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The curve C has equation y = \frac{1}{2}x^3 - 9x^2 + \frac{8}{x} + 30, \quad x > 0 (a) Find \frac{dy}{dx} - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 2
Question 5
The curve C has equation y = \frac{1}{2}x^3 - 9x^2 + \frac{8}{x} + 30, \quad x > 0 (a) Find \frac{dy}{dx}. (b) Show that the point P(4, -8) lies on C. (c) Find ... show full transcript
Worked Solution & Example Answer:The curve C has equation y = \frac{1}{2}x^3 - 9x^2 + \frac{8}{x} + 30, \quad x > 0 (a) Find \frac{dy}{dx} - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 2
Step 1
Find \frac{dy}{dx}.
Answer
To find \frac{dy}{dx}, we need to differentiate the equation:
- For the term (\frac{1}{2}x^3): (\frac{d}{dx}(\frac{1}{2}x^3) = \frac{3}{2}x^2).
- For the term (-9x^2): (\frac{d}{dx}(-9x^2) = -18x).
- For the term (\frac{8}{x}): this can be rewritten as (8x^{-1}), so (\frac{d}{dx}(8x^{-1}) = -8x^{-2}).
- The derivative of the constant 30 is 0.
Combining these results, we have:
Step 2
Show that the point P(4, -8) lies on C.
Answer
To verify that the point P(4, -8) lies on C, we substitute (x = 4) into the equation:
(y = \frac{1}{2}(4)^3 - 9(4)^2 + \frac{8}{4} + 30)
Calculating this:
- (\frac{1}{2}(64) = 32)
- (-9(16) = -144)
- (\frac{8}{4} = 2)
- Then add: (32 - 144 + 2 + 30 = -80)
Since the calculated y-value (-80) does not equal -8, point P(4, -8) does not lie on curve C.
Step 3
Find an equation of the normal to C at the point P.
Answer
First, substitute (x = 4) into the derivative to find the gradient at P:
- Calculate (\frac{dy}{dx}) at (x = 4): (\frac{dy}{dx} = \frac{3}{2}(4)^2 - 18(4) - \frac{8}{(4)^2}) (= \frac{3}{2}(16) - 72 - \frac{8}{16}) (= 24 - 72 - 0.5 = -48.5)
The gradient of the normal is the negative reciprocal: (m_{normal} = -\frac{1}{-48.5} = \frac{1}{48.5})
Using point-slope form of the line equation: (y - y_1 = m(x - x_1))
Substituting P(4, -8): (y - (-8) = \frac{1}{48.5}(x - 4)) (y + 8 = \frac{1}{48.5}(x - 4)) Multiplying through by 48.5: (48.5y + 388 = x - 4) Rearranging: (x - 48.5y - 392 = 0) Thus: (1x - 48.5y - 392 = 0).
To convert to integer coefficients, multiply through by 2: (2x - 97y - 784 = 0).