Question

The volume of the cap of a sphere of radius \(r\) and thickness \(h\) is \(V=\frac{\pi}{3} h^{2}(3 r-h),\) for \(0 \leq h \leq r.\) a. Compute the partial derivatives \(V_{h}\) and \(V_{r}\) b. For a sphere of any radius, is the rate of change of volume with respect to \(r\) greater when \(h=0.2 r\) or when \(h=0.8 r ?\) c. For a sphere of any radius, for what value of \(h\) is the rate of change of volume with respect to \(r\) equal to \(1 ?\) d. For a fixed radius \(r,\) for what value of \(h(0 \leq h \leq r)\) is the rate of change of volume with respect to \(h\) the greatest?

Short answer

b. Which value of \(h\) has a greater rate of change of volume with respect to \(r\): \(h=0.2r\) or \(h=0.8r\)? c. What is the value of \(h\) where the rate of change of volume with respect to \(r\) is equal to 1? d. For a fixed radius, what is the value of \(h\) where the rate of change of volume with respect to \(h\) is the greatest?

Step-by-step solution

a. Compute the partial derivatives \(V_{h}\) and \(V_{r}\)

We are given the volume formula for a cap of sphere: \(V=\frac{\pi}{3} h^{2}(3 r-h).\) To compute the partial derivatives, we differentiate the volume formula with respect to \(h\) and \(r\) separately. First, let's find the partial derivative with respect to \(h\): \(V_{h}=\frac{\partial V}{\partial h}=\frac{\partial}{\partial h}\left(\frac{\pi}{3} h^{2}(3 r-h)\right)\). Using product rule and applying the derivative, we get: \(V_{h}=\frac{\pi}{3}\left(2h(3r-h) - h^{2}\right)\). Next, let's find the partial derivative with respect to \(r\): \(V_{r}=\frac{\partial V}{\partial r}=\frac{\partial}{\partial r}\left(\frac{\pi}{3} h^{2}(3 r-h)\right)\). Applying the derivative, we get: \(V_{r}=\frac{\pi}{3}(3h^{2})\).

b. Comparing the rate of change of volume with respect to \(r\) for different values of \(h\)

Now let's evaluate the rate of change of volume with respect to \(r\) at \(h=0.2r\) and \(h=0.8r\). We'll use the \(V_{r}\) we just found. For \(h=0.2r\), \(V_{r}(0.2r)=\frac{\pi}{3}(3(0.2r)^{2})=\frac{12\pi r^{2}}{30}=\frac{2\pi r^{2}}{5}\). For \(h=0.8r\), \(V_{r}(0.8r)=\frac{\pi}{3}(3(0.8r)^{2})=\frac{12\pi r^{2}}{5}\). Comparing these values, we can see that the rate of change of volume with respect to \(r\) is greater when \(h=0.8r.\)

c. Finding the value of \(h\) where the rate of change of volume with respect to \(r\) is 1

We want to find the value of \(h\) for which \(V_{r}(h)=1\). \(\frac{\pi}{3}(3h^{2})=1\) To find the value of \(h\), we first solve for \(h^{2}\): \(3h^{2}=\frac{3}{\pi}\), \(h^{2}=\frac{1}{\pi}\). Taking the square root of both sides: \(h=\sqrt{\frac{1}{\pi}}\) or \(h=-\sqrt{\frac{1}{\pi}}\). Since \(0 \leq h \leq r\), we choose the positive value for \(h\): \(h=\sqrt{\frac{1}{\pi}}\).

d. The value of \(h\) for the greatest rate of change of volume with respect to \(h\)

Given a fixed radius \(r\), we want to find the value of \(h\) for which the rate of change of volume with respect to \(h\) is the greatest. Using the \(V_{h}\) we found earlier: \(V_{h}=\frac{\pi}{3}\left(2h(3r-h) - h^{2}\right)\). To find the greatest rate of change, we set the second derivative equal to 0: \(V_{hh}=\frac{\partial^{2} V}{\partial h^{2}}=\frac{\partial}{\partial h}(V_{h})=0\). Taking the derivative of \(V_{h}\) with respect to \(h\), we get: \(V_{hh}=\frac{\pi}{3}\left(2(3r-3h)-4h\right)\) Setting \(V_{hh}\) to 0 and solving for \(h\): \(2(3r-3h)-4h=0\) \(6r-6h=4h\) \(6r=10h\) Finally, we find the value of \(h\): \(h=\frac{3}{5}r\). So, for a fixed radius \(r\), the rate of change of volume with respect to \(h\) is the greatest when \(h=\frac{3}{5}r\).

Key concepts

Volume of a Sphere

The volume of a sphere is a fundamental geometric concept that helps in understanding three-dimensional shapes. For a complete sphere with radius \( r \), the volume is calculated using the formula: \[ V = \frac{4}{3} \pi r^3 \]. However, the problem addressing the volume here is a specific portion of a sphere known as the cap. The volume of this cap segment is slightly different as it depends not only on the radius \( r \) of the sphere but also on the height \( h \) of the cap segment. The formula provided is: \[ V = \frac{\pi}{3} h^2 (3r-h) \], where \( 0 \leq h \leq r \). In essence, this formula calculates how much space the cap, or part of the sphere, occupies in a 3D space. The reliance on \( h \), or height, indicates that as the cap grows taller, the volume generally increases up to a point.

Rate of Change

The rate of change is crucial in understanding how a quantity alters when we change one of the components on which that quantity depends. In calculus, the rate of change with respect to a variable is expressed as a derivative.
The exercise requires computing the partial derivatives to evaluate how the volume changes with respect to the sphere's radius \( r \) and the cap height \( h \).

• Partial derivative with respect to \( h \): \( V_h = \frac{\pi}{3} \left( 2h(3r-h) - h^2 \right) \)
• Partial derivative with respect to \( r \): \( V_r = \frac{\pi}{3} (3h^2) \)

By understanding these derivatives, we can evaluate scenarios such as whether the volume increases faster when \( h = 0.2r \) or \( h = 0.8r \). The derivative \( V_r \) reveals the volume change concerning the radius, demonstrating that the volume alters more substantially at the increased cap height \( h = 0.8r \).

Product Rule

The product rule is a fundamental concept in calculus used to differentiate functions that are products of two or more sub-functions. It states that the derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by: \[\frac{d}{dx}(uv) = u'v + uv'\].
In the context of the volume of the sphere cap, this rule helps in finding the derivative of terms like \( h^2(3r-h) \) with respect to \( h \). By using the product rule, we correctly compute the derivative which involves separating \( h^2 \) as one function and \( 3r-h \) as another. This results in: \( V_h = \frac{\pi}{3} \left( 2h(3r-h) - h^2 \right) \). The rule is crucial because, without it, breaking down the differentiations in such complex expressions becomes far more challenging. Therefore, mastering the product rule empowers students to tackle complex differentiation tasks fluently.