Question

The volume of a torus (doughnut or bagel) with an inner radius of \(a\) and an outer radius of \(b\) is \(V=\pi^{2}(b+a)(b-a)^{2} / 4\) a. Find \(d b / d a\) for a torus with a volume of \(64 \pi^{2}\). b. Evaluate this derivative when \(a=6\) and \(b=10\)

Short answer

Based on the given information, the derivative of the outer radius (b) with respect to the inner radius (a), or (db/da), when the volume of the torus is 64π^2 with a=6 and b=10, is -1/12.

Step-by-step solution

Write down the given formula for the volume of the torus

The volume of the torus is given by the formula \(V=\pi^{2}(b+a)(b-a)^{2} / 4\)

Apply implicit differentiation

To find \(db/da\), first take the derivative of both sides of the volume equation with respect to \(a\), using implicit differentiation: \(\frac{dV}{da} = \frac{d}{da}\left(\pi^{2}(b+a)(b-a)^{2} / 4\right)\)

Use the product rule and chain rule

The expression on the right-hand side is a product of two functions, so we'll use the product rule. Also, \((b-a)^{2}\) requires the chain rule to differentiate. Applying those rules, we get: \(\frac{dV}{da} = \pi^{2}\left[\frac{d}{da}(b+a)\cdot\frac{(b-a)^{2}}{4} + \frac{d}{da}\left(\frac{(b-a)^{2}}{4}\right)\cdot(b+a)\right]\)

Simplify the expression

Differentiate and simplify the expression: \(\frac{dV}{da} = \pi^{2}\left[1\cdot\frac{(b-a)^{2}}{4} - \frac{2(b-a)}{4}\cdot(b+a)\right]\) \(\frac{dV}{da} = \pi^{2}\left[\frac{(b-a)^{2}}{4} - \frac{(b-a)(b+a)}{2}\right]\)

Solve for \(db/da\) at a given volume

We are asked to find \(db/da\) at \(V = 64\pi^2\). Substitute in the given volume and solve for \(db/da\) : \(64\pi^{2}= \pi^{2}\left[\frac{(b-a)^{2}}{4} - \frac{(b-a)(b+a)}{2}\right]\) We can divide both sides by \(\pi^2\): \(64 = \frac{(b-a)^{2}}{4} - \frac{(b-a)(b+a)}{2}\). Now, we need to find \(db/da\) by rearranging and using implicit differentiation. But since it is hard to find \(db/da\) from this equation, it is better if we find \(da/db\) first then take its reciprocal. So now, we differentiate with respect to \(b\) : \(\frac{da}{db} = \frac{d}{db}\left(\frac{(b-a)^{2}}{4} - \frac{(b-a)(b+a)}{2} -64\right)\)

Differentiate and simplify

Apply the product and chain rule, and simplify the expression: \(\frac{da}{db} = \frac{2(b-a)}{4} - \frac{1\cdot(b+a) + (b-a)\cdot1}{2}\) \(\frac{da}{db} = \frac{1}{2}(b-a) - (b+a) + \frac{1}{2}(b-a)\) \(\frac{da}{db} = b-a - (b+a) + b-a\) \(\frac{da}{db} = -2a\)

Find \(db/da\)

Now that we have \(da/db\), we can find \(db/da\) by taking the reciprocal: \(\frac{db}{da} = -\frac{1}{2a}\)

Evaluate the derivative when a=6 and b=10

Finally, substitute the given values of \(a\) and \(b\) to find the specified value: \(\frac{db}{da} = -\frac{1}{2\cdot6} = -\frac{1}{12}\)

Key concepts

Implicit Differentiation

Implicit differentiation is a powerful technique to find derivatives of equations that are not easily solved for one variable in terms of another. Unlike regular differentiation, where you explicitly solve for one variable before differentiating, implicit differentiation allows the simultaneous differentiation of both sides of an equation with respect to one variable.
This method is particularly useful when dealing with equations where one variable cannot be easily isolated. In our exercise, implicit differentiation helps us find the derivative of the torus volume equation with respect to the variable \(a\) without having to explicitly solve for \(b\) in terms of \(a\). This is done by treating all occurrences of the variable as functions and applying differentiation rules accordingly.

Chain Rule

The chain rule is an essential principle in calculus for finding the derivative of composite functions. A composite function is a function that involves inserting one function inside another.
The chain rule states that if you have a composite function \(f(g(x))\), the derivative is given by the derivative of the outer function evaluated at the inner function times the derivative of the inner function: \[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\] In the context of our exercise, we use the chain rule to differentiate the expression \((b-a)^2\) as we treat \(b-a\) as an inner function. This allows us to correctly find the implicit differentiation of the torus's volume function by recognizing how changes in \(b\) and \(a\) affect its value.

Product Rule

The product rule is another fundamental tool in calculus that helps us differentiate products of two functions. When faced with a situation where two functions are multiplied together, the product rule states that the derivative is: \[\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)\] Where \(u(x)\) and \(v(x)\) are the two functions being multiplied, and \(u'(x)\) and \(v'(x)\) are their respective derivatives.
In our problem, the product rule is applied to the expression \((b+a) \cdot (b-a)^2/4\). This expression consists of two main parts: \((b+a)\) and \((b-a)^2/4\). By applying the product rule, and then using the chain rule for the component \((b-a)^2\), we systematically find the derivative needed to assess how the volume of a torus changes with a change in the radius \(a\).

Volume of a Torus

A torus is a three-dimensional geometric shape that resembles a doughnut or a bagel. The volume of a torus is determined by its radii – the inner radius \(a\) and the outer radius \(b\). The formula to compute the volume \(V\) of a torus is: \[V = \frac{\pi^2 (b+a)(b-a)^2}{4}\] This formula takes into account the entire space enclosed by the torus using the regions defined by \(a\) and \(b\).
The exercise specifically deals with deriving the relationship between the radii that affect this volume when it is held constant, showing how each radius's change impacts the other's rate of change. Understanding the volume formula is crucial because it forms the basis of applying various calculus techniques like implicit differentiation to investigate how these dynamic variables interact in complex geometries like a torus.