Question

A square of side length s is inscribed in a circle of radius r. (a) Write the area A of the square as a function of the radius r of the circle. (b) Find the (instantaneous) rate of change of the area A with respect to the radius r of the circle. (c) Evaluate the rate of change of \(A\) at \(r=1\) and \(r=8\) (d) If \(r\) is measured in inches and \(A\) is measured in square inches, what units would be appropriate for \(d A / d r ?\)

Short answer

a) \(A = 2 * r^2\), b) \(dA/dr = 4*r\), c) For r=1, \(dA/dr=4\) and for r=8, \(dA/dr=32\), d) 'square inches per inch' would be the appropriate units for \(dA/dr\).

Step-by-step solution

Express the Area of the Square as a Function of the Radius of the Circle

Drawing a diagonal of the square cuts the circle into two, forming two identical right-angled triangles. The hypotenuse of each triangle is the diameter of the circle, and the sides of the triangle are the lengths of the square. Given that the diameter of the circle is \(2r\), then from Pythagoras' theorem, we derive that the length of the sides of the square \(s = \sqrt{2} * r\). The area of the square \(A = s^2\), substituting \(s\) for \( \sqrt{2} * r\), we get \(A = 2 * r^2\).

Find the Rate of Change of Area, 'A'

The rate of change of 'A' with respect to 'r' is given by \(dA/dr\). So, differentiating \(A = 2*r^2\) with respect to r, gives \(dA/dr = 4*r\).

Evaluate the Rate of Change at r=1 and r=8

Substituting r=1 into \(dA/dr\), we get: \(dA/dr = 4*1 = 4\). And substituting r=8 into \(dA/dr\) gives: \(dA/dr = 4*8 = 32\).

Determine the Appropriate Units for the Rate of Change

Given that 'r' is in inches and 'A' is in square inches, the units of \(dA/dr\) would be 'square inches per inch'.

Key concepts

Inscribed Square

When we talk about an inscribed square, we are referring to a square that is drawn inside a circle such that all four corners of the square touch the circle. Imagine a picture frame that fits snugly inside a dinner plate; that frame is to the plate what an inscribed square is to its circle. The relationship between the square and the circle is geometrically special because it sets the stage for applying the powerful Pythagoras' theorem.

To find the side length of the inscribed square, one must remember that the diagonal of the square is equal to the diameter of the circle. This relationship allows us to use the properties of right triangles to derive the formula for the side length in terms of the circle's radius, which then leads to determining the area of the square.

Pythagoras' Theorem

Pythagoras' theorem is a fundamental principle in geometry that relates the sides of a right-angled triangle. According to the theorem, in any right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, if 'c' represents the length of the hypotenuse, and 'a' and 'b' represent the lengths of the other two sides, the theorem is expressed as:\[c^2 = a^2 + b^2\].

In the context of an inscribed square in a circle, we use this theorem to determine the side length of the square from the radius of the circle. Applying this principle effectively ties together algebraic expressions and geometric properties, facilitating deeper understanding not just for this specific problem, but for a broad range of applications in mathematics.

Derivative of Area

Mathematically, the derivative of a function represents the rate at which the function's output changes in relation to changes in its input. In the case of the area of a shape, taking the derivative of the area function with respect to one of its dimensions provides the rate of change of the area with respect to that dimension. For instance, if we have a square whose area 'A' is given as a function of its side length 's', the derivative of 'A' with respect to 's' (denoted as \(\frac{dA}{ds}\)) would tell us how much the area of the square changes for a small increase in the side length.

In tutorial exercises involving an inscribed square, calculating the area's derivative concerning the circle's radius (instead of the side length) gives us powerful insights into how the square's area expands or contracts as the circle's size changes. The direct functional relationship established between area and radius through differentiation is essential for exploring and solving problems related to rates of change in calculus.

Functional Relationship

In mathematics, a functional relationship is the connection between two variables such that each input value corresponds with exactly one output value. For example, when we express the area 'A' of a square as a function of its side length 's', we define a specific dependence where changing 's' will yield a particular change in 'A'. This relationship is simple and direct; as 's' increases, so does 'A', following the rule \(A = s^2\).

However, when a square is inscribed in a circle, and we wish to express the square's area as a function of the circle's radius, the functional relationship becomes slightly more complex. The side length 's' of the square is not independent of the circle's radius 'r', which requires us to use geometric theorems to link 's' and 'r'. Once we understand how 's' is related to 'r', we can then rewrite the area 'A' of the square as a function of 'r'. This concept is a beautiful illustration of how different domains of mathematics, such as geometry and algebra, interconnect to solve problems.