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Edexcel GCSE Maths (Higher) Study Notes

2.10.1 Estimating Gradients of Curves

Understanding how to estimate gradients by drawing tangents on curves is a fundamental aspect of calculus that allows us to analyse the behaviour of functions at specific points. This section is dedicated to mastering these techniques, providing a clear pathway to comprehend the rate of change of functions graphically.

Introduction to Gradient Estimation

The gradient of a curve at a given point reflects the slope of the tangent at that point. It's an essential concept, especially when we delve into rates of change in various scientific disciplines.

What is a Tangent?

  • A tangent is a straight line that lightly touches a curve at one point, representing the direction in which the curve is heading at that point.

Estimating Gradients: The Process

Estimating a curve's gradient involves drawing a tangent at the point of interest and calculating this tangent's slope using the formula:

Gradient=ΔyΔx\text{Gradient} = \frac{\Delta y}{\Delta x}

Drawing Tangents: A Step-by-Step Guide

1. Identify the Point of Interest: Choose the specific point on the curve where you want to estimate the gradient.

2. Draw the Tangent: Sketch a straight line that just touches the curve at the chosen point.

3. Select Points on the Tangent: Pick two points on the tangent line, ideally far apart, for a more accurate gradient calculation.

4. Calculate the Gradient: Use the chosen points' coordinates to calculate the tangent's slope, providing an estimate of the curve's gradient at your point of interest.

Example: Estimating the Gradient of y=x2y = x^2 at x=2x = 2

Let's estimate the gradient of the curve y=x2y = x^2 at the point where x=2x = 2.

Drawing the Curve and Tangent

1. We first sketch the curve of y=x2y = x^2.

Graph of y = x^2

2. At x=2x = 2, we draw a tangent to the curve. Upon drawing the tangent, we select two points on this line, for instance, x=3x = -3 and x=5x = 5, and calculate the gradient of the tangent. For our example, the calculated gradient at x=2x = 2 is 4, and the equation of the tangent line can be described by y=4x4y = 4x - 4.

2. This tangent touches the curve precisely at the point (2,4)(2, 4). The slope of the tangent, which represents the estimated gradient of the curve at x=2x = 2, is 4, and the y-intercept of the tangent line is -4.

Estimating the gradient of y = x^2 at x = 2

Practice Problems

Problem 1:

Estimating the Gradient of y=3x32x2+x5y = 3x^3 - 2x^2 + x - 5 at x=1x = 1.

Solution:

1. Sketch the Curve: Draw the curve y=3x32x2+x5y = 3x^3 - 2x^2 + x - 5.

Graph of y = 3x^3 - 2x^2 + x - 5

2. Tangent at x=1x = 1: Draw a tangent at the point x=1x = 1, and calculate its slope.

  • The derivative of yy with respect to xx at x=1x = 1 gives the slope of the tangent.
  • Use the derivative formula to find the exact slope and proceed with the gradient calculation similarly to the example above.
Estimating the Gradient of (y = 3x^3 - 2x^2 + x - 5) at (x = 1)

Problem 2:

Determine the Gradient of y=x at x=4y = \sqrt{x} \text{ at } x = 4

Solution:

1. Graph the Function: Plot y=xy = \sqrt{x} on a graph.

Graph of y = \sqrt{x}

2. Draw and Calculate: At x=4x = 4, draw a tangent and estimate its gradient.

  • Again, the slope of the tangent at x=4x = 4 can be found by deriving yy with respect to xx and evaluating at x=4x = 4.
  • Follow the steps to calculate the gradient as demonstrated.
Estimating the Gradient of (y = \sqrt{x}) at (x = 4)

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