The gradient of the curve C is given by dy/dx = (3x - 1)². The point P(1, 4) lies on C. (a) Find an equation of the normal to C at P. (b) Find an equation for the curve C in the form y = f(x). (c) Using dy/dx = (3x - 1)², show that there is no point on C at which the tangent is parallel to the line y = 1 - 2x.

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The gradient of the curve C is given by dy/dx = (3x - 1)² - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 2

Question 10

The gradient of the curve C is given by dy/dx = (3x - 1)². The point P(1, 4) lies on C. (a) Find an equation of the normal to C at P. (b) Find an equation for th... show full transcript

Worked Solution & Example Answer:The gradient of the curve C is given by dy/dx = (3x - 1)² - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 2

Step 1

Find an equation of the normal to C at P.

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Answer

To find the equation of the normal at point P(1, 4), we first need to evaluate the gradient (slope) at this point.

Substitute x = 1 into the gradient equation: $\frac{dy}{dx} = (3(1) - 1)² = (2)² = 4$
The slope of the normal line is the negative reciprocal of the tangent's gradient: $m_{normal} = -\frac{1}{4}$
Use the point-slope form of the line equation, which is: y - y_1 = m(x - x_1): $y - 4 = -\frac{1}{4}(x - 1)$
Rearranging gives: $y = -\frac{1}{4}x + 1 + 4$ $y = -\frac{1}{4}x + 5$

Thus, the equation of the normal is: $y = -\frac{1}{4}x + 5$

Step 2

Find an equation for the curve C in the form y = f(x).

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Answer

Given the gradient function: $\frac{dy}{dx} = (3x - 1)²$

To find the equation of the curve, we need to integrate the gradient function: $y = \int (3x - 1)² dx$
Expanding the integrand: $(3x - 1)² = 9x² - 6x + 1$
Integrating term by term: $y = \int (9x² - 6x + 1) dx = 3x³ - 3x² + x + C$
Now, we use point P(1, 4) to find C: $4 = 3(1)³ - 3(1)² + 1 + C$ $4 = 3 - 3 + 1 + C$ $C = 0$

Thus, the equation for curve C is: $y = 3x³ - 3x² + x$

Step 3

Using dy/dx = (3x - 1)², show that there is no point on C at which the tangent is parallel to the line y = 1 - 2x.

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Answer

The line y = 1 - 2x has a slope of -2. For a tangent to be parallel to this line, we need: $\frac{dy}{dx} = -2$

Set the gradient function equal to -2: $(3x - 1)² = -2$
Note that a square of a real number is always non-negative. Hence: $(3x - 1)² \geq 0$
Since -2 is negative, this results in no solution. Consequently, there is no point on curve C at which the tangent is parallel to the given line.

The gradient of the curve C is given by dy/dx = (3x - 1)² - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 2

Worked Solution & Example Answer:The gradient of the curve C is given by dy/dx = (3x - 1)² - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 2

Find an equation of the normal to C at P.

Find an equation for the curve C in the form y = f(x).

Using dy/dx = (3x - 1)², show that there is no point on C at which the tangent is parallel to the line y = 1 - 2x.

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