A curve C has the equation $y^3 - 3y = x^3 + 8$ - Edexcel - A-Level Maths Pure - Question 3 - 2009 - Paper 3
Question 3
A curve C has the equation $y^3 - 3y = x^3 + 8$.
(a) Find \( \frac{dy}{dx} \) in terms of $x$ and $y$.
(b) Hence find the gradient of C at the point where $y = 3... show full transcript
Worked Solution & Example Answer:A curve C has the equation $y^3 - 3y = x^3 + 8$ - Edexcel - A-Level Maths Pure - Question 3 - 2009 - Paper 3
Step 1
Find \( \frac{dy}{dx} \) in terms of $x$ and $y$
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Answer
To find ( \frac{dy}{dx} ), we will differentiate the given equation implicitly.
Starting from the equation: y3−3y=x3+8
we differentiate both sides with respect to x.
Using the product and chain rules: 3y2dxdy−3dxdy=3x2
Now, factor out ( \frac{dy}{dx} ): (3y2−3)dxdy=3x2
Next, we solve for ( \frac{dy}{dx} ): dxdy=3y2−33x2=y2−1x2.
Step 2
Hence find the gradient of C at the point where $y = 3$
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Answer
To find the gradient of the curve at the point where y=3, we will substitute ( y = 3 ) into our expression for ( \frac{dy}{dx} ).
Substituting, we compute:
dxdy=32−1x2=9−1x2=8x2
Next, we need to find the corresponding x value when y=3.
Plugging y=3 back into the original equation:
27 - 9 = x^3 + 8\
18 = x^3 + 8\
x^3 = 10\
x = \sqrt[3]{10}$$
Finally, substituting this value into the gradient:
$$\frac{dy}{dx} = \frac{(\sqrt[3]{10})^2}{8} = \frac{10^{2/3}}{8}$$
Thus, the gradient of C at the point where $y = 3$ is \( \frac{10^{2/3}}{8} \).
