🚀 Level 3 - Topic 1: Applications of Derivatives (Part 1)

1) Introduction: Putting Derivatives to Work

Now that we've built a strong toolkit for finding derivatives (basic rules, product rule, quotient rule, chain rule, implicit differentiation, higher-order derivatives), it's time to explore their power in applications. Derivatives are not just abstract mathematical concepts; they are fundamental tools for solving problems in various fields.

In this topic, we will cover some fundamental applications:

Tangent Lines: Finding equations of tangent lines to curves at specific points. This revisits the core geometric interpretation of the derivative.
Motion Along a Line: Analyzing motion (position, velocity, acceleration) of objects moving along a straight line using derivatives.
Related Rates (Introduction): Introducing the concept of related rates problems, where we find the rate of change of one quantity by relating it to the rates of change of other related quantities. We will start with simpler related rates problems in this topic.

These applications showcase the versatility of derivatives in describing change and solving real-world problems.

2) Tangent Lines Revisited: Equations Using Derivatives

Recall that the derivative \( f'(a) \) at a point \( x = a \) represents the slope of the tangent line to the graph of \( y = f(x) \) at the point \( (a, f(a)) \). We can use this fact, along with the point-slope form of a line, to find the equation of the tangent line.

Equation of the Tangent Line

The equation of the tangent line to the curve \( y = f(x) \) at the point \( (a, f(a)) \) is given by the point-slope form:

\( y - f(a) = f'(a)(x - a) \)

where:

\( (a, f(a)) \) is the point of tangency (point on the curve).
\( f'(a) \) is the slope of the tangent line at \( x = a \) (found by evaluating the derivative at \( x = a \)).

Examples: Finding Tangent Line Equations

Example 1: Find the equation of the tangent line to \( y = x^2 + 3x \) at \( x = 1 \).
1. Find the \( y \)-coordinate of the point of tangency.
  When \( x = 1 \), \( y = f(1) = (1)^2 + 3(1) = 1 + 3 = 4 \). So the point is \( (1, 4) \). Here \( a = 1 \) and \( f(a) = f(1) = 4 \).
2. Find the derivative \( f'(x) \).
  \( f'(x) = \frac{d}{dx}[x^2 + 3x] = 2x + 3 \).
3. Find the slope \( f'(a) = f'(1) \).
  \( f'(1) = 2(1) + 3 = 5 \). So the slope of the tangent line at \( x = 1 \) is 5.
4. Use the point-slope form to write the equation.
  Using \( y - f(a) = f'(a)(x - a) \) with \( (a, f(a)) = (1, 4) \) and \( f'(a) = 5 \):
  
  \( y - 4 = 5(x - 1) \)
  
  We can rewrite in slope-intercept form: \( y - 4 = 5x - 5 \Rightarrow y = 5x - 1 \).
  
  Thus, the equation of the tangent line is \( y = 5x - 1 \).
Example 2: Find the tangent line to \( y = \frac{x}{x + 2} \) at \( x = 2 \).
1. Find the \( y \)-coordinate.
  When \( x = 2 \), \( y = f(2) = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2} \). Point is \( (2, \frac{1}{2}) \). Here \( a = 2 \) and \( f(a) = f(2) = \frac{1}{2} \).
2. Find the derivative \( f'(x) \) using the Quotient Rule.
  \( f'(x) = \frac{d}{dx}\left[\frac{x}{x + 2}\right] = \frac{(1)(x + 2) - x(1)}{(x + 2)^2} = \frac{2}{(x + 2)^2} \).
3. Find the slope \( f'(a) = f'(2) \).
  \( f'(2) = \frac{2}{(2 + 2)^2} = \frac{2}{4^2} = \frac{2}{16} = \frac{1}{8} \). Slope is \( \frac{1}{8} \).
4. Write the tangent line equation in point-slope form.
  \( y - \frac{1}{2} = \frac{1}{8}(x - 2) \).
  
  Slope-intercept form (optional simplification): \( y - \frac{1}{2} = \frac{1}{8}x - \frac{2}{8} = \frac{1}{8}x - \frac{1}{4} \Rightarrow y = \frac{1}{8}x - \frac{1}{4} + \frac{1}{2} = \frac{1}{8}x + \frac{1}{4} \).
  
  The equation of the tangent line is \( y = \frac{1}{8}x + \frac{1}{4} \) or \( y - \frac{1}{2} = \frac{1}{8}(x - 2) \).

3) Motion Along a Line: Position, Velocity, and Acceleration

Derivatives are fundamental in describing motion. Consider an object moving along a straight line (say, the x-axis), and let \( s(t) \) be its position at time \( t \).

Definitions: Position, Velocity, Acceleration

Position: \( s(t) \) = position of the object at time \( t \).
Velocity: \( v(t) = s'(t) = \frac{ds}{dt} \) = the instantaneous rate of change of position with respect to time. Velocity is the derivative of position. It indicates speed and direction of motion.
Acceleration: \( a(t) = v'(t) = s''(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} \) = the instantaneous rate of change of velocity with respect to time. Acceleration is the derivative of velocity (and the second derivative of position). It indicates how quickly velocity is changing.
Speed: \( |v(t)| \) = the magnitude of velocity, always non-negative. Speed is how fast the object is moving, regardless of direction.

Units are important. If position is in meters (m) and time in seconds (s):

Velocity is in meters per second (m/s).
Acceleration is in meters per second squared (m/s²).
Jerk (third derivative, although less commonly used here) would be in meters per second cubed (m/s³).

Examples: Analyzing Motion

Example 1: Particle Motion \( s(t) = t^3 - 6t^2 + 9t \) (for \( t \geq 0 \))

Given the position function \( s(t) = t^3 - 6t^2 + 9t \) (in meters, time in seconds), analyze the motion for \( t \geq 0 \).

Find the velocity and acceleration functions.
Velocity: \( v(t) = s'(t) = \frac{d}{dt}[t^3 - 6t^2 + 9t] = 3t^2 - 12t + 9 \)

Acceleration: \( a(t) = v'(t) = \frac{d}{dt}[3t^2 - 12t + 9] = 6t - 12 \)
When is the particle at rest?
Particle is at rest when velocity is zero: \( v(t) = 3t^2 - 12t + 9 = 0 \). Divide by 3: \( t^2 - 4t + 3 = 0 \). Factor: \( (t - 1)(t - 3) = 0 \). So \( t = 1 \) s and \( t = 3 \) s.
When is the particle moving in the positive direction (right)? Negative direction (left)?
Positive direction when \( v(t) > 0 \). \( 3t^2 - 12t + 9 > 0 \Rightarrow t^2 - 4t + 3 > 0 \Rightarrow (t - 1)(t - 3) > 0 \). This is true when both factors are positive ( \( t > 3 \) ) or both are negative ( \( t < 1 \) ). So, positive direction for \( 0 \leq t < 1 \) and \( t > 3 \). Negative direction when \( v(t) < 0 \). \( (t - 1)(t - 3) < 0 \). This is true when one factor is positive and the other is negative, which occurs for \( 1 < t < 3 \).

When is the particle speeding up? Slowing down?

Speeding up when velocity and acceleration have the same sign (both positive or both negative). Slowing down when they have opposite signs.

We need to analyze signs of \( v(t) = 3(t - 1)(t - 3) \) and \( a(t) = 6(t - 2) \).

Critical points for \( v(t) \) are \( t = 1, 3 \). Critical point for \( a(t) \) is \( t = 2 \).

Test intervals: \( [0, 1), (1, 2), (2, 3), (3, \infty) \).

Interval	\( v(t) = 3(t-1)(t-3) \) Sign	\( a(t) = 6(t-2) \) Sign	Speeding Up / Slowing Down
\( [0, 1) \) (e.g., \( t=0.5 \))	\( 3(-0.5)(-2.5) = + \)	\( 6(-1.5) = - \)	Slowing Down
\( (1, 2) \) (e.g., \( t=1.5 \))	\( 3(0.5)(-1.5) = - \)	\( 6(-0.5) = - \)	Speeding Up
\( (2, 3) \) (e.g., \( t=2.5 \))	\( 3(1.5)(-0.5) = - \)	\( 6(0.5) = + \)	Slowing Down
\( (3, \infty) \) (e.g., \( t=4 \))	\( 3(3)(1) = + \)	\( 6(2) = + \)	Speeding Up

So, speeding up on \( (1, 2) \) and \( (3, \infty) \). Slowing down on \( [0, 1) \) and \( (2, 3) \).

4) Introduction to Related Rates

Related rates problems involve situations where several quantities are changing with time, and these quantities are related to each other by some equation(s). The goal is typically to find the rate of change of one quantity in terms of the rates of change of others. Derivatives with respect to time \( t \) are central to solving these problems.

A key step is to identify the relationships between the variables involved, often using geometric formulas, trigonometric relationships, or other given equations. Once we have an equation relating the quantities, we can use implicit differentiation with respect to time \( t \) to relate their rates of change.

General Strategy for Related Rates Problems

Identify all given quantities and quantities to be found. Draw a diagram if possible. Assign variables to all quantities that are functions of time \( t \). State the given rates (derivatives) and the rate you need to find.
Find a relationship (equation) between the variables whose rates of change are related. This may come from geometry, trigonometry, or the problem description.
Differentiate both sides of the equation implicitly with respect to time \( t \). Use the Chain Rule where needed. Remember that all variables are functions of \( t \).
Substitute in the given values for variables and rates at the specific instant in time. Solve for the unknown rate.
State your answer with correct units. Check if your answer is reasonable in the context of the problem.

Examples: Introduction to Related Rates

Example 1: Expanding Circle
A circle's radius is increasing at a rate of 3 cm/s. At what rate is the area of the circle increasing when the radius is 6 cm?
1. Identify variables and rates.
  Variables: \( r \) = radius, \( A \) = area. Both are functions of time \( t \).
  
  Given rate: \( \frac{dr}{dt} = 3 \) cm/s (positive because radius is increasing).
  
  Rate to find: \( \frac{dA}{dt} \) when \( r = 6 \) cm.
2. Find a relationship between variables.
  Area of a circle formula: \( A = \pi r^2 \).
3. Differentiate with respect to time \( t \).
  \( \frac{dA}{dt} = \frac{d}{dt}[\pi r^2] = \pi \frac{d}{dt}[r^2] = \pi (2r) \frac{dr}{dt} = 2\pi r \frac{dr}{dt} \) (using Chain Rule with \( r \) as a function of \( t \)).
4. Substitute given values and solve for \( \frac{dA}{dt} \).
  We are given \( r = 6 \) cm and \( \frac{dr}{dt} = 3 \) cm/s. Substitute these into \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \):
  
  \( \frac{dA}{dt} = 2\pi (6 \text{ cm}) (3 \text{ cm/s}) = 36\pi \) cm²/s.
5. State the answer with units.
  The area of the circle is increasing at a rate of \( 36\pi \) cm²/s when the radius is 6 cm.
Example 2: Moving Ladder
A 10-foot ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet from the wall?
1. Identify variables and rates. Draw a diagram.
  Let \( x \) be the distance from the bottom of the ladder to the wall, and \( y \) be the distance from the top of the ladder to the ground. Both \( x \) and \( y \) are functions of time \( t \). The ladder length is constant at 10 ft.
  
  Given rate: \( \frac{dx}{dt} = 1 \) ft/s (positive because \( x \) is increasing).
  
  Rate to find: \( \frac{dy}{dt} \) when \( x = 6 \) ft.
  
  (We expect \( \frac{dy}{dt} \) to be negative because \( y \) is decreasing as the top slides down).
2. Find a relationship between variables.
  Pythagorean theorem relates \( x \), \( y \), and the ladder length (hypotenuse): \( x^2 + y^2 = 10^2 = 100 \).
3. Differentiate with respect to time \( t \).
  \( \frac{d}{dt}[x^2 + y^2] = \frac{d}{dt}[100] \)
  
  \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \) (using Chain Rule for both \( x^2 \) and \( y^2 \) as functions of \( t \), and Constant Rule for 100).
4. Substitute given values and solve for \( \frac{dy}{dt} \).
  We are given \( x = 6 \) ft and \( \frac{dx}{dt} = 1 \) ft/s. We also need to find \( y \) when \( x = 6 \). Use \( x^2 + y^2 = 100 \): \( 6^2 + y^2 = 100 \Rightarrow 36 + y^2 = 100 \Rightarrow y^2 = 64 \Rightarrow y = 8 \) ft (take positive root since \( y \) is a length).
  
  Substitute \( x = 6 \), \( y = 8 \), \( \frac{dx}{dt} = 1 \) into \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \):
  
  \( 2(6)(1) + 2(8) \frac{dy}{dt} = 0 \Rightarrow 12 + 16 \frac{dy}{dt} = 0 \Rightarrow 16 \frac{dy}{dt} = -12 \Rightarrow \frac{dy}{dt} = -\frac{12}{16} = -\frac{3}{4} \).
5. State the answer with units.
  The top of the ladder is sliding down the wall at a rate of \( \frac{3}{4} \) ft/s when the bottom of the ladder is 6 feet from the wall. The negative sign indicates that \( y \) is decreasing (top sliding down).

5) Practice Questions - Applications of Derivatives (Tangent Lines, Motion, Related Rates)

Test your skills in applying derivatives to find tangent lines, analyze motion, and solve basic related rates problems.

Fundamental Practice Questions

Instructions: Solve the following problems.

Q1. Find the equation of the tangent line to the curve \( y = x^3 - 2x^2 + 5 \) at \( x = -1 \).
Q2. Find the equation of the tangent line to \( y = \sqrt{x + 1} \) at \( x = 3 \).
Q3. A particle moves along a line with position function \( s(t) = 2t^3 - 9t^2 + 12t + 1 \) (meters, seconds, \( t \geq 0 \)).
1. Find the velocity and acceleration functions, \( v(t) \) and \( a(t) \).
2. Find the times when the particle is at rest.
3. Determine the time intervals when the particle is moving in the positive direction and when in the negative direction.
4. Find the intervals when the particle is speeding up and slowing down.
Q4. Water is dripping into a cylindrical tank of radius 4 meters at a rate of 2 cubic meters per minute. How fast is the water level rising? (Volume of cylinder \( V = \pi r^2 h \)).
Q5. A car is moving west at 50 mph and a truck is moving north at 60 mph. Both are headed toward an intersection. At what rate are they approaching each other when the car is 0.3 miles and the truck is 0.4 miles from the intersection? (Use Pythagorean theorem).
Q6. Find the derivative of the function \( f(x) = 5x^4 - 3x^2 + 2x - 7 \).
Q7. Find the derivative of the function \( g(x) = \sin(x) + 4\cos(x) \).
Q8. Find the derivative of the function \( h(x) = 3e^x - \ln(x) + x^{10} \).
Q9. Find the equation of the tangent line to the curve \( y = \frac{2}{x} \) at \( x = 2 \).
Q10. A particle moves along a line with position function \( s(t) = t^2 - 4t + 6 \) for \( t \geq 0 \). Find the velocity and acceleration of the particle at time \( t = 1 \).
Q11. Find the derivative of \( y = \frac{x^5}{3} - \frac{x^3}{2} + 5x + \sqrt{2} \).
Q12. Find the derivative of \( f(x) = 6\sin(x) - 7e^x + \frac{4}{x^3} \).

Challenging Practice Questions

Instructions: These problems may involve more complex functions, motion analysis, or related rates scenarios.

Q1. Find the equation of the tangent line to the curve \( x^2 + y^2 = 25 \) at the point \( (4, 3) \). (Use implicit differentiation to find the slope).
Q2. A ball is thrown vertically upward with initial velocity 64 ft/s from a height of 80 ft. Its height at time \( t \) is \( h(t) = -16t^2 + 64t + 80 \) (feet, seconds).
1. Find the maximum height reached by the ball.
2. What is the velocity of the ball when it hits the ground (height = 0)?
3. What is the acceleration at any time \( t \)? Is it constant?
Q3. A conical water tank is 10 ft tall and has a radius of 4 ft at the top. Water is pouring in at a rate of 10 cubic ft/min. How fast is the water level rising when the water is 5 ft deep? (Volume of cone \( V = \frac{1}{3}\pi r^2 h \). Use similar triangles to relate \( r \) and \( h \)).
Q4. Two cars start moving from the same point. One travels south at 60 mph and the other travels west at 25 mph. At what rate is the distance between the cars increasing two hours later?
Q5. Explain in your own words why the slope of the tangent line at a point represents the instantaneous rate of change of the function at that point. How is this concept used in motion problems?

6) Summary & Cheat Sheet - Applications of Derivatives (Part 1)

Quick recap of the key applications covered.

6.1) Tangent Line Equation

Equation of tangent line to \( y = f(x) \) at \( (a, f(a)) \): \( y - f(a) = f'(a)(x - a) \), where \( f'(a) \) is the slope.

6.2) Motion Along a Line

Position: \( s(t) \)
Velocity: \( v(t) = s'(t) \)
Acceleration: \( a(t) = v'(t) = s''(t) \)
Speed: \( |v(t)| \)

6.3) Related Rates - General Strategy

Identify quantities, rates, and unknowns. Draw a diagram.
Find an equation relating the variables.
Differentiate implicitly with respect to time \( t \).
Substitute known values and solve for the unknown rate.
State answer with units.

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🚀 Level 3 - Topic 1: Applications of Derivatives (Part 1) ⚙️