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The gradient of the curve C is given by dy/dx = (3x - 1)² - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 2
Question 2
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 2 - 2005 - Paper 2
Step 1
Find an equation of the normal to C at P.
Answer
To find the equation of the normal, we first need to evaluate the gradient at the point P(1, 4).
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Calculate the gradient at P:
Substitute x = 1 into the gradient equation:
rac{dy}{dx} = (3(1) - 1)^2 = (3 - 1)^2 = 2^2 = 4.The gradient at P is 4.
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Determine the slope of the normal: The slope of the normal is the negative reciprocal of the gradient:
ext{slope of normal} = - rac{1}{4}.
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Use point-slope form to find the equation of the normal: The equation is given by where (x1, y1) is the point P(1, 4) and m is the slope from above. Thus:
y - 4 = - rac{1}{4}(x - 1) Simplifying yields: y - 4 = - rac{1}{4}x + rac{1}{4} y = - rac{1}{4}x + rac{17}{4}.
Step 2
Find an equation for the curve C in the form y = f(x).
Answer
To find the equation of the curve C, we need to integrate the gradient.
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Rewrite the gradient equation:
The gradient is given by:
rac{dy}{dx} = (3x - 1)^2
- Integrate the gradient:
Performing the integration:
ightarrow du = 3 dx ightarrow dx = rac{du}{3}$$
Substituting gives: = rac{1}{3} rac{u^3}{3} + C = rac{(3x - 1)^3}{9} + C. Thus: y = rac{(3x - 1)^3}{9} + C.
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Determine the constant C using point P(1, 4): Substitute (1, 4):
4 = rac{(3(1) - 1)^3}{9} + C
4 = rac {2^3}{9} + C = rac{8}{9} + C Thus: C = 4 - rac{8}{9} = rac{36}{9} - rac{8}{9} = rac{28}{9}.
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Final equation: The equation of the curve is: y = rac{(3x - 1)^3}{9} + rac{28}{9}.
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.
Answer
To show that there is no point on C with a tangent parallel to the line, we need to find the gradient of the line.
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Identify the gradient of the line: The equation of the line is given by: The gradient of this line is -2.
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Set the gradient of the curve equal to -2: We have: rac{dy}{dx} = (3x - 1)^2 = -2. However, since the left-hand side, , is always non-negative, we analyze further:
The right-hand side must be greater than or equal to 0, therefore for any x. Consequently, there are no values of x where the tangent is parallel to the given line.