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 \(s=\frac{1}{2}(a+b+c)\) is the semiperimeter of the triangle. a. Find the partial derivatives \(A_{\sigma}, A_{b},\) and \(A_{c}\) b. A triangle has sides of length \(a=2, b=4, c=5 .\) 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%. Solution: First, we will find the ratio between the areas of the original (\(A_1\)) and new triangle (\(A_2\)). We are given a percent increase of 5% in the side length, so we have: $$L' = L(1 + \frac{5}{100}) = L(1.05)$$ Now, we need to compute the areas of the original (\(A_1\)) and new triangle (\(A_2\)) using the formula \(A = \frac{\sqrt{3}}{4}L^2\). $$A_1 = \frac{\sqrt{3}}{4}L^2$$ $$A_2 = \frac{\sqrt{3}}{4}(L')^2 = \frac{\sqrt{3}}{4}(L(1.05))^2$$ Next, we will determine the percent change in area: $$\text{Percent change} = \frac{A_2-A_1}{A_1} \times 100\%$$ Substituting the expressions for \(A_1\) and \(A_2\), we get: $$\text{Percent change} = \frac{\frac{\sqrt{3}}{4}(L(1.05))^2-\frac{\sqrt{3}}{4}L^2}{\frac{\sqrt{3}}{4}L^2} \times 100\%$$ Simplify the expression: $$\text{Percent change} = \frac{(1.05)^2-1}{1} \times 100\%$$ Calculate the result to get: $$\text{Percent change} \approx 10.25\%$$ Hence, the percent change in the area of an equilateral triangle when all sides increase in length by 5% is approximately 10.25%.

Step-by-step solution

Compute s

First, we need to compute the semiperimeter \(s\). Recall that \(s = \frac{1}{2}(a+b+c)\). We will then use \(s\) to find the area \(A\) using Heron's formula.

Finding the partial derivatives

Now we need to find the partial derivatives \(A_{a}\), \(A_{b}\), and \(A_{c}\). Using Heron's formula: $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ Taking the partial derivatives with respect to \(a\), \(b\), and \(c\) respectively, we get: $$A_a = \frac{dA}{da} = \frac{\partial}{\partial a} \sqrt{s(s-a)(s-b)(s-c)}$$ $$A_b = \frac{dA}{db} = \frac{\partial}{\partial b} \sqrt{s(s-a)(s-b)(s-c)}$$ $$A_c = \frac{dA}{dc} = \frac{\partial}{\partial c} \sqrt{s(s-a)(s-b)(s-c)}$$ Applying the chain rule, we find the partial derivatives: $$A_a = \frac{-(s-b)(s-c)}{2\sqrt{s(s-a)(s-b)(s-c)}}$$ $$A_b = \frac{-(s-a)(s-c)}{2\sqrt{s(s-a)(s-b)(s-c)}}$$ $$A_c = \frac{-(s-a)(s-b)}{2\sqrt{s(s-a)(s-b)(s-c)}}$$ Part b:

Triangle with given side lengths and increments

We are given a triangle with side lengths \(a = 2\), \(b = 4\), \(c = 5\), and respective increments \(\Delta a = 0.03\), \(\Delta b = -0.08\), \(\Delta c = 0.6\). We will now estimate the change in the area by evaluating the partial derivatives of A at the given side lengths and multiplying them by the increments. First, compute \(s\): \(s = \frac{1}{2}(2+4+5) = 5.5\). Now, evaluate the partial derivatives at the given side lengths: $$A_a = \frac{-(s-b)(s-c)}{2\sqrt{s(s-a)(s-b)(s-c)}} = \frac{-3.5\cdot1.5}{2\sqrt{5.5\cdot3.5\cdot1.5\cdot0.5}}$$ $$A_b = \frac{-(s-a)(s-c)}{2\sqrt{s(s-a)(s-b)(s-c)}} = \frac{-4.5\cdot1.5}{2\sqrt{5.5\cdot3.5\cdot1.5\cdot0.5}}$$ $$A_c = \frac{-(s-a)(s-b)}{2\sqrt{s(s-a)(s-b)(s-c)}} = \frac{-3.5\cdot0.5}{2\sqrt{5.5\cdot3.5\cdot1.5\cdot0.5}}$$

Estimate the change in area

To estimate the change in area, we will use the linear approximation, which states that: $$\Delta A \approx A_a \Delta a + A_b \Delta b + A_c \Delta c$$ Substituting the values from the previous steps, we get: $$\Delta A \approx \frac{-3.5\cdot1.5}{2\sqrt{5.5\cdot3.5\cdot1.5\cdot0.5}}(0.03) + \frac{-4.5\cdot1.5}{2\sqrt{5.5\cdot3.5\cdot1.5\cdot0.5}}(-0.08) + \frac{-3.5\cdot0.5}{2\sqrt{5.5\cdot3.5\cdot1.5\cdot0.5}}(0.6)$$ Calculate the result to get the estimated change in the area. Part c:

Finding the percent change for an equilateral triangle with p% increase in side length

For an equilateral triangle, we have \(a=b=c\). Let's say the side length is \(L\). When all sides of the triangle increase in length by \(p\%\), each side becomes \(L(1+\frac{p}{100})\). Let the new side length be \(L'\). Now we will find the percent change in area. We need to find the ratio between the areas of the original (\(A_1\)) and new triangle (\(A_2\)). For an equilateral triangle, we use the formula \(A = \frac{\sqrt{3}}{4}L^2\). $$A_1 = \frac{\sqrt{3}}{4}L^2$$ $$A_2 = \frac{\sqrt{3}}{4}(L')^2 = \frac{\sqrt{3}}{4}(L(1+\frac{p}{100}))^2$$ Now, find the percent change in area: $$\text{Percent change} = \frac{A_2-A_1}{A_1} \times 100\%$$ Calculate the result to get the percent change in the area when all sides of the equilateral triangle increase in length by \(p \%\).

Key concepts

Partial Derivatives

Partial derivatives play a crucial role in understanding how a multivariable function, like the area of a triangle in terms of its sides, changes when one variable is altered while others remain constant. Think of partial derivatives as a way to peek into the behavior of a function, scrutinizing the rate of change one slice at a time.

For instance, in Heron's formula, the partial derivatives with respect to each side of the triangle provide insight into how the area 'reacts' when that specific side is extended or shortened, while the other sides stay the same. This information could be valuable, for example, when trying to determine how sensitive the area of a land plot is to changes in one of its boundary lengths.

When we calculate the partial derivatives of the area (\( A \)) using Heron's formula with respect to the sides (\(a, b, c\)), we apply the chain rule from calculus. This rule allows us to break down the derivative of the complex square root expression by treating the other variables as constants and focusing on each side independently. By understanding these derivatives, we are better equipped to predict minor adjustments in the triangle's geometry and their implications on its area.

Semiperimeter

The semiperimeter of a triangle, represented by the variable \( s \), is essentially half of the triangle's perimeter. The semiperimeter is not just a smaller version of the perimeter but a bridge to various geometric properties and formulas like Heron's formula.

Understanding the concept of the semiperimeter can reveal fascinating insights into triangle geometry. It is used in Heron's formula because of the magical way it simplifies the expression for the area. By using the semiperimeter, we can elegantly express the area of a triangle in a single formula without knowing anything about its angles or height.

To calculate the semiperimeter, you add the lengths of all three sides and then divide by two. This simple yet powerful step is the starting point in Heron's formula, connecting the lengths of the sides of a triangle directly to the magnitude of its area.

Area of a Triangle

The area of a triangle (\( A \) for short) is a measure of the two-dimensional space enclosed by the three sides of the triangle. There are various formulas to calculate the area depending on the information available, but Heron's formula provides a direct method when the lengths of all three sides are known.

Heron’s formula is remarkable because it doesn't require an altitude or angle measurements. It states that the area of a triangle can be determined by knowing just the lengths of its sides and the semi-perimeter, unlocking a new dimension of simplicity and elegance in triangle geometry. In the context of this problem, mastery of this area calculation allows one to deduce the impact of side length variations on the overall shape and size of the triangle. This can be especially useful in fields such as architecture and engineering, where precise measurements are paramount.

Remember, each time we speak of the area of a triangle, we are referring to a two-dimensional concept that plays a fundamental role in geometry and everyday applications. Whether it is a simple garden plot or a complex architectural component, the principles for finding the area remain grounded in methods like Heron’s formula.