Question

Heron's formula The area of a triangle with sides of length \(a, b\), and \(c\) is given by a formula from antiquity called Heron's formula: $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ where \(s=(a+b+c) / 2\) is the semiperimeter of the triangle. a. Find the partial derivatives \(A_{a}, A_{b},\) and \(A_{c}\). b. A triangle has sides of length \(a=2, b=4\), and \(c=5\). Estimate the change in the area when \(a\) increases by \(0.03, b\) decreases by \(0.08\), and \(c\) increases by \(0.6\). c. For an equilateral triangle with \(a=b=c\), estimate the percent change in the area when all sides increase in length by \(p \%\).

Short answer

Question: Compute the percent change in the area of an equilateral triangle when all sides increase in length by 5%. Answer: To compute the percent change in the area, we'll use the formula we derived earlier: $$ = (\sqrt{(\frac{3}{2})(\frac{p}{100})^3} - 1)\times 100\% $$ In this case, \(p = 5\), so we have: $$ = (\sqrt{(\frac{3}{2})(\frac{5}{100})^3} - 1)\times 100\% $$ Calculate the value inside the square root: $$ = (\sqrt{(\frac{3}{2})(\frac{1}{8})^3} - 1)\times 100\% $$ Calculate the square root: $$ = (\sqrt{(\frac{3}{2})(\frac{1}{512})} - 1)\times 100\% $$ $$ = (\sqrt{\frac{3}{1024}} - 1)\times 100\% $$ Now, calculate the value of the square root: $$ = (0.0862 - 1)\times 100\% $$ Finally, compute the percent change: $$ = (-0.9138)\times 100\% \approx -91.4\% $$ The percent change in the area of an equilateral triangle when all sides increase in length by 5% is approximately -91.4%. Note: There is a mistake in the derived formula and the answer, as the value cannot be negative, and it should be a positive value indicating an increase in area. The correct formula (derived from the correct steps in the Solution) should be: $$ = (\sqrt{(\frac{3}{2})(1+\frac{p}{100})^2} - 1)\times 100\% $$ With this correct formula, the percent change in the area would be: $$ = (\sqrt{(\frac{3}{2})(1+\frac{5}{100})^2} - 1)\times 100\% \approx 10.3\% $$ The correct percent change in the area of an equilateral triangle when all sides increase in length by 5% is approximately 10.3%.

Step-by-step solution

Find Partial Derivatives

To find the partial derivatives \(A_a, A_b,\) and \(A_c\), we'll differentiate Heron's formula with respect to each corresponding variable. Note that \(A=\sqrt{s(s-a)(s-b)(s-c)}\) and \(s=\frac{a+b+c}{2}\). First, we'll need to find the derivatives \(\frac{ds}{da}, \frac{ds}{db}, \) and \(\frac{ds}{dc}\): $$\frac{ds}{da}=\frac{d\frac{a+b+c}{2}}{da}=\frac{1}{2}$$ $$\frac{ds}{db}=\frac{d\frac{a+b+c}{2}}{db}=\frac{1}{2}$$ $$\frac{ds}{dc}=\frac{d\frac{a+b+c}{2}}{dc}=\frac{1}{2}$$ Now, we'll use the chain rule to differentiate Heron's formula: $$\frac{dA}{da} = \frac{dA}{ds}\frac{ds}{da}$$ $$\frac{dA}{db} = \frac{dA}{ds}\frac{ds}{db}$$ $$\frac{dA}{dc} = \frac{dA}{ds}\frac{ds}{dc}$$ Then, $$ \frac{dA}{ds} = \frac{s(s-a)(s-b)(s-c)}{\sqrt{s(s-a)(s-b)(s-c)}}=\frac{s(s-a)(s-b)(s-c)}{A} $$We have to compute the derivative of the numerator and denominator. Note that: $$ \frac{d}{ds}(s(s-a)(s-b)(s-c)) = 1(s-a)(s-b)(s-c)+s(-1)(s-b)(s-c)+s(s-a)(-1)(s-c)+s(s-a)(s-b)(-1) $$Now, we're ready to compute the partial derivatives: $$A_a = \frac{dA}{da} = \frac{dA}{ds}\frac{ds}{da} = \frac{1}{2}\frac{- s(s-b)(s-c)+s(s-a)(s-c)+s(s-a)(s-b)}{A}$$ $$A_b = \frac{dA}{db} = \frac{dA}{ds}\frac{ds}{db} = \frac{1}{2}\frac{s(s-a)(s-c)-s(s-a)(s-b)+s(s-b)(s-c)}{A}$$ $$A_c = \frac{dA}{dc} = \frac{dA}{ds}\frac{ds}{dc} = \frac{1}{2}\frac{s(s-a)(s-b)+s(s-b)(s-c)-s(s-a)(s-b)}{A}$$

Estimate Change in Area

Now that we have found the partial derivatives, we can estimate the change in the area of the triangle given by $$ \Delta A \approx A_a\Delta a + A_b\Delta b + A_c\Delta c $$For a triangle with sides of length \(a=2, b=4\), and \(c=5\), we'll use the given increments: \(\Delta a = 0.03, \Delta b = -0.08\), and \(\Delta c = 0.6\). First, we need to find the value of \(s\) and \(A\) for the given triangle: $$s = \frac{a+b+c}{2}=\frac{2+4+5}{2}=5.5$$ $$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5.5(5.5-2)(5.5-4)(5.5-5)} = \sqrt{5.5\cdot3.5\cdot1.5\cdot0.5} = 3.5$$Now, we can estimate the change in the area: $$ \Delta A \approx (0.5)\left(\frac{-5.5(3.5)(0.5)+3.5(1.5)(0.5)+3.5(1.5)(3.5)}{3.5}\right)(0.03) + (0.5)\left(\frac{5.5(1.5)(0.5)-3.5(3.5)(1.5)+3.5(3.5)(0.5)}{3.5}\right)(-0.08) + (0.5)\left(\frac{5.5(3.5)(1.5)+3.5(3.5)(0.5)-3.5(1.5)(3.5)}{3.5}\right)(0.6) $$After doing the calculations, we find that $$ \Delta A \approx 0.276 $$So, the change in the area of the triangle is approximately \(0.276\) square units.

Percent Change in Equilateral Triangle Area

For an equilateral triangle with \(a=b=c\), we'll estimate the percent change in the area when all sides increase in length by \(p\%\). Let the original side length be \(x\). Then, the new side length after increasing by \(p\%\) is \(x(1+p/100)\). Since \(A = \sqrt{s(s-a)(s-b)(s-c)}\), we'll compute $$ \frac{A_{new} - A_{old}}{A_{old}} \times 100\% = \frac{\sqrt{s'(s'-x(1+p/100))^3} - \sqrt{s(s-x)^3}}{\sqrt{s(s-x)^3}}\times 100\% $$where \(s' = (3x(1+p/100))/2\) and \(s = \frac{3x}{2}\). Now, we can express it in term of \(x\) and \(p\), and notice that the \(x\) terms in the numerator will cancel out: $$ \frac{\sqrt{(3/2)x(1+\frac{p}{100})(1+\frac{p}{100}-1)^3} - \sqrt{(3/2)x(1)^3}}{(3/2)x\cdot1^3}\times 100\% = \frac{\sqrt{(3/2)x(1+\frac{p}{100})(\frac{p}{100})^3} - (3/2)x}{(3/2)x}\times 100\% $$After simplification, we get $$ \frac{\sqrt{x(\frac{3}{2})(\frac{p}{100})^3} - x}{x}\times 100\% = \sqrt{(\frac{3}{2})(\frac{p}{100})^3} - 1\times 100\% $$The percent change in the area of an equilateral triangle when all sides increase in length by \(p\%\) is approximately $$ = (\sqrt{(\frac{3}{2})(\frac{p}{100})^3} - 1)\times 100\% $$

Key concepts

Heron's Formula

Heron's formula is a beautiful mathematical expression used to calculate the area of a triangle when the lengths of its three sides are known. Unlike other methods for finding a triangle's area, you don't need to know the height. This formula is especially useful when dealing with irregularly shaped triangles where determining the height might be complex or infeasible. The formula is given by \(A = \sqrt{s(s-a)(s-b)(s-c)}\), where \(s = \frac{a+b+c}{2}\) is the semiperimeter of the triangle. The semiperimeter is simply half the sum of the three side lengths of the triangle. This simplicity in its use allows anyone to quickly compute the area if they have the side lengths at their disposal.
Heron’s formula is a valuable tool for both theoretical proofs and solving practical problems in geometry. It reflects the intrinsic relationship between a triangle's sides and its area without needing additional size determinations such as altitude.

Partial Derivatives

Partial derivatives are fundamental in multivariable calculus, helping us understand how a function changes as each input variable changes. In the context of Heron's formula, we’re interested in \(A_a, A_b,\) and \(A_c\), which represent how the area \(A\) changes with respect to changes in the side lengths \(a, b,\), and \(c\).
To find these derivatives, we first identify the dependencies of the formula: area \(A\) depends on the semiperimeter \(s\) and the side lengths \(a, b,\), and \(c\). By applying the chain rule from calculus, which allows us to handle functions of functions, we can pinpoint the sensitivity of the area to each side length. Each partial derivative combines a differential contribution from altering \(s\) with a direct effect from changing each side length.
Understanding partial derivatives here helps in precisely estimating area changes when side lengths vary slightly, a crucial step when precise calculations or optimizations are needed in fields like engineering and physics.

Triangle Area Estimation

Estimating the area of a triangle becomes important when exact calculations are either impractical or unnecessary. By applying the partial derivatives we derived earlier, small changes in the triangle's sides result in rough but valuable approximations of how the area is affected. This process involves calculating \(\Delta A \approx A_a \Delta a + A_b \Delta b + A_c \Delta c\), where each term accounts for a side's contribution to the overall change.
This estimation method is particularly effective in scenarios where incremental changes occur, such as structural adjustments in engineering or iterative designs in architecture. For example, given a triangle with sides \(a = 2, b = 4,\), and \(c = 5\), we used the computed derivatives to determine the approximate area change when the sides vary by small amounts.
The strategy of estimating versus calculating exact areas saves time and resources, especially when changes are minor. By understanding how each dimension influences the triangle's area, professionals can make informed decisions swiftly and effectively.