in the question might indicate a broader context where the radius is changing over time, but the derivative \(\frac{dA}{dr}\) is calculated as shown.
• Other Hemisphere Properties:
  • Volume of a hemisphere: \(V = \frac{2}{3}\pi r^3\). The rate of change of volume with respect to radius would be \(\frac{dV}{dr} = \frac{d}{dr}(\frac{2}{3}\pi r^3) = \frac{2}{3}\pi (3r^2) = 2\pi r^2\).
  • Curved surface area: \(A_{curved} = 2\pi r^2\). The rate of change of curved surface area with respect to radius is \(\frac{dA_{curved}}{dr} = \frac{d}{dr}(2\pi r^2) = 2\pi (2r) = 4\pi r\).

Understanding these different rates of change helps in solving various problems involving geometry and calculus.