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The curve C has equation y = 2x - 8 \, \sqrt{x} + 5, \quad x > 0 a) Find \frac{dy}{dx}, giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3
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The curve C has equation y = 2x - 8 \, \sqrt{x} + 5, \quad x > 0 a) Find \frac{dy}{dx}, giving each term in its simplest form. The point P on C has x-coordinate e... show full transcript
Worked Solution & Example Answer:The curve C has equation y = 2x - 8 \, \sqrt{x} + 5, \quad x > 0 a) Find \frac{dy}{dx}, giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3
Step 1
Find \frac{dy}{dx}, giving each term in its simplest form.
Answer
To find \frac{dy}{dx}, we differentiate each term of the function (y = 2x - 8\sqrt{x} + 5).
The differentiation is as follows:
- The derivative of (2x) is (2).
- Using the power rule on (-8\sqrt{x} = -8x^{1/2}):
- The derivative is (-8 \cdot \frac{1}{2}x^{-1/2} = -4x^{-1/2}) or (-\frac{4}{\sqrt{x}}).
- The derivative of the constant (+5) is (0).
So, we have:
[ \frac{dy}{dx} = 2 - \frac{4}{\sqrt{x}} ]
Step 2
Find the equation of the tangent to C at the point P, giving your answer in the form y = ax + b, where a and b are constants.
Answer
First, we substitute the x-coordinate of point P, which is (x = \frac{1}{4}), into (\frac{dy}{dx}):
[ \frac{dy}{dx} \bigg|_{x=\frac{1}{4}} = 2 - \frac{4}{\sqrt{\frac{1}{4}}} = 2 - \frac{4}{\frac{1}{2}} = 2 - 8 = -6. ]
Thus, the slope (m) of the tangent at point P is (-6).
Next, we find the coordinates of point P by substituting (x = \frac{1}{4}) into the original equation:
[ y = 2\left(\frac{1}{4}\right) - 8\sqrt{\frac{1}{4}} + 5 = \frac{1}{2} - 8(\frac{1}{2}) + 5 = \frac{1}{2} - 4 + 5 = \frac{1}{2} + 1 = \frac{3}{2}. ]
Now we have the point P as (\left(\frac{1}{4}, \frac{3}{2}\right)).
Using the point-slope form for the equation of a line: [ y - y_1 = m(x - x_1) ] Substituting the values gives: [ y - \frac{3}{2} = -6\left(x - \frac{1}{4}\right) ]
Simplifying this: [ y - \frac{3}{2} = -6x + \frac{3}{2} \quad \Rightarrow \quad y = -6x + 3. ]
Step 3
Find the coordinates of Q.
Answer
The tangent to C at point Q is parallel to the line 2x - 3y + 18 = 0.
First, we find the gradient of the given line: Rearranging gives: [ 3y = 2x + 18\quad \Rightarrow \quad y = \frac{2}{3}x + 6 ] The gradient is thus (\frac{2}{3}), which means the tangent at point Q has the same gradient: (\frac{dy}{dx} = \frac{2}{3}).
Setting (\frac{dy}{dx}) equal to (\frac{2}{3}): [ 2 - \frac{4}{\sqrt{x}} = \frac{2}{3} ]
Solving for (x):
- Add (\frac{4}{\sqrt{x}} ) to both sides: [ 2 = \frac{2}{3} + \frac{4}{\sqrt{x}} ]
- Subtract (\frac{2}{3} ) from both sides: [ 2 - \frac{2}{3} = \frac{4}{\sqrt{x}} \quad \Rightarrow \quad \frac{6}{3} - \frac{2}{3} = \frac{4}{\sqrt{x}}\quad \Rightarrow \quad \frac{4}{3} = \frac{4}{\sqrt{x}} ]
- Cross multiplying gives: [ 4\sqrt{x} = 4 \quad \Rightarrow \quad \sqrt{x} = 1 \quad \Rightarrow \quad x = 1. ]
To find the corresponding value of (y):
Substituting (x = 1) into the original equation:
[ y = 2(1) - 8\sqrt{1} + 5 = 2 - 8 + 5 = -1. ]
Thus, the coordinates of Q are ((1, -1)).