Problem 8 - Calculus Early Transcendentals

Question

Shrinking isosceles triangle The hypotenuse of an isosceles right triangle decreases in length at a rate of \(4 \mathrm{m} / \mathrm{s}\). a. At what rate is the area of the triangle changing when the legs are \(5 \mathrm{m}\) long? b. At what rate are the lengths of the legs of the triangle changing? c. At what rate is the area of the triangle changing when the area is \(4 \mathrm{m}^{2} ?\)

Step-by-Step Solution

Answer

a. The legs are 5 meters long. b. The area is 4 square meters. Answer: a. When the legs are 5 meters long, the rate at which the lengths of the legs are changing is -2√2 meters per second, and the rate at which the area is changing is -20 square meters per second. b. When the area is 4 square meters, the rate at which the lengths of the legs are changing is -√2 meters per second, and the rate at which the area is changing is -4 square meters per second.

Step 1: Understand the problem and variables

Recall that an isosceles right triangle has two legs of equal length and a hypotenuse. So, let the length of one leg be "a" meters, the length of the other leg be "b" meters, and the length of the hypotenuse be "c" meters. If the triangle is isosceles and right-angled, then \(a=b\). The area of the triangle (A) will be half of the product of the legs, which means \(A=\frac{1}{2}ab\). Let's also denote the rates at which the hypotenuse (dc/dt), legs (da/dt = db/dt), and area (dA/dt) are changing with respect to time (t).

Step 2: Find the relationship between the variables

Since we have a right-angled triangle, we can use the Pythagorean theorem to find the relationship between the sides: \(a^2 + b^2 = c^2\). Since \(a=b\), we can rewrite the equation as: \(2a^2 = c^2\).

Step 3: Differentiate the equations with respect to time

Now we have to differentiate the equation from Step 2 with respect to time: \(\frac{d}{dt}(2a^2) = \frac{d}{dt}(c^2)\). Then, we get \(4a\frac{da}{dt}=2c\frac{dc}{dt}\). We also need to differentiate the area equation with respect to time: \(\frac{d}{dt}(A)=\frac{d}{dt}\big(\frac{1}{2}ab\big)\). This gives us \(\frac{dA}{dt}=\frac{1}{2}(a\frac{db}{dt}+b\frac{da}{dt})\). a. To find the rate at which the area is changing when the legs are 5 meters long, we will use the information given: \(a=b=5\,\mathrm{m}\) and \(\frac{dc}{dt}=-4\,\mathrm{m/s}\), and then find \(\frac{dA}{dt}\) using the equations.

Step 4a: Apply the given conditions and solve for dA/dt

Given \(a=b=5\,\mathrm{m}\) and \(\frac{dc}{dt}=-4\,\mathrm{m/s}\). Now, we have to calculate c using Pythagorean theorem: \(c = \sqrt{a^2 + b^2} = \sqrt{2a^2}=a\sqrt{2}=5\sqrt{2}\,\mathrm{m}\). Then, replace the values in the differentiated equation of step 3 to get: \(4(5)\frac{da}{dt}=2(5\sqrt{2})(-4)\). Solving this equation for da/dt, we get \(\frac{da}{dt}=-2\sqrt{2}\,\mathrm{m/s}\). Since \(a=b\), we have \(\frac{db}{dt}=-2\sqrt{2}\,\mathrm{m/s}\). Therefore, substitute these values in the area rate equation: \(\frac{dA}{dt}=\frac{1}{2}(5(-2\sqrt{2})+5(-2\sqrt{2}))\). Solving this, we find \(\frac{dA}{dt}=-20\,\mathrm{m^2/s}\). b. The rate at which the lengths of the legs are changing is the same as what we found in Part a: \(\frac{da}{dt}=\frac{db}{dt}=-2\sqrt{2}\,\mathrm{m/s}\). c. To find the rate at which the area is changing when the area is 4 square meters, we will use the information given: \(A = 4\,\mathrm{m^2}\) and \(\frac{dc}{dt}=-4\,\mathrm{m/s}\), and then find \(\frac{dA}{dt}\) using the equations.

Step 4c: Calculate the lengths of the legs when the area is 4 square meters

When the area is \(4\,\mathrm{m^2}\), we can use the area equation to find the lengths of the legs: \(4=\frac{1}{2}(a^2)\). Solving this, we get \(a=2\sqrt{2}\,\mathrm{m}\). Since \(a=b\), we have \(b=2\sqrt{2}\,\mathrm{m}\).

Step 5c: Apply the given conditions and solve for dA/dt

We already know \(\frac{dc}{dt}=-4\,\mathrm{m/s}\), and from step 4c, we know \(a=b=2\sqrt{2}\,\mathrm{m}\). Now, calculate c as well: \(c=a\sqrt{2}=2\sqrt{2}\sqrt{2}=4\,\mathrm{m}\). Then, replace the values in the differentiated equation from Step 3 to get: \(4(2\sqrt{2})\frac{da}{dt}=2(4)(-4)\). Solving this equation for da/dt, we get \(\frac{da}{dt}=-\sqrt{2}\,\mathrm{m/s}\). Since \(a=b\), we have \(\frac{db}{dt}=-\sqrt{2}\,\mathrm{m/s}\). Therefore, substitute these values in the area rate equation: \(\frac{dA}{dt}=\frac{1}{2}(2\sqrt{2}(-\sqrt{2})+2\sqrt{2}(-\sqrt{2}))\). Solving this, we find \(\frac{dA}{dt}=-4\,\mathrm{m^2/s}\). So, the solutions are: a. The rate at which the area is changing when the legs are 5 meters long is -20 square meters per second. b. The rate at which the lengths of the legs are changing is -2√2 meters per second. c. The rate at which the area is changing when the area is 4 square meters is -4 square meters per second.

Key Concepts

Isosceles Right Triangle

An isosceles right triangle is a special type of triangle where two sides, known as the legs, are equal in length. The third side is the hypotenuse, which is opposite the right angle. This triangle has unique properties that make it a compelling subject in geometry and calculus.

In the context of related rates and this exercise, the isosceles right triangle allows us to explore how changing dimensions affect area and side lengths. This is particularly interesting because the symmetry of the triangle simplifies the calculations. For example, if the legs are both labeled as "a," the formula for the hypotenuse becomes quite straightforward: since it's derived using the Pythagorean Theorem, the hypotenuse will always be \(a\sqrt{2}\) for an isosceles right triangle.

Understanding this foundational concept is critical when moving into related rates problems because any change in the hypotenuse directly affects both legs equally.

Pythagorean Theorem

The Pythagorean Theorem is pivotal in scenarios involving right triangles. It states that for any right triangle with legs "a" and "b" and hypotenuse "c," the relationship \(a^2 + b^2 = c^2\) holds true.

In an isosceles right triangle, this theorem simplifies to \(2a^2 = c^2\) because both legs are equal, as previously discussed. This simplification makes it easier to calculate how one side changes as the triangle's dimensions shift over time.

This theorem is especially useful in related rates problems where we track how quickly the dimensions change. By differentiating this equation with respect to time, we derive a formula that reveals the rate at which the hypotenuse and the legs change. This differential relationship helps us understand interdependency between rates of change in the sides of the triangle.

Differentiation with Respect to Time

Differentiation with respect to time is a powerful calculus technique used to determine how a quantity changes as time progresses. It's particularly useful in related rates problems, where multiple variables are interdependent.

When differentiating the equation of a triangle's sides, \(2a^2 = c^2\), with respect to time \(t\), we apply the chain rule to each term. This means we'll take the derivative of each side of the equation separately, resulting in an expression like \(4a\frac{da}{dt} = 2c\frac{dc}{dt}\).

Here, \(\frac{da}{dt}\) and \(\frac{dc}{dt}\) represent the rates at which the legs and hypotenuse are changing, respectively. Solving these derivatives in practice gives us a clear view of how each variable's rate changes in response to adjustments in others. This is essential for determining output rates, such as how fast area changes.

Area of a Triangle

The area of an isosceles right triangle is calculated as \(A = \frac{1}{2}ab\). Since in our scenario, \(a = b\), it simplifies to \(A = \frac{1}{2}a^2\).

To determine how the area changes over time, you take the derivative with respect to time, resulting in \(\frac{dA}{dt} = \frac{1}{2}(a\frac{db}{dt} + b\frac{da}{dt})\). Given that \(a = b\) in our triangle, this can further reduce to \(\frac{dA}{dt} = a\frac{da}{dt}\).

This differentiation allows us to compute the rate of change of the triangle's area as its dimensions change dynamically. Understanding the dynamics of how triangle areas change helps in a variety of real-world applications, especially in situations where dimensions are not constant.