Exam 3 Notes

Derivatives and Their Applications

3.7 The Rate of Change in Natural and Social Sciences

Example: Particle Motion

• The position of a particle is given by the function
  
  $S(t) = 3t^3 - 6t^2 + 9t$
  
  
  where 't' is time in seconds, and 's' is position in meters.

A. Find the Velocity

• The velocity function is the derivative of the position function:
  
  $v(t) = S'(t) = 3t^2 - 12t + 9$

• Units of velocity: meters per second (m/s).

B. Find Velocity at Specific Times

• For 2 seconds:
  
  $v(2) = 3(2)^2 - 12(2) + 9 = -3 \text{ m/s}$

• For 4 seconds:
  
  $v(4) = 3(4)^2 - 12(4) + 9 = 9 \text{ m/s}$

C. When is the Particle at Rest?

• Set the velocity equal to zero: $0 = 3t^2 - 12t + 9$

• Factoring yields:
  
  $0 = 3(t^2 - 4t + 3) = 3(t - 3)(t - 1)$

• Thus, the particle is at rest at:
  
  $t = 3 \text{ sec}, \text{ and } t = 1 \text{ sec}$

D. When is the Particle Moving Forward?

• To find when the particle is moving forward, determine when the velocity is positive:
  
  $0 < t < 1 \quad \text{ or } \quad t > 3$

E. Diagram Representation of Motion

• Create a diagram to represent particle position vs. time. (Include illustrative details such as direction and velocity at specific points).

F. Find Total Distance Traveled in First Five Seconds

• Total distance for time intervals: ### $\text{Distance} = S(2) - S(0) + S(4) - S(3)$

• Calculations involve evaluating the position function at given times:

• For t = 1:
  $S(1) = 3(1)^3 - 6(1^2) + 9 = 6 \text{ m}$

• For t = 3:
  $S(3) = 3(3^3) - 6(3^2) + 9 \text{ = } 36 - 54 + 9 = -9$

• For t = 5: ### $S(5) = 3(5^3) - 6(5^2) + 9 = 48$
  Distance traveled = $\text{Distance} = | S(5) - S(3) | + | S(3) - S(1) |$

G. Find the Acceleration at Any Time t

• The acceleration function is the derivative of the velocity function:
  $a(t) = S''(t) = 6t - 12$

• Example Calculation: For time t=4
  $a(4) = 6(4) - 12 = 12 \text{ m/s}^2$

3.8 Maximum and Minimum Values

A. Definitions of Extrema

• Absolute Maximum: The largest value of $f(x)$ on a given interval. Also known as global maximum.

• Absolute Minimum: The smallest value of $f(x)$ on a given interval. Also known as global minimum.

• Local Maximum: A value $f(c)$ such that $f(c) \ge f(x)$ for all ''x'' in an open interval around ''c''.

• Local Minimum: A value $f(c)$ such that $f(c) \le f(x)$ for all ''x'' in an open interval around ''c''.

• Extrema: Refers to either maximum or minimum values.

B. Fermat's Theorem

• If $f$ has a local maximum or minimum at ''c'', and if $f'(c)$ exists, then $f'(c) = 0$.

• Important Note: The converse is not necessarily true (i.e., if $f'(c) = 0$, it doesn't guarantee a local extremum).

C. Critical Numbers (or Critical Points)

• A critical number of a function $f$ is a number ''c'' in the domain of $f$ such that either $f'(c) = 0$ or $f'(c)$ does not exist.

• Significance: By Fermat's Theorem, local extrema can only occur at critical numbers or endpoints of an interval.

3.9 The Mean Value Theorem

A. Rolle's Theorem

• Let $f$ be a function that satisfies three hypotheses:

  1. $f$ is continuous on the closed interval $[a, b]$.

  2. $f$ is differentiable on the open interval $(a, b)$.

  3. $f(a) = f(b)$.

• Then there is a number $c$ in $(a, b)$ such that $f'(c) = 0$.

B. The Mean Value Theorem (MVT)

• Let $f$ be a function that satisfies two hypotheses:

  1. $f$ is continuous on the closed interval $[a, b]$.

  2. $f$ is differentiable on the open interval $(a, b)$.

• Then there is a number $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

• Interpretation: The MVT states that there is at least one point where the instantaneous rate of change (derivative) equals the average rate of change over the interval.

3.10 How Derivatives Affect the Shape of a Graph

A. Increasing and Decreasing Functions

• Definition: A function $f$ is increasing on an interval if, for any two numbers $x<em>1$ and $x</em>2$ in the interval, $x<em>1 < x</em>2$ implies $f(x<em>1) < f(x</em>2)$.

• Definition: A function $f$ is decreasing on an interval if, for any two numbers $x<em>1$ and $x</em>2$ in the interval, $x<em>1 < x</em>2$ implies $f(x<em>1) > f(x</em>2)$.

• The Increasing/Decreasing Test (First Derivative Test for Monotonicity):

  • If $f'(x) > 0$ on an interval, then $f$ is increasing on that interval.

  • If $f'(x) < 0$ on an interval, then $f$ is decreasing on that interval.

B. The First Derivative Test

• Suppose ''c'' is a critical number of a continuous function $f$.

  1. If $f'$ changes from positive to negative at ''c'', then $f$ has a local maximum at ''c''.

  2. If $f'$ changes from negative to positive at ''c'', then $f$ has a local minimum at ''c''.

  3. If $f'$ does not change sign at ''c'' (e.g., $f'$ is positive on both sides of ''c'' or negative on both sides), then $f$ has no local maximum or minimum at ''c''.

C. Concavity

• Definition: If the graph of $f$ lies above all of its tangent lines on an interval, it is called concave upward on that interval.

• Definition: If the graph of $f$ lies below all of its tangent lines on an interval, it is called concave downward on that interval.

• The Concavity Test (Second Derivative Test for Concavity):

  • If $f''(x) > 0$ for all ''x'' in an interval, then the graph of $f$ is concave upward on that interval.

  • If $f''(x) < 0$ for all ''x'' in an interval, then the graph of $f$ is concave downward on that interval.

D. Inflection Points

• Definition: An inflection point is a point where a curve changes its concavity (from concave upward to concave downward or vice versa).

• Condition: If ($c, f(c)$) is an inflection point, then either $f''(c) = 0$ or $f''(c)$ does not exist.

E. The Second Derivative Test (for Local Extrema)

• Suppose $f''(c)$ is continuous near ''c'':

  1. If $f'(c) = 0$ and $f''(c) > 0$, then $f$ has a local minimum at ''c''.

  2. If $f'(c) = 0$ and $f''(c) < 0$, then $f$ has a local maximum at ''c''.

  3. If $f'(c) = 0$ and $f''(c) = 0$, the test is inconclusive (the first derivative test must be used).

3.11 Related Rates

• Relate two or more quantities that are changing with respect to time.

Example: Volume Expansion in a Balloon

• Air is being pumped into a spherical balloon at a rate of $5 \text{ cm}^3/\text{min}$. Find the rate at which the radius increases when the diameter is $20 \text{ cm}$.

• The volume of a sphere is given by:
  
  $V = \frac{4}{3} \pi r^3$

A. Implicit Relation for Volume and Radius

• Given the rate of volume change:
  
  $\frac{dV}{dt} = 5 \text{ cm}^3/\text{min}$

B. Chain Rule Application

• Differentiate the volume equation with respect to time:
  
  $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

• Rearranging gives the rate of radius increase: $\frac{dr}{dt} = \frac{5}{4 \pi r^2}$

• Substituting diameter:

  1. Diameter = $20 \text{ cm}$

  2. Radius = $10 \text{ cm}$

  3. Rate of Radius Change:
     
     $\frac{dr}{dt} = \frac{5}{4 \pi \times (10)^2} = \frac{5}{400 \pi} \text{ cm/min}$

• Which simplifies to approximately:
  
  $\frac{dr}{dt} \text{ = 0.004 cm/min}$

Example: Ladder Problem

• A 16-ft long ladder rests along a wall with the foot 10 ft away. The bottom is pushed towards the wall at a rate of 1 ft/sec. Determine the rate at which the top is moving up the wall after 2 seconds.

A. Establish Relations Using Right Triangle Geometry

• Let:

  • x = distance from wall

  • y = height up wall

  • L = length of ladder = $16 \text{ ft}$

• Relation: $x^2 + y^2 = L^2$

• Differentiating both sides:
  
  $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

• Substitute known values at t=2:

  1. x = $10 - 2(1) = 8 \text{ ft}$

  2. Find corresponding 'y' from Pythagorean:
     
     $y^2 = 16^2 - 8^2 \ y^2 = 256 - 64 = 192 \ y = 4 \sqrt{3} \text{ ft}$

• At this point, substitute all values into differentiated equation.
  
  $2(8)(1) + 2(4 \sqrt{3}) \times \frac{dy}{dt} = 0 \ 16 + 8 \sqrt{3} \frac{dy}{dt} = 0 \ \frac{dy}{dt} = - \frac{16}{8 \sqrt{3}} = - \frac{2}{\sqrt{3}} \text{ ft/sec}$

Further Examples of Applications

• Cone-shaped tank leakage and depth rate change calculations are practical exercises using the formulas for volume and integrative rates.

• Other practical scenarios like car speeds approaching intersections or rates of change in economic models serve as excellent demonstration for the application of related rates in real-life contexts.

3.12 General Tips for Problem Solving with Derivatives

Problem Solving Strategy

1. Read the Problem Carefully: Understand the context and parameters.

2. Draw Diagrams: Visual aids can clarify relationships.

3. Introduce Notation: Clearly define what each variable represents.

4. Express Information: Relate given data mathematically.

5. Write Relations: Identify and write equations linking quantities.

6. Differentiate: Use the chain rule or product rule as appropriate, then substitute known values.