M1 Calculus: Study Notes for Applications of Differentiation
Hello there! Welcome to one of the most powerful and exciting parts of calculus. Ever wondered how businesses find the perfect price to maximize their profit, or how engineers design a container that holds the most volume using the least material? The answer is differentiation!
In this chapter, we're going to see how the derivative, which you know as the slope of a function, is much more than just a number. It's a tool that helps us understand how things change, find the highest and lowest points, and solve real-world problems. Don't worry if this sounds complicated – we'll break it all down step-by-step. Let's dive in!
Section 1: The Geometry of a Curve - Tangents
Remember the first thing you learned about derivatives? The derivative of a function at a point gives you the slope of the tangent line at that exact point. A tangent is a straight line that just "touches" a curve at one point without crossing through it (at least at that point).
How to Find the Equation of a Tangent Line
Finding the equation of a tangent is a classic exam question, and it's super easy if you follow a recipe! We just need two things for a straight line: a point and a slope.
Let's find the equation of the tangent to the curve $$y = f(x)$$ at the point where $$x = a$$.
- Find the Point (a, f(a)): If you only have the x-value, plug it into the original function $$f(x)$$ to get the y-value. So your point is $$(a, f(a))$$.
- Find the Slope (m): The slope is the derivative! First, find the derivative function, $$f'(x)$$. Then, plug your x-value into the derivative to get the slope: $$m = f'(a)$$.
- Use the Point-Slope Formula: Now you have a point $$(x_1, y_1) = (a, f(a))$$ and a slope $$m$$. Just use the good old formula for a straight line: $$ y - y_1 = m(x - x_1) $$
Step-by-Step Example:
Find the equation of the tangent to the curve $$f(x) = x^3 - 2x + 5$$ at $$x = 2$$.
Step 1: Find the point.
The x-coordinate is $$x=2$$.
The y-coordinate is $$f(2) = (2)^3 - 2(2) + 5 = 8 - 4 + 5 = 9$$.
So, our point is (2, 9).
Step 2: Find the slope.
First, find the derivative: $$f'(x) = 3x^2 - 2$$.
Now, plug in $$x=2$$ to get the slope: $$m = f'(2) = 3(2)^2 - 2 = 3(4) - 2 = 10$$.
The slope is 10.
Step 3: Use the point-slope formula.
We have the point (2, 9) and slope m = 10.
$$ y - 9 = 10(x - 2) $$
$$ y - 9 = 10x - 20 $$
$$ y = 10x - 11 $$
And that's our final answer! See? It's just a 3-step process.
Key Takeaway for Section 1:
The derivative $$f'(a)$$ is the slope of the tangent at $$x=a$$. To find the tangent's equation, you need a point and this slope.
Section 2: The Speed of Change - Rates of Change
The derivative doesn't just tell us the slope of a graph; it tells us the instantaneous rate of change. Think of it like this: your car's average speed for a trip is the total distance divided by total time. But the number on your speedometer at any moment is your instantaneous speed. The derivative is the speedometer for any changing quantity!
The notation $$\frac{dy}{dx}$$ literally means "the rate of change of y with respect to x". If the variable is time, `t`, then $$\frac{dy}{dt}$$ means "how fast y is changing over time".
Real-World Examples:
- If $$s(t)$$ is the displacement (position) of an object at time $$t$$, then $$s'(t) = \frac{ds}{dt}$$ is its velocity.
- If $$v(t)$$ is the velocity of an object at time $$t$$, then $$v'(t) = \frac{dv}{dt}$$ is its acceleration.
- If $$V(r)$$ is the volume of a sphere with radius $$r$$, then $$\frac{dV}{dr}$$ tells us how much the volume changes for a tiny change in the radius.
Step-by-Step Example:
The volume of a sphere is given by $$V = \frac{4}{3}\pi r^3$$. Find the rate of change of the volume with respect to the radius when the radius is 5 cm.
Step 1: Identify what you need to find.
"Rate of change of volume with respect to radius" means we need to find $$\frac{dV}{dr}$$.
Step 2: Differentiate the function.
The function is $$V = \frac{4}{3}\pi r^3$$. Remember that $$\frac{4}{3}$$ and $$\pi$$ are just constants.
$$ \frac{dV}{dr} = \frac{4}{3}\pi \cdot (3r^2) $$
$$ \frac{dV}{dr} = 4\pi r^2 $$
Did you know? The rate of change of a sphere's volume is its surface area! Cool, right?
Step 3: Plug in the given value.
We need to find the rate when the radius is 5 cm, so we plug in $$r=5$$.
$$ \frac{dV}{dr} |_{r=5} = 4\pi (5)^2 = 4\pi(25) = 100\pi $$
So, when the radius is 5 cm, the volume is changing at a rate of $$100\pi$$ cm³/cm.
Key Takeaway for Section 2:
The derivative measures the instantaneous rate of change. When you see "rate of change of A with respect to B", you should immediately think "find $$\frac{dA}{dB}$$".
Section 3: Uphills, Downhills, and Flat Spots - Finding Maximums and Minimums
This is where differentiation becomes a super-tool for optimization. We can use it to find the peaks and valleys of any function.
Imagine walking along a curve from left to right:
- When you're going uphill, the slope is positive, so $$f'(x) > 0$$. The function is increasing.
- When you're going downhill, the slope is negative, so $$f'(x) < 0$$. The function is decreasing.
- When you're at the very top of a hill or the bottom of a valley, the ground is momentarily flat. The slope is zero, so $$f'(x) = 0$$. These important points are called stationary points.
Types of Stationary Points
Stationary points (where $$f'(x) = 0$$) can be:
- Local Maximum: The top of a "hill". The function goes from increasing to decreasing.
- Local Minimum: The bottom of a "valley". The function goes from decreasing to increasing.
- Stationary Point of Inflection: A flat spot that isn't a peak or a valley. The function flattens out and then continues in the same direction.
The First Derivative Test
This test helps us classify stationary points. It's like sending a scout just before and just after the flat spot to check the slope.
How to do it: 1. Find the x-values where $$f'(x) = 0$$. These are your stationary points. 2. Create a sign table for $$f'(x)$$. 3. Test a value of $$x$$ in each interval (before, between, and after your stationary points) to see if $$f'(x)$$ is positive (+) or negative (-). 4. Analyse the signs around each stationary point:
- Sign changes from + to - ($$\nearrow \searrow$$): You have a local maximum.
- Sign changes from - to + ($$\searrow \nearrow$$): You have a local minimum.
- Sign does not change (+ to + or - to -): You have a stationary point of inflection.
Step-by-Step Example:
Find and classify the stationary points of $$f(x) = x^3 - 3x^2 + 1$$.
Step 1: Find $$f'(x)$$ and set it to zero.
$$f'(x) = 3x^2 - 6x$$
$$3x^2 - 6x = 0$$
$$3x(x - 2) = 0$$
So, the stationary points are at $$x = 0$$ and $$x = 2$$.
Step 2: Create a sign table for $$f'(x)$$.
We test values in the intervals $$x<0$$, $$0
- Test $$x = -1$$: $$f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9$$ (Positive)
- Test $$x = 1$$: $$f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3$$ (Negative)
- Test $$x = 3$$: $$f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9$$ (Positive)
The Sign Table:
x x < 0 x = 0 0 < x < 2 x = 2 x > 2
Sign of f'(x) + 0 - 0 +
Shape $$\nearrow$$ (Max) $$\searrow$$ (Min) $$\nearrow$$
Step 3: Conclude and find the coordinates.
At $$x=0$$, the sign changes from + to -, so it's a local maximum. The point is $$(0, f(0)) = (0, 1)$$.
At $$x=2$$, the sign changes from - to +, so it's a local minimum. The point is $$(2, f(2)) = (2, 8 - 12 + 1) = (2, -3)$$.
Quick Review Box
Increasing function: $$f'(x) > 0$$ (slope is positive)
Decreasing function: $$f'(x) < 0$$ (slope is negative)
Stationary point: $$f'(x) = 0$$ (slope is zero)
Section 4: The Shape of the Curve - Concavity and the Second Derivative Test
The first derivative tells us if a function is going up or down. The second derivative, $$f''(x)$$, tells us about the function's shape or curvature. This is called concavity.
- Concave Up: The curve is shaped like a 'U' (it holds water). The slope is increasing (e.g., from -2 to -1 to 0 to 1). This happens when $$f''(x) > 0$$.
- Concave Down: The curve is shaped like an 'n' (it spills water). The slope is decreasing (e.g., from 2 to 1 to 0 to -1). This happens when $$f''(x) < 0$$.
The Second Derivative Test (A Handy Shortcut!)
The Second Derivative Test is a faster way to classify stationary points, without needing a sign table.
How to do it: 1. Find the x-values where $$f'(x) = 0$$ (your stationary points). 2. Find the second derivative, $$f''(x)$$. 3. Plug each stationary point's x-value into $$f''(x)$$:
- If $$f''(x) > 0$$ (positive), the curve is concave up ('U' shape), so you've found a local minimum.
- If $$f''(x) < 0$$ (negative), the curve is concave down ('n' shape), so you've found a local maximum.
- If $$f''(x) = 0$$, the test is inconclusive. It fails! You must go back and use the First Derivative Test.
Memory Aid:
- $$f''(x)$$ is Positive --> Think of a plus sign --> Smiley face :) --> It's a Minimum.
- $$f''(x)$$ is Negative --> Think of a minus sign --> Frowny face :( --> It's a Maximum.
Step-by-Step Example (using the same function):
Use the second derivative test to classify the stationary points of $$f(x) = x^3 - 3x^2 + 1$$.
Step 1: Find stationary points.
We already found them: $$f'(x) = 3x^2 - 6x = 0$$ at $$x = 0$$ and $$x = 2$$.
Step 2: Find the second derivative.
$$f''(x) = 6x - 6$$
Step 3: Test each point in $$f''(x)$$.
- For $$x=0$$: $$f''(0) = 6(0) - 6 = -6$$. This is negative, so $$x=0$$ is a local maximum.
- For $$x=2$$: $$f''(2) = 6(2) - 6 = 6$$. This is positive, so $$x=2$$ is a local minimum.
This matches what we found with the First Derivative Test, but it was much faster!
Key Takeaway for Section 4:
$$f''(x)$$ describes concavity. The Second Derivative Test is a fast way to classify stationary points: $$f''(x) > 0 \Rightarrow$$ Minimum, $$f''(x) < 0 \Rightarrow$$ Maximum.
Section 5: Putting It All Together - Optimization Problems
This is the ultimate goal! We want to find the absolute biggest or smallest value of a function. This could be maximum profit, minimum cost, maximum area, etc.
Local vs. Global Extrema
- Local Extrema (or Relative Extrema): These are the local hills and valleys we've been finding. A point is a local max if it's higher than all the points immediately surrounding it.
- Global Extrema (or Absolute Extrema): This is the absolute highest or lowest point on the entire function, or over a specific interval.
Important! The global maximum or minimum on a closed interval $$[a, b]$$ can occur at a stationary point OR at one of the endpoints ($$x=a$$ or $$x=b$$). You always have to check the endpoints!
The Ultimate Recipe for Solving Optimization Problems:
1. Understand the Problem: Read carefully. Identify the quantity you want to maximize or minimize (e.g., Area, Volume, Cost). 2. Write the Equations: Write down a formula for the quantity you're optimizing (the 'objective function'). You might also have a 'constraint' equation (e.g., fixed perimeter). Use the constraint to write the objective function in terms of a single variable. 3. Find the Derivative: Differentiate your single-variable objective function. 4. Find Stationary Points: Set the derivative to zero and solve. 5. Test and Justify: Use the First or Second Derivative Test to confirm that you've found a maximum or a minimum. 6. Check Endpoints: If the problem has a closed interval (e.g., "x must be between 0 and 10"), you must also calculate the value of the function at the endpoints. 7. Find the Final Answer: Compare the values from the stationary points and the endpoints. The biggest is the global max, the smallest is the global min. Make sure you answer the actual question asked!
Step-by-Step Example:
You have 40 metres of fence to make a rectangular garden. What is the maximum possible area of the garden?
Step 1: Understand. We want to MAXIMIZE the AREA.
Step 2: Equations. Let the length be $$L$$ and width be $$W$$. - Objective function (Area): $$A = LW$$. - Constraint (Perimeter): $$2L + 2W = 40$$. Let's get $$A$$ in terms of one variable. From the constraint, $$2L = 40 - 2W$$, so $$L = 20 - W$$. Substitute this into the Area formula: $$ A(W) = (20 - W)W = 20W - W^2 $$
Step 3: Derivative. $$ A'(W) = 20 - 2W $$
Step 4: Stationary Points. $$ 20 - 2W = 0 $$ $$ 2W = 20 $$ $$ W = 10 $$
Step 5: Test. Let's use the Second Derivative Test. $$ A''(W) = -2 $$ Since $$A''(W)$$ is always -2 (which is negative), our stationary point at $$W=10$$ must be a maximum. Success!
Step 6: Endpoints. The width $$W$$ must be greater than 0, and also less than 20 (otherwise L would be 0 or negative). So the interval is $$(0, 20)$$. Since this is an open interval, we don't have endpoints to check. Our stationary point is the only candidate.
Step 7: Final Answer. The question asks for the MAXIMUM AREA, not just the width. If $$W = 10$$, then $$L = 20 - 10 = 10$$. Maximum Area $$ = L \times W = 10 \times 10 = 100$$ m².
The maximum possible area is 100 square metres (which happens when the garden is a square!).
Common Mistakes to Avoid:
- Forgetting to check the endpoints on a closed interval. This is a very common way to lose marks!
- Solving for `x` but then forgetting to plug it back in to find the actual maximum/minimum value.
- Mixing up the First and Second Derivative Tests. Remember, $$f'(x)=0$$ finds the point, $$f''(x)$$ classifies it.
Key Takeaway for Section 5:
To find the global max/min on an interval, find the function's value at all stationary points AND at the endpoints. The biggest/smallest value is your answer.