Edexcel IAL June 2022 P4 (WMA14/01) Q4, Differentiation, Implicit Differentiation
In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
A curve has equation
\( 16x^3 – 9kx^2y + 8y^3 = 875 \)
where \( k \) is a constant.
(a) Show that
\( \frac{dy}{dx} = \frac{6kxy – 16x^2}{8y^2 – 3kx^2} \)
(4)
Given that the curve has a turning point at \( x = \frac{5}{2} \)
(b) find the value of \( k \)(4)
(b) find the value of \( k \)(4)
Solution to Part a: Deriving the Derivative \(\frac{dy}{dx}\)
Key Concepts Used:
- Implicit differentiation
Step-by-Step Working:
We need to use implicit differentiation to find \(\frac{dy}{dx}\) in terms of \( x, y, and k \).
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Differentiate Both Sides with Respect to \( x \)
Given equation:
\( 16x^3 – 9kx^2y + 8y^3 = 875 \)
Differentiate term-by-term: -
First term (\( 16x^3 \)):
\( \frac{d}{dx} (16x^3) = 48x^2 \)
-
Second term (\( -9kx^2y \)):
Apply the product rule: \(\frac{d}{dx}(uv) = u’v + uv’,\) where \(u = x^2, v = y\):
\( \frac{d}{dx}(-9kx^2y) = -9k\left(2xy + x^2\frac{dy}{dx}\right) \)
-
Third term (\( 8y^3 \)):
Apply the chain rule: \(\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}\):
\( \frac{d}{dx}(8y^3) = 24y^2\frac{dy}{dx} \)
-
Right-hand side (875):
\( \frac{d}{dx}(875) = 0 \)
-
Combined Result:
\( 48x^2 – 18kxy – 9kx^2\frac{dy}{dx} + 24y^2\frac{dy}{dx} = 0 \)
-
Collect Terms with \(\frac{dy}{dx}\)
Group terms containing \(\frac{dy}{dx}\):
\( -9kx^2\frac{dy}{dx} + 24y^2\frac{dy}{dx} = 18kxy – 48x^2 \)
-
Factor out \(\frac{dy}{dx}\):
\( \frac{dy}{dx}(24y^2 – 9kx^2) = 18kxy – 48x^2 \)
-
Solve for \(\frac{dy}{dx}\)
Divide both sides by 3 to simplify:
\( \frac{dy}{dx}(8y^2 – 3kx^2) = 6kxy – 16x^2 \)
\( \frac{dy}{dx} = \frac{6kxy – 16x^2}{8y^2 – 3kx^2} \)
Final Answer:
\( \frac{dy}{dx} = \frac{6kxy – 16x^2}{8y^2 – 3kx^2} \)
Solution to Part b: Value of Constant k for Turning Point at \( x = \frac{5}{2} \)
Key Concepts Used:
- Implicit differentiation
- Turning point condition \( \frac{dy}{dx} = 0 \)
Step-by-Step Working:
Let’s determine the value of k given a turning point at \( x = \frac{5}{2} \):
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Understand Turning Points
At a turning point:
\( \frac{dy}{dx} = 0 \)
From part (a), set numerator equal to 0:
\( 6kxy – 16x^2 = 0 \)
-
Substitute \( x = \frac{5}{2} \)
\( 6k \left( \frac{5}{2} \right) y – 16 \left( \frac{5}{2} \right)^2 = 0 \)
Simplify:
\( 15ky – 100 = 0 \)
Solve for y:
\( y = \frac{100}{15k} = \frac{20}{3k} \)
-
Substitute \( (x, y) \) into Original Equation
Original equation at \( x = \frac{5}{2} \):
\( 16 \left( \frac{5}{2} \right)^3 – 9k \left( \frac{5}{2} \right)^2 y + 8y^3 = 875 \)
Simplify x-terms:
\( 16 \times \frac{125}{8} – 9k \times \frac{25}{4} y + 8y^3 = 875 \)
\( 250 – \frac{225k}{4} y + 8y^3 = 875 \)
-
Substitute \( y = \frac{20}{3k} \)
\( 250 – \frac{225k}{4} \left( \frac{20}{3k} \right) + 8 \left( \frac{20}{3k} \right)^3 = 875 \)
Simplify:
Second term:
\( \frac{225k}{4} \times \frac{20}{3k} = \frac{4500}{12} = 375 \)
Third term:
\( 8 \times \frac{8000}{27k^3} = \frac{64000}{27k^3} \)
Equation becomes:
\( 250 – 375 + \frac{64000}{27k^3} = 875 \)
\( -125 + \frac{64000}{27k^3} = 875 \)
\( \frac{64000}{27k^3} = 1000 \)
-
Solve for k
\( 64000 = 27000k^3 \)
\( k^3 = \frac{64000}{27000} = \frac{64}{27} \)
\( k = \frac{64}{\sqrt{27}} = \frac{4}{3} \)
Final Answer:
\( k = \frac{4}{3} \)