Detour of Variable Changes: A Complete Exploration of Related Rates

What are related rates like?

Related rates are used by first identifying the relationship between variables in a situation, like the radius of a balloon and its volume. Then, understanding how one variable changes (like the rate of air entering the balloon) helps predict the change in another variable (like the balloon’s radius). This approach is widely applied in physics to predict movements, in engineering for design dynamics, and in everyday problem-solving where different factors are interconnected and change simultaneously. For additional educational resources,.

Example 1:

A circular oil spill is expanding on water. The radius of the oil spill is increasing at a constant rate of \( 2 \) meters per minute. Find the rate at which the area of the spill is increasing when the radius is \( 5 \) meters.

Solution:

to define the variable, let \( r \) be the radius of the oil spill and \( A \) be the area of the spill. Given that \( \frac{dr}{dt} = 2 \) meters/minute. The area of a circle is given by \( A = \pi r^2 \). Differentiate both sides of the equation with respect to time \( t \) to find \( \frac{dA}{dt} \):

\( \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \)

Applying the chain rule, \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \).

Now, substituting values, when \( r=5 \) meters and \( \frac{dr}{dt} = 2 \) meters per minute:

\( \frac{dA}{dt} = 2\pi \times 5 \times 2 \)

\( \frac{dA}{dt} = 20\pi \) \( \frac{meters^2}{minutes} \)

So, The area of the oil spill is increasing at a rate of \( 20\pi \) square meters per minute when the radius is 5 meters.

Example 2:

A gas cloud diffuses so that its radius increases at a rate of \( 3 \frac{cm}{s} \). Determine the rate at which the volume of the cloud is increasing when the radius is \( 10 \) cm.

Solution:

\( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)

\( \frac{dV}{dt} = 4\pi \times 10^2 \times 3 = 1200\pi \) cm³/s.

Understanding Related Rates Problems and Variable Changes

Related rates problems involve quantities that change with respect to time, and we’re asked to find the rate of change of one quantity given the rate of change of another. The key strategy is to set up a relationship (usually geometric or physical) between the variables, then differentiate with respect to time using implicit differentiation. When a problem involves a change of variables (like switching from one coordinate system to another), we apply the chain rule carefully.

General Methodology for Related Rates with Variable Changes

Step 1: Identify the relationship. Determine the geometric or physical equation linking the variables. For example, the Pythagorean theorem for a ladder sliding down a wall, or \(V = \frac{1}{3}\pi r^2 h\) for a cone.

Step 2: Differentiate with respect to time. Use implicit differentiation to get rates of change. Apply the chain rule: \(\frac{d}{dt}[x^2] = 2x\frac{dx}{dt}\).

Step 3: Substitute known values. Plug in the given rates (\(\frac{dx}{dt}\), etc.) and any specific values at the moment in question.

Step 4: Solve for the unknown rate. Algebraically isolate the rate you’re looking for.

Step 5: Interpret the sign. A positive rate means increasing; negative means decreasing.

Classic Example 1: The Sliding Ladder

Problem: A 13-foot ladder leans against a wall. The bottom slides away at 2 feet per second. How fast is the top sliding down the wall when the bottom is 5 feet from the wall?

Step 1 Relationship: By the Pythagorean theorem: \(x^2 + y^2 = 13^2 = 169\), where \(x\) is the distance from the wall to the base, and \(y\) is the height on the wall.

Step 2 Differentiate: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)

Step 3 Substitute known values: We know \(\frac{dx}{dt} = 2\) ft/s and \(x = 5\) ft. Find \(y\): \(5^2 + y^2 = 169\), so \(y^2 = 144\), thus \(y = 12\) ft.

Step 4 Solve: \(2(5)(2) + 2(12)\frac{dy}{dt} = 0\) → \(20 + 24\frac{dy}{dt} = 0\) → \(\frac{dy}{dt} = -\frac{5}{6}\) ft/s.

Interpretation: The top is sliding down at \(\frac{5}{6}\) feet per second (negative indicates downward motion).

Classic Example 2: The Expanding Balloon

Problem: A spherical balloon inflates at a rate of 500 cubic inches per minute. How fast is the radius increasing when the radius is 10 inches?

Step 1 Relationship: Volume of a sphere: \(V = \frac{4}{3}\pi r^3\)

Step 2 Differentiate: \(\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}\)

Step 3 & 4 Substitute and solve: \(500 = 4\pi(10)^2\frac{dr}{dt}\) → \(500 = 400\pi\frac{dr}{dt}\) → \(\frac{dr}{dt} = \frac{5}{4\pi}\) in/min ≈ 0.398 in/min.

Strategy for Variable Changes

When variables change (e.g., radius shrinking as height increases), express one in terms of the other using the geometric constraints before differentiating. This reduces the number of variables and simplifies the algebra. In cone problems, use the ratio of dimensions to eliminate variables.

Related Resources and Learning

Master calculus fundamentals to strengthen your implicit differentiation skills. Our AP Calculus BC course covers related rates in depth. For geometric intuition, review geometry properties of cones, spheres, and rectangles.

Practice Problems

Problem 1: A rectangular tank is 4 meters long and 2 meters wide. Water fills it at 0.5 cubic meters per minute. How fast is the water level rising?
Answer: \(\frac{dh}{dt} = \frac{0.5}{4 \times 2} = 0.0625\) m/min

Problem 2: A circle’s area increases at 3 square inches per second. How fast is the radius increasing when the radius is 2 inches?
Answer: Using \(A = \pi r^2\), we get \(\frac{dA}{dt} = 2\pi r\frac{dr}{dt}\). So \(3 = 2\pi(2)\frac{dr}{dt}\), giving \(\frac{dr}{dt} = \frac{3}{4\pi}\) in/s.

Understanding Related Rates Problems and Variable Changes

Related rates problems involve quantities that change with respect to time, and we’re asked to find the rate of change of one quantity given the rate of change of another. The key strategy is to set up a relationship (usually geometric or physical) between the variables, then differentiate with respect to time using implicit differentiation. When a problem involves a change of variables (like switching from one coordinate system to another), we apply the chain rule carefully.

General Methodology for Related Rates with Variable Changes

Step 1: Identify the relationship. Determine the geometric or physical equation linking the variables. For example, the Pythagorean theorem for a ladder sliding down a wall, or \(V = \frac{1}{3}\pi r^2 h\) for a cone.

Step 2: Differentiate with respect to time. Use implicit differentiation to get rates of change. Apply the chain rule: \(\frac{d}{dt}[x^2] = 2x\frac{dx}{dt}\).

Step 3: Substitute known values. Plug in the given rates (\(\frac{dx}{dt}\), etc.) and any specific values at the moment in question.

Step 4: Solve for the unknown rate. Algebraically isolate the rate you’re looking for.

Step 5: Interpret the sign. A positive rate means increasing; negative means decreasing.

Classic Example 1: The Sliding Ladder

Problem: A 13-foot ladder leans against a wall. The bottom slides away at 2 feet per second. How fast is the top sliding down the wall when the bottom is 5 feet from the wall?

Step 1 Relationship: By the Pythagorean theorem: \(x^2 + y^2 = 13^2 = 169\), where \(x\) is the distance from the wall to the base, and \(y\) is the height on the wall.

Step 2 Differentiate: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)

Step 3 Substitute known values: We know \(\frac{dx}{dt} = 2\) ft/s and \(x = 5\) ft. Find \(y\): \(5^2 + y^2 = 169\), so \(y^2 = 144\), thus \(y = 12\) ft.

Step 4 Solve: \(2(5)(2) + 2(12)\frac{dy}{dt} = 0\) → \(20 + 24\frac{dy}{dt} = 0\) → \(\frac{dy}{dt} = -\frac{5}{6}\) ft/s.

Interpretation: The top is sliding down at \(\frac{5}{6}\) feet per second (negative indicates downward motion).

Classic Example 2: The Expanding Balloon

Problem: A spherical balloon inflates at a rate of 500 cubic inches per minute. How fast is the radius increasing when the radius is 10 inches?

Step 1 Relationship: Volume of a sphere: \(V = \frac{4}{3}\pi r^3\)

Step 2 Differentiate: \(\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}\)

Step 3 & 4 Substitute and solve: \(500 = 4\pi(10)^2\frac{dr}{dt}\) → \(500 = 400\pi\frac{dr}{dt}\) → \(\frac{dr}{dt} = \frac{5}{4\pi}\) in/min ≈ 0.398 in/min.

Strategy for Variable Changes

When variables change (e.g., radius shrinking as height increases), express one in terms of the other using the geometric constraints before differentiating. This reduces the number of variables and simplifies the algebra. In cone problems, use the ratio of dimensions to eliminate variables.

Related Resources and Learning

Master calculus fundamentals to strengthen your implicit differentiation skills. Our AP Calculus BC course covers related rates in depth. For geometric intuition, review geometry properties of cones, spheres, and rectangles.

Practice Problems

Problem 1: A rectangular tank is 4 meters long and 2 meters wide. Water fills it at 0.5 cubic meters per minute. How fast is the water level rising?
Answer: \(\frac{dh}{dt} = \frac{0.5}{4 \times 2} = 0.0625\) m/min

Problem 2: A circle’s area increases at 3 square inches per second. How fast is the radius increasing when the radius is 2 inches?
Answer: Using \(A = \pi r^2\), we get \(\frac{dA}{dt} = 2\pi r\frac{dr}{dt}\). So \(3 = 2\pi(2)\frac{dr}{dt}\), giving \(\frac{dr}{dt} = \frac{3}{4\pi}\) in/s.

Comprehensive Methodology for Related Rates with Variable Changes

Related rates problems involve quantities that change over time. We seek to find how one quantity changes when another’s rate is known. The methodology has five systematic steps: (1) Identify the relationship between variables using geometry, physics, or chemistry. (2) Differentiate this relationship with respect to time using implicit differentiation and the chain rule. (3) Substitute all known rates (derivatives) and the specific values at the moment of interest. (4) Solve algebraically for the unknown rate. (5) Interpret the sign and units: positive means increasing; negative means decreasing.

Classic Problem 1: The Sliding Ladder

Scenario: A 13-foot ladder leans against a vertical wall. Its base slides away from the wall at 2 feet per second. How fast is the top descending when the base is 5 feet away?

Setup: Let \(x\) = distance from wall to base, \(y\) = height on wall. By Pythagorean theorem: \(x^2 + y^2 = 13^2 = 169\).

Differentiate: \(\frac{d}{dt}[x^2 + y^2] = \frac{d}{dt}[169]\) → \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\).

Given Information: \(\frac{dx}{dt} = 2\) ft/s, \(x = 5\) ft at the moment in question. Find \(y\): \(5^2 + y^2 = 169\) → \(y^2 = 144\) → \(y = 12\) ft.

Substitute and Solve: \(2(5)(2) + 2(12)\frac{dy}{dt} = 0\) → \(20 + 24\frac{dy}{dt} = 0\) → \(\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6}\) ft/s.

Interpretation: The top is sliding down at \(\frac{5}{6}\) feet per second (the negative sign indicates downward motion).

Classic Problem 2: The Inflating Spherical Balloon

Scenario: A spherical balloon inflates at 500 cubic inches per minute. How fast is the radius increasing when the radius is 10 inches?

Setup: Volume of sphere: \(V = \frac{4}{3}\pi r^3\).

Differentiate: \(\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}\).

Given Information: \(\frac{dV}{dt} = 500\) in³/min, \(r = 10\) in.

Solve: \(500 = 4\pi(10)^2\frac{dr}{dt}\) → \(500 = 400\pi\frac{dr}{dt}\) → \(\frac{dr}{dt} = \frac{500}{400\pi} = \frac{5}{4\pi} \approx 0.398\) in/min.

Advanced Example: Conical Tank with Variable Change

Problem: Water fills a cone (apex downward) at 1 ft³/s. The cone has height 10 ft and base radius 5 ft. How fast is the water level rising when depth is 4 ft?

Key Insight: The water surface is also conical, maintaining the same proportions as the tank. From similarity: \(\frac{r}{h} = \frac{5}{10} = \frac{1}{2}\), so \(r = \frac{h}{2}\).

Volume Equation: \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{h}{2})^2 h = \frac{\pi h^3}{12}\).

Differentiate: \(\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2\frac{dh}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}\).

At \(h = 4\): \(1 = \frac{\pi(16)}{4}\frac{dh}{dt}\) → \(\frac{dh}{dt} = \frac{1}{4\pi} \approx 0.0796\) ft/s.

Mastery Resources and Advanced Study

Strengthen your implicit differentiation with comprehensive calculus training. The AP Calculus BC course dedicates substantial time to related rates. Review geometry to strengthen understanding of cones, cylinders, spheres, and rectangles essential for problem setup.

Extended Practice Problems

Problem 1: A rectangular trough is 4 m long and 2 m wide. Water flows in at 0.5 m³/min. How fast does the water level rise? Answer: Area = 8 m², so \(\frac{dh}{dt} = \frac{0.5}{8} = 0.0625\) m/min.

Problem 2: A circular oil spill expands at 2 ft²/s. How fast is the radius growing when the radius is 3 ft? From \(A = \pi r^2\): \(\frac{dA}{dt} = 2\pi r\frac{dr}{dt}\). So \(2 = 2\pi(3)\frac{dr}{dt}\) → \(\frac{dr}{dt} = \frac{1}{3\pi}\) ft/s.

Problem 3: A 25-meter ladder leans against a wall. The base moves away at 0.5 m/s. How fast is the top sliding down when the base is 7 m away? Answer: \(y = 24\) m, \(\frac{dy}{dt} = -\frac{7}{48}\) m/s.