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The curve C has equation y = 2x - 8 adical{x} + 5, x > 0 a) Find \( \frac{dy}{dx} \), giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 1
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The curve C has equation y = 2x - 8 adical{x} + 5, x > 0 a) Find \( \frac{dy}{dx} \), giving each term in its simplest form. (3) The point P on C has x-coordinat... show full transcript
Worked Solution & Example Answer:The curve C has equation y = 2x - 8 adical{x} + 5, x > 0 a) Find \( \frac{dy}{dx} \), giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 1
Step 1
Find \( \frac{dy}{dx} \)
Answer
To find ( \frac{dy}{dx} ), we differentiate each term of the equation:
- The derivative of ( 2x ) is ( 2 ).
- Next, we differentiate ( -8\sqrt{x} ), using the power rule. Recall that ( \sqrt{x} = x^{1/2} ), so: [ \frac{d}{dx}(-8x^{1/2}) = -8 \cdot \frac{1}{2}x^{-1/2} = -4x^{-1/2} = -\frac{4}{\sqrt{x}}. ]
- The derivative of the constant ( 5 ) is ( 0 ).
Thus, [ \frac{dy}{dx} = 2 - \frac{4}{\sqrt{x}}. ]
Step 2
Find the equation of the tangent to C at the point P
Answer
First, we calculate ( \frac{dy}{dx} ) at the point where ( x = \frac{1}{4} ). Substituting this value:
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[ \frac{dy}{dx} = 2 - \frac{4}{\sqrt{\frac{1}{4}}} = 2 - 4 = -2. ]
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Now, we find the y-coordinate of point P by substituting ( x = \frac{1}{4} ) into the original equation: [ y = 2(\frac{1}{4}) - 8\sqrt{\frac{1}{4}} + 5 = \frac{1}{2} - 4 + 5 = 3. ] So, point P is ( (\frac{1}{4}, 3) ).
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Using the point-slope form of a line, the equation of the tangent line is: [ y - 3 = -2( x - \frac{1}{4} ). ] Simplifying this gives: [ y = -2x + \frac{1}{2} + 3 = -2x + \frac{7}{2}. ]
Step 3
Find the coordinates of Q
Answer
Since the tangent at Q is parallel to the line ( 2x - 3y + 18 = 0 ), we first find the gradient of this line by rearranging it:
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Rearranging gives: [ 3y = 2x + 18 \Rightarrow y = \frac{2}{3}x + 6. ] So, the gradient is ( \frac{2}{3} ).
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We set ( \frac{dy}{dx} ) equal to this gradient to find ( x ): [-2 + \frac{4}{\sqrt{x}} = \frac{2}{3} \Rightarrow \frac{4}{\sqrt{x}} = \frac{2}{3} + 2. ] Solving gives: [ \frac{4}{\sqrt{x}} = \frac{2}{3} + \frac{6}{3} = \frac{8}{3} \Rightarrow \sqrt{x} = \frac{4 \cdot 3}{8} = \frac{3}{2}. ] Therefore, ( x = \left( \frac{3}{2} \right)^{2} = \frac{9}{4}. )
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To find the corresponding y-coordinate, substitute this value into the original equation: [ y = 2(\frac{9}{4}) - 8\sqrt{\frac{9}{4}} + 5 = \frac{9}{2} - 12 + 5 = \frac{9}{2} - \frac{24}{2} + \frac{10}{2} = -\frac{5}{2}. ]
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Therefore, the coordinates of Q are ( \left( \frac{9}{4}, -\frac{5}{2} \right) ).