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The curve C has equation y = \frac{1}{2}x^3 - 9x^2 + \frac{8}{x} + 30, x > 0 (a) Find \frac{dy}{dx} - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 1
Question 2
The curve C has equation y = \frac{1}{2}x^3 - 9x^2 + \frac{8}{x} + 30, x > 0 (a) Find \frac{dy}{dx}. (4) (b) Show that the point P(4, -8) lies on C. (2) (c) Fin... show full transcript
Worked Solution & Example Answer:The curve C has equation y = \frac{1}{2}x^3 - 9x^2 + \frac{8}{x} + 30, x > 0 (a) Find \frac{dy}{dx} - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 1
Step 1
Find \frac{dy}{dx}
Answer
To find \frac{dy}{dx}, we differentiate the given equation. Using the power rule and quotient rule, we get:
- For the term \frac{1}{2}x^3, the derivative is \frac{3}{2}x^2.
- For the term -9x^2, the derivative is -18x.
- For the term \frac{8}{x}, we can rewrite this as 8x^{-1}. Thus, the derivative is -8x^{-2}.
Combining these results, we have:
Step 2
Show that the point P(4, -8) lies on C
Answer
To check if 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 ] [ y = \frac{1}{2}(64) - 9(16) + 2 + 30 ] [ y = 32 - 144 + 2 + 30 = -8 ]
Since the calculated y-value is -8, the point P(4, -8) indeed lies on C.
Step 3
Find an equation of the normal to C at the point P
Answer
First, we calculate the gradient of the tangent at P(4, -8) using \frac{dy}{dx}:
Substituting x = 4:
- (\frac{dy}{dx} = \frac{3}{2}(4^2) - 18(4) - \frac{8}{(4)^2} = 24 - 72 - \frac{1}{2} = -48.5)
The gradient of the normal is the negative reciprocal:
[ \text{Gradient of Normal} = -\frac{1}{\left(-48.5\right)} = \frac{1}{48.5} ]
Now we can use the point-slope form of a line:
[ y - (-8) = \frac{1}{48.5}(x - 4) ]
Rearranging into the form ax + by + c = 0 results in:
[ 7y - 2x + 64 = 0 ]
where a = -2, b = 7, and c = 64.