Tangents and Curves Revision Notes for Leaving Cert Mathematics

Tangents and Curves

Understanding tangents to curves

A tangent to a curve is a straight line that touches the curve at exactly one point and has the same slope as the curve at that point. The slope of this tangent line is found using differentiation.

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Key concept: $\frac{dy}{dx}$ represents the slope of the tangent to a curve at any point $(x, y)$ on the curve.

This graph shows how a tangent line touches the curve at point P, demonstrating the relationship between the curve and its tangent.

Method for finding tangent equations

To find the equation of a tangent to the curve $y = f(x)$ at point $(x_1, y_1)$ , follow these steps:

Find $\frac{dy}{dx}$ by differentiating the function
Find the slope by substituting the x-coordinate into $\frac{dy}{dx}$
Use the point-slope form $y - y_1 = m(x - x_1)$ to find the tangent equation

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The point-slope form is: $y - y_1 = m(x - x_1)$

Where:

$(x_1, y_1)$ is the given point on the curve
$m$ is the slope of the tangent

Worked example 1: Finding tangent equation at a given point

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Worked Example: Finding tangent equation at a given point

Problem: Find the slope of the tangent to the curve $y = 3x^2 + 4x - 5$ at the point $(1, 2)$ , then find the equation of the tangent.

Solution:

Step 1: Find the derivative $\frac{dy}{dx} = 6x + 4$

Step 2: Find the slope at $x = 1$ $\frac{dy}{dx} = 6(1) + 4 = 10$

Therefore, the slope of the tangent = 10

Step 3: Use point-slope form with $(x_1, y_1) = (1, 2)$ and $m = 10$ $y - 2 = 10(x - 1)$ $y - 2 = 10x - 10$ $10x - y - 8 = 0$

The equation of the tangent is $10x - y - 8 = 0$ .

Worked example 2: Finding coordinates with a given slope

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Worked Example: Finding coordinates with a given slope

Problem: Find the coordinates of the point on the curve $y = x^2 + 2x + 1$ at which the slope of the tangent is 4.

Solution:

Step 1: Find the derivative $\frac{dy}{dx} = 2x + 2$

Step 2: Set the derivative equal to the given slope Since the slope of the tangent = 4: $2x + 2 = 4$ $2x = 2$ $x = 1$

Step 3: Find the y-coordinate by substituting $x = 1$ into the original equation $y = (1)^2 + 2(1) + 1 = 4$

Therefore, the point on the curve is $(1, 4)$ .

Worked example 3: Parallel tangents

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Worked Example: Parallel tangents

Problem: Find the coordinates of the point on the curve $y = x^2 - 3x + 7$ at which the tangent is parallel to the line $y = 3x + 4$ .

Solution:

Step 1: Find the slope of the given line The line $y = 3x + 4$ has slope = 3

Step 2: Since parallel lines have equal slopes, the tangent slope must also be 3 $\frac{dy}{dx} = 2x - 3$

Step 3: Set the derivative equal to 3 $2x - 3 = 3$ $2x = 6$ $x = 3$

Step 4: Find the y-coordinate $y = (3)^2 - 3(3) + 7 = 9 - 9 + 7 = 7$

Therefore, the point on the curve is $(3, 7)$ .

Key rules and formulas

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Key Points to Remember:

Tangent slope formula: $\frac{dy}{dx}$ gives the slope at any point
Point-slope form: $y - y_1 = m(x - x_1)$
Parallel lines: Have equal slopes
Perpendicular lines: Have slopes that multiply to give -1

Common exam tips

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Essential Exam Tips:

Always differentiate first to find $\frac{dy}{dx}$
Substitute the x-coordinate carefully to find the numerical slope
Remember to use the point-slope form correctly
When finding coordinates with a given slope, set $\frac{dy}{dx}$ equal to that slope
For parallel tangents, the slopes must be equal
Check your final answer by substituting back into the original equation

Remember!

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Summary - Key Concepts:

$\frac{dy}{dx}$ is the slope of the tangent to the curve at any point $(x, y)$
Use the three-step method: differentiate, substitute, apply point-slope form
For parallel lines, slopes are equal
Always substitute coordinates back into the original equation to find missing values
The point-slope form $y - y_1 = m(x - x_1)$ is essential for finding tangent equations

Tangents and Curves (Leaving Cert Mathematics): Revision Notes