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Ažurirano Apr 14, 2026

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Mastering Differentiation: Tangents, Normals, and Curve Sketching

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# Applications of Differentiation

An overview of applications

Differentiation isn't just about finding the derivative of a function. It's

Applications Overview and Key Concepts

Understanding differentiation gives you the power to solve problems that matter in the real world. The derivative tells you how steep a curve is at any point, which translates to finding maximum profits, minimum costs, or optimal designs.

When you see $\frac{dy}{dx}$ or $f'(x)$ , you're looking at the instantaneous rate of change - basically the gradient of the tangent line at any point. This is your foundation for everything else.

Stationary points occur where $f'(x) = 0$ , meaning the gradient is zero and you've got a horizontal tangent. These points are crucial because they're often where maximum and minimum values occur - exactly what you need for optimisation problems.

Remember: A tangent touches the curve at one point with the same gradient, while a normal is perpendicular to the tangent at that same point.

Finding Tangent and Normal Lines Plus Rates of Change

Getting the equation of a tangent follows a straightforward process: find $f'(x)$ , substitute your x-coordinate to get the gradient, then use $y - y_1 = m(x - x_1)$ . For the normal line, use $m_N = -\frac{1}{m_T}$ since perpendicular lines have gradients that multiply to give -1.

Rates of change connect maths to physics beautifully. If you've got displacement $s(t)$ , then velocity is $v = \frac{ds}{dt}$ and acceleration is $a = \frac{d^2s}{dt^2}$ . It's all about how quickly things change over time.

The real power comes when you realise that any rate of change problem follows the same pattern. Whether it's water flowing from a tank or profit changing with production levels, the derivative gives you the rate.

Top Tip: Always check your perpendicular gradients multiply to give -1 - it's an easy way to catch mistakes!

Classifying Stationary Points

The second derivative test is your best friend for determining whether stationary points are maximums, minimums, or points of inflection. Once you've found where $f'(x) = 0$ , substitute those x-values into $f''(x)$ .

If $f''(x) > 0$ , you've got a local minimum - think of a smile shape. If $f''(x) < 0$ , it's a local maximum - like a frown. When $f''(x) = 0$ , the test is inconclusive and you'll need to check the behaviour on either side.

Points of inflection occur where the curve changes from concave up to concave down (or vice versa). These might also be stationary points, but not always.

Memory Trick: Positive second derivative = minimum (like a positive, happy smile ☺). Negative second derivative = maximum (like a negative, sad frown ☹).

Curve Sketching Techniques

Curve sketching brings together everything you know about a function into one clear picture. Start with the y-intercept $let x = 0$ , find any obvious x-intercepts, then locate and classify all stationary points.

Consider what happens as x approaches positive and negative infinity - for polynomials, the highest power term dominates the behaviour. This tells you how the curve behaves at the extremes.

Plot your key points (intercepts and stationary points) and connect them with smooth curves that respect the nature of each point. Maximums create peaks, minimums create troughs.

Pro Tip: Always sketch a rough version first to check your curve makes sense before drawing the final version!

Worked Example: Tangent and Normal Lines

Let's work through finding tangent and normal equations for $y = x^2 - 4x + 1$ at point (1, -2). First, differentiate to get $\frac{dy}{dx} = 2x - 4$ .

At x = 1, the gradient of the tangent is $m_T = 2(1) - 4 = -2$ . Using the point-slope form: $y - (-2) = -2(x - 1)$ , which simplifies to $2x + y = 0$.

For the normal, the gradient is $m_N = -\frac{1}{-2} = \frac{1}{2}$ . Using the same point: $y + 2 = \frac{1}{2}(x - 1)$ , which gives us $x - 2y - 5 = 0$ .

Check Your Work: Verify that $m_T \times m_N = (-2) \times \frac{1}{2} = -1$ ✓

Optimisation Example: Maximum Area Problem

Optimisation problems are where differentiation really shines. Consider a rectangular garden against a wall, using 80m of fencing for three sides. Let the parallel side be l and the other sides be w.

Since fencing covers $l + 2w = 80$ , we get $l = 80 - 2w$ . The area function becomes $A = lw = (80 - 2w)w = 80w - 2w^2$ .

To maximise area, find $\frac{dA}{dw} = 80 - 4w$ and set it to zero: $80 - 4w = 0 $gives$ w = 20m $. Therefore$ l = 80 - 2(20) = 40m $. Since$ \frac{d^2A}{dw^2} = -4 < 0$, this confirms a maximum.

Real-World Check: Always verify your answer makes physical sense - negative dimensions would be impossible!

Essential Tips and Quick Reference

Common mistakes to avoid: Always substitute x-values back into the original function $f(x)$ for coordinates, not into the derivative. When the second derivative test gives zero, check the sign of $f'(x)$ on either side of the stationary point.

Read optimisation questions carefully - are you finding the maximum value itself or the conditions that create it? Context matters enormously.

Quick reference for revision: Stationary points occur when $f'(x) = 0$ . Use $f''(x) > 0$ for minimums, $f''(x) < 0$ for maximums. For motion problems: velocity is $\frac{ds}{dt}$ and acceleration is $\frac{d^2s}{dt^2}$ .

Success Strategy: Practice identifying what type of problem you're dealing with first - this determines which technique to use!

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Mathematics

•

Ažurirano Apr 14, 2026

•

7 stranice

Mastering Differentiation: Tangents, Normals, and Curve Sketching

Differentiation isn't just abstract maths - it's your toolkit for solving real-world problems like finding the steepest point on a road or calculating maximum profit. You'll use derivatives to analyse how functions behave and find optimal solutions to practical situations.

Registruj se da vidiš sadržajBesplatno je!

Pristup svim dokumentima

Poboljšaj svoje ocene

Pridruži se milionima učenika

Applications Overview and Key Concepts

Remember: A tangent touches the curve at one point with the same gradient, while a normal is perpendicular to the tangent at that same point.

Registruj se da vidiš sadržajBesplatno je!

Pristup svim dokumentima

Poboljšaj svoje ocene

Pridruži se milionima učenika

Finding Tangent and Normal Lines Plus Rates of Change

Top Tip: Always check your perpendicular gradients multiply to give -1 - it's an easy way to catch mistakes!

Registruj se da vidiš sadržajBesplatno je!

Pristup svim dokumentima

Poboljšaj svoje ocene

Pridruži se milionima učenika

Classifying Stationary Points

Points of inflection occur where the curve changes from concave up to concave down (or vice versa). These might also be stationary points, but not always.

Memory Trick: Positive second derivative = minimum (like a positive, happy smile ☺). Negative second derivative = maximum (like a negative, sad frown ☹).

Registruj se da vidiš sadržajBesplatno je!

Pristup svim dokumentima

Poboljšaj svoje ocene

Pridruži se milionima učenika

Curve Sketching Techniques

Consider what happens as x approaches positive and negative infinity - for polynomials, the highest power term dominates the behaviour. This tells you how the curve behaves at the extremes.

Plot your key points (intercepts and stationary points) and connect them with smooth curves that respect the nature of each point. Maximums create peaks, minimums create troughs.

Pro Tip: Always sketch a rough version first to check your curve makes sense before drawing the final version!

Registruj se da vidiš sadržajBesplatno je!

Pristup svim dokumentima

Poboljšaj svoje ocene

Pridruži se milionima učenika

Worked Example: Tangent and Normal Lines

Let's work through finding tangent and normal equations for $y = x^2 - 4x + 1$ at point (1, -2). First, differentiate to get $\frac{dy}{dx} = 2x - 4$ .

At x = 1, the gradient of the tangent is $m_T = 2(1) - 4 = -2$ . Using the point-slope form: $y - (-2) = -2(x - 1)$ , which simplifies to $2x + y = 0$.

For the normal, the gradient is $m_N = -\frac{1}{-2} = \frac{1}{2}$ . Using the same point: $y + 2 = \frac{1}{2}(x - 1)$ , which gives us $x - 2y - 5 = 0$ .

Check Your Work: Verify that $m_T \times m_N = (-2) \times \frac{1}{2} = -1$ ✓

Registruj se da vidiš sadržajBesplatno je!

Pristup svim dokumentima

Poboljšaj svoje ocene

Pridruži se milionima učenika

Optimisation Example: Maximum Area Problem

Since fencing covers $l + 2w = 80$ , we get $l = 80 - 2w$ . The area function becomes $A = lw = (80 - 2w)w = 80w - 2w^2$ .

To maximise area, find $\frac{dA}{dw} = 80 - 4w$ and set it to zero: $80 - 4w = 0 $gives$ w = 20m $. Therefore$ l = 80 - 2(20) = 40m $. Since$ \frac{d^2A}{dw^2} = -4 < 0$, this confirms a maximum.

Real-World Check: Always verify your answer makes physical sense - negative dimensions would be impossible!

Registruj se da vidiš sadržajBesplatno je!

Pristup svim dokumentima

Poboljšaj svoje ocene

Pridruži se milionima učenika

Essential Tips and Quick Reference

Read optimisation questions carefully - are you finding the maximum value itself or the conditions that create it? Context matters enormously.

Success Strategy: Practice identifying what type of problem you're dealing with first - this determines which technique to use!

Mislili smo da nikad nećeš pitati...

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