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The gradient of a curve C is given by $$\frac{dy}{dx} = \frac{(x^2 + 3)^3}{x^3}, \quad x \neq 0.$$ (a) Show that $$\frac{dy}{dx} = x^2 + 6 + 9x^{-2}.$$ The point (3, 20) lies on C - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 1
Question 1
The gradient of a curve C is given by $$\frac{dy}{dx} = \frac{(x^2 + 3)^3}{x^3}, \quad x \neq 0.$$ (a) Show that $$\frac{dy}{dx} = x^2 + 6 + 9x^{-2}.$$ The poin... show full transcript
Worked Solution & Example Answer:The gradient of a curve C is given by $$\frac{dy}{dx} = \frac{(x^2 + 3)^3}{x^3}, \quad x \neq 0.$$ (a) Show that $$\frac{dy}{dx} = x^2 + 6 + 9x^{-2}.$$ The point (3, 20) lies on C - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 1
Step 1
Show that $$\frac{dy}{dx} = x^2 + 6 + 9x^{-2}.$$
Answer
To show that we start with expanding ((x^2 + 3)^3):
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Expand:
[(x^2 + 3)^3 = (x^2)^3 + 3 \cdot (x^2)^2 \cdot 3 + 3 \cdot x^2 \cdot 3^2 + 3^3 = x^6 + 9x^4 + 27x^2 + 27.] -
Divide by (x^3):
[\frac{(x^2 + 3)^3}{x^3} = \frac{x^6 + 9x^4 + 27x^2 + 27}{x^3} = x^3 + 9x + \frac{27}{x} + \frac{27}{x^3}.] -
Rewrite:
Therefore:
[\frac{dy}{dx} = x^2 + 6 + 9x^{-2}.]
This matches the form required.
Step 2
Find an equation for the curve C in the form $y = f(x)$.
Answer
Given that the point (3, 20) lies on the curve, we can find the equation as follows:
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Integrate:
We start with (\frac{dy}{dx} = x^2 + 6 + 9x^{-2}.)
Integrating both sides gives:
[y = \int (x^2 + 6 + 9x^{-2}) , dx = \frac{x^3}{3} + 6x - 9x^{-1} + c.] -
Apply the point (3, 20):
Substituting in (x = 3) and (y = 20):
[20 = \frac{3^3}{3} + 6(3) - 9(\frac{1}{3}) + c.]
This simplifies to:
[20 = 9 + 18 - 3 + c]
[20 = 24 + c \Rightarrow c = -4.] -
Final equation:
Substituting (c) back into the equation, we get:
[y = \frac{x^3}{3} + 6x - 9x^{-1} - 4.]