M2 Chapter: Applications of Differentiation
Hey everyone! Welcome to one of the most powerful chapters in M2 Calculus: Applications of Differentiation. You might be wondering, "Why did we spend all that time learning how to find derivatives?" Well, this is why!
Think of differentiation as getting a special power: the ability to see the "rate of change" of anything. In this chapter, we'll use that power to:
- Find the exact slope of a curve at any point to determine the equation of its tangent.
- Find the highest and lowest points of a graph, which helps us solve problems like maximising profit or minimising cost. These are called maximum and minimum values.
- Become masters at sketching complex curves, revealing their hidden shapes and behaviours.
- Solve real-world problems involving how things change over time, known as rates of change.
Don't worry if it sounds like a lot. We'll break it down step-by-step. Let's get started!
1. Finding the Equation of a Tangent
Remember from your previous studies that a derivative, $$ \frac{dy}{dx} $$ or $$ f'(x) $$, gives you the gradient (slope) of the tangent line to a curve at any point x.
A tangent is just a straight line that "kisses" the curve at a single point. And what do you need to find the equation of any straight line?
- A point on the line, $$ (x_1, y_1) $$
- The gradient of the line, $$ m $$
Step-by-Step Guide to Finding the Tangent Equation:
Let's find the tangent to a curve $$ y = f(x) $$ at the point where $$ x = a $$.
Step 1: Find the point $$ (x_1, y_1) $$
We know $$ x_1 = a $$. To find $$ y_1 $$, just substitute $$ a $$ into the original function: $$ y_1 = f(a) $$. So our point is $$ (a, f(a)) $$.
Step 2: Find the gradient function $$ f'(x) $$
Differentiate the function $$ f(x) $$ to get $$ f'(x) $$.
Step 3: Find the specific gradient $$ m $$ at the point
Substitute $$ x = a $$ into the derivative: $$ m = f'(a) $$.
Step 4: Use the point-slope form
Plug your point and gradient into the equation for a straight line:
$$ y - y_1 = m(x - x_1) $$
Example Walkthrough
Find the equation of the tangent to the curve $$ y = x^3 - 2x + 5 $$ at $$ x = 2 $$.
Step 1: Find the point.
When $$ x = 2 $$, $$ y = (2)^3 - 2(2) + 5 = 8 - 4 + 5 = 9 $$.
So, our point $$ (x_1, y_1) $$ is $$ (2, 9) $$.
Step 2: Find the derivative.
$$ y = x^3 - 2x + 5 $$
$$ \frac{dy}{dx} = 3x^2 - 2 $$
Step 3: Find the gradient at the point.
At $$ x = 2 $$, the gradient $$ m = 3(2)^2 - 2 = 3(4) - 2 = 10 $$.
Step 4: Use the point-slope form.
$$ y - y_1 = m(x - x_1) $$
$$ y - 9 = 10(x - 2) $$
$$ y - 9 = 10x - 20 $$
$$ y = 10x - 11 $$
And that's our tangent equation!
Key Takeaway
To find a tangent's equation, you need a point (from the original function) and the gradient (from the derivative at that point's x-value).
2. Maximum and Minimum Values (Extrema)
Imagine a rollercoaster. The tops of the hills are "local maximums" and the bottoms of the valleys are "local minimums". At the very peak of a hill or the very bottom of a valley, the track is perfectly flat for an instant. This means its gradient is zero!
Stationary Points: The Candidates
A point on a curve where the derivative is zero ($$ f'(x) = 0 $$) is called a stationary point. These are our candidates for being maximum or minimum points.
There are three types of stationary points:
- Local Maximum: The curve goes from increasing (positive slope) to decreasing (negative slope). It's a peak.
- Local Minimum: The curve goes from decreasing (negative slope) to increasing (positive slope). It's a valley.
- Stationary Point of Inflexion: The curve flattens out but then continues in the same direction. The slope is zero but doesn't change sign.
How to Test Stationary Points: The Second Derivative Test
Once you've found the x-values where $$ f'(x) = 0 $$, how do you know if it's a max, min, or something else? The easiest way is the Second Derivative Test.
Memory Aid: Think of a face!
- If $$ f''(x) > 0 $$ (positive), the curve is "concave up", like a happy face ☺. This means you've found a local minimum.
- If $$ f''(x) < 0 $$ (negative), the curve is "concave down", like a sad face ☹. This means you've found a local maximum.
- If $$ f''(x) = 0 $$, the test fails! You must use the First Derivative Test (checking the sign of $$ f'(x) $$ just before and after the point) to classify it.
Example Walkthrough
Find and classify the stationary points of $$ f(x) = 2x^3 - 3x^2 - 12x + 1 $$.
Step 1: Find the first derivative and set it to zero.
$$ f'(x) = 6x^2 - 6x - 12 $$
Set $$ f'(x) = 0 $$:
$$ 6x^2 - 6x - 12 = 0 $$
$$ x^2 - x - 2 = 0 $$
$$ (x - 2)(x + 1) = 0 $$
So, stationary points occur at $$ x = 2 $$ and $$ x = -1 $$.
Step 2: Find the second derivative.
$$ f''(x) = 12x - 6 $$
Step 3: Test each point using the second derivative.
For $$ x = 2 $$:
$$ f''(2) = 12(2) - 6 = 18 $$. This is POSITIVE (> 0), so it's a happy face ☺.
Therefore, at $$ x = 2 $$ we have a local minimum. The point is ($$ 2, f(2) $$) = ($$ 2, -19 $$).
For $$ x = -1 $$: $$ f''(-1) = 12(-1) - 6 = -18 $$. This is NEGATIVE (< 0), so it's a sad face ☹. Therefore, at $$ x = -1 $$ we have a local maximum. The point is ($$ -1, f(-1) $$) = ($$ -1, 8 $$).
Global vs. Local Extrema
A local max/min is the highest/lowest point in its immediate neighbourhood. A global (or absolute) max/min is the absolute highest/lowest point over a specific interval.
To find the global extrema on a closed interval [a, b]:
- Find all stationary points inside the interval.
- Calculate the y-values at these stationary points.
- Calculate the y-values at the endpoints, i.e., find $$ f(a) $$ and $$ f(b) $$.
- Compare all the y-values you've found. The biggest is the global max, the smallest is the global min.
Common Mistake: Forgetting to check the endpoints! Sometimes the highest or lowest point isn't a stationary point at all, but one of the ends of the interval.
Key Takeaway
Set $$ f'(x)=0 $$ to find potential max/min points. Use the sign of $$ f''(x) $$ (the "face test") to classify them. For global extrema, always remember to check the endpoints of the interval!
3. Curve Sketching
This is where we become detectives and put all our clues together to draw an accurate sketch of a polynomial or rational function. Follow this checklist, and you can't go wrong!
The Ultimate 7-Step Sketching Checklist
For a function $$ y = f(x) $$:
1. Domain: Are there any x-values that are not allowed?
For polynomials, the domain is all real numbers. For rational functions $$ \frac{P(x)}{Q(x)} $$, the domain is all real numbers except where $$ Q(x)=0 $$.
2. Intercepts:
y-intercept: Set $$ x=0 $$, find y.
x-intercept(s): Set $$ y=0 $$, solve for x.
3. Symmetry: (A quick check)
Even function? If $$ f(-x) = f(x) $$, it's symmetric about the y-axis (like $$ y=x^2 $$).
Odd function? If $$ f(-x) = -f(x) $$, it has rotational symmetry about the origin (like $$ y=x^3 $$).
4. Asymptotes (The "Invisible Fences" for Rational Functions):
An asymptote is a line that the curve gets closer and closer to but never touches.
- Vertical Asymptotes: Occur where the denominator is zero. Find x-values that make the function undefined.
- Horizontal Asymptotes: What happens as $$ x \to \infty $$ and $$ x \to -\infty $$? For rational functions, compare the degree (highest power) of the numerator (N) and denominator (D):
- deg(N) < deg(D): HA is $$ y = 0 $$.
- deg(N) = deg(D): HA is $$ y = \frac{\text{leading coefficient of N}}{\text{leading coefficient of D}} $$.
- deg(N) > deg(D): No HA. Check for an oblique asymptote.
- Oblique (Slant) Asymptotes: These exist only if deg(N) is exactly one more than deg(D). To find it, perform polynomial long division. The equation of the asymptote is the quotient part (ignore the remainder).
5. Stationary Points:
Find where $$ f'(x)=0 $$. Find the full coordinates ($$ x, y $$) of these points and use the Second Derivative Test ($$ f''(x) $$) to classify them as local max or min.
6. Points of Inflexion:
A point of inflexion is where the concavity changes (from ☺ to ☹, or vice versa).
Find potential points by solving $$ f''(x)=0 $$. Then, check that the sign of $$ f''(x) $$ actually changes around that point.
7. Sketch!
Draw your axes. Add asymptotes as dashed lines. Plot your intercepts and stationary points. Then, connect the dots, making sure your curve follows the "invisible fences" and has the correct shape (concave up/down) in each region.
Quick Example: Sketching $$ y = \frac{x^2}{x-2} $$
- Domain: $$ x \neq 2 $$.
- Intercepts: y-int is (0,0). x-int is (0,0).
- Symmetry: None.
- Asymptotes:
- Vertical: Denominator is zero at $$ x=2 $$. So, VA is $$ x=2 $$.
- Horizontal: deg(N)=2, deg(D)=1. Since deg(N) > deg(D), no HA.
- Oblique: Since deg(N) is one more than deg(D), yes! Long division of $$ x^2 $$ by $$ (x-2) $$ gives a quotient of $$ x+2 $$. So, OA is $$ y=x+2 $$.
- Stationary Points: Use the quotient rule to find $$ f'(x) = \frac{x(x-4)}{(x-2)^2} $$. Setting to 0 gives $$ x=0 $$ and $$ x=4 $$. These are a local max at (0,0) and a local min at (4,8).
- Sketch: Draw the asymptotes $$ x=2 $$ and $$ y=x+2 $$. Plot the points (0,0) and (4,8). Draw the two branches of the curve approaching the asymptotes.
Key Takeaway
Curve sketching is a process. Follow the checklist every time. Asymptotes are your guide rails, and stationary points are your key landmarks.
4. Real-World Applications (Optimization and Rates of Change)
This is where we use all our skills to solve practical problems.
Optimization Problems (Max/Min)
The goal is to find the best possible value, like the maximum volume of a box or the minimum cost to build a fence.
Problem-Solving Strategy:
- Understand and Diagram: Read the problem carefully. Draw a picture! Label variables. Identify the quantity you want to maximize or minimize (e.g., Area A, Volume V).
- Formulate an Equation: Write an equation for the quantity you want to optimize. It must be in terms of a single variable. You may need a second "constraint" equation to substitute and eliminate other variables.
- Differentiate: Find the derivative of your equation (e.g., $$ \frac{dA}{dx} $$).
- Find Stationary Points: Set the derivative to zero and solve.
- Test and Justify: Use the second derivative test to prove it's a maximum or a minimum.
- Answer the Question: Reread the question. Are you asked for the dimension (like x) or the actual maximum/minimum value (like A)? Don't forget units!
Example:
You have 40m of fencing to build a rectangular garden. What is the maximum possible area of the garden?
1. Diagram: A rectangle with length L and width W.
Quantity to maximize: Area, $$ A = LW $$.
Constraint: Perimeter, $$ 2L + 2W = 40 $$, which simplifies to $$ L+W=20 $$ or $$ L = 20 - W $$.
2. Equation in one variable: Substitute L into the Area formula:
$$ A = (20 - W)W = 20W - W^2 $$.
3. Differentiate: $$ \frac{dA}{dW} = 20 - 2W $$.
4. Set to zero: $$ 20 - 2W = 0 \implies W = 10 $$m.
5. Test: The second derivative is $$ \frac{d^2A}{dW^2} = -2 $$. Since this is negative, it confirms a maximum.
6. Answer: The width is 10m. The length is $$ L = 20 - 10 = 10 $$m. The maximum area is $$ A = 10 \times 10 = 100 \text{ m}^2 $$.
Rates of Change Problems
The derivative $$ \frac{dy}{dt} $$ means "the rate at which y is changing with respect to time t".
Often, we use the Chain Rule to connect different rates. For example, if we want to find how the area (A) of a circle is changing over time ($$ \frac{dA}{dt} $$), we can use its radius (r):
$$ \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} $$This formula connects the rate of change of area to the rate of change of the radius.
Key Takeaway
For optimization, turn the word problem into a function of one variable, then find its maximum or minimum. For rates of change, identify the rates you know and the rate you need, and connect them with the Chain Rule.
Did you know? The methods of finding maxima and minima were first developed by Pierre de Fermat, a lawyer and amateur mathematician, even before Newton and Leibniz formalized calculus. He used it to figure out the path light takes, which is always the path that takes the minimum amount of time!