A-Level Edexcel Maths Pure Differentiation: Given that $ y = 37 $ at \( x

Given that $ y = 37 $ at $ x = 4 $, find $ y $ in terms of $ x $, giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Question 9

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Given that $ y = 37 $ at $ x = 4 $, find $ y $ in terms of $ x $, giving each term in its simplest form.

Worked Solution & Example Answer:Given that $ y = 37 $ at $ x = 4 $, find $ y $ in terms of $ x $, giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Step 1

Integrate $ \frac{dy}{dx} = 6x^2 + \frac{1}{4} + \sqrt{x} $

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Answer

To find ( y ), we start by integrating the expression:

[ y = \int \left( 6x^2 + \frac{1}{4} + \sqrt{x} \right) dx ]

Performing the integration yields:

[ y = 2x^3 + \frac{1}{4}x + \frac{2}{3}x^{\frac{3}{2}} + c ]

Step 2

Substituting the initial condition

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Answer

We use the given condition ( y = 37 ) at ( x = 4 ) to find the constant ( c ):

[ 37 = 2(4)^3 + \frac{1}{4}(4) + \frac{2}{3}(4)^{\frac{3}{2}} + c ]

Simplifying each term:
[ 37 = 2(64) + 1 + \frac{32}{3} + c ]
[ 37 = 128 + 1 + \frac{32}{3} + c ]
[ 37 = 129 + \frac{32}{3} + c ]

Rearranging gives:
[ c = 37 - 129 - \frac{32}{3} ]

Calculate ( c ):
[ c = -\frac{389}{3} ]

Step 3

Final Equation for $ y $

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Answer

Substituting ( c ) back into the equation:

[ y = 2x^3 + \frac{1}{4}x + \frac{2}{3}x^{\frac{3}{2}} - \frac{389}{3} ]

This gives the final form of ( y ) in terms of ( x ).

Given that \( y = 37 \) at \( x = 4 \), find \( y \) in terms of \( x \), giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Worked Solution & Example Answer:Given that \( y = 37 \) at \( x = 4 \), find \( y \) in terms of \( x \), giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Integrate \( \frac{dy}{dx} = 6x^2 + \frac{1}{4} + \sqrt{x} \)

Substituting the initial condition

Final Equation for \( y \)

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