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The curve C has equation $$ y = 12igg( ext{√}(x)igg) - x^{ rac{3}{2}} - 10, \, x > 0 $$ (a) Use calculus to find the coordinates of the turning point on C - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 4

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Question 1

The-curve-C-has-equation---$$-y-=-12igg(-ext{√}(x)igg)---x^{-rac{3}{2}}---10,-\,-x->-0-$$--(a)-Use-calculus-to-find-the-coordinates-of-the-turning-point-on-C-Edexcel-A-Level Maths Pure-Question 1-2009-Paper 4.png

The curve C has equation $$ y = 12igg( ext{√}(x)igg) - x^{ rac{3}{2}} - 10, \, x > 0 $$ (a) Use calculus to find the coordinates of the turning point on C. (b)... show full transcript

Worked Solution & Example Answer:The curve C has equation $$ y = 12igg( ext{√}(x)igg) - x^{ rac{3}{2}} - 10, \, x > 0 $$ (a) Use calculus to find the coordinates of the turning point on C - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 4

Step 1

Use calculus to find the coordinates of the turning point on C.

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Answer

To find the turning point, we first need to differentiate the function:

  1. Differentiate the equation:
    dydx=12⋅12x−12−32x12\frac{dy}{dx} = 12 \cdot \frac{1}{2} x^{-\frac{1}{2}} - \frac{3}{2} x^{\frac{1}{2}}
    Simplifying this gives: dydx=6x−32x\frac{dy}{dx} = \frac{6}{\sqrt{x}} - \frac{3}{2} \sqrt{x}.

  2. Set the derivative equal to zero to find critical points: 6x−32x=0\frac{6}{\sqrt{x}} - \frac{3}{2} \sqrt{x} = 0.

    To solve this, multiply through by (\sqrt{x}):
    6−32x=06 - \frac{3}{2} x = 0
    Rearranging gives:
    32x=6\frac{3}{2} x = 6, hence
    x=4x = 4.

  3. Calculate the corresponding y-coordinate by substituting x back into the original equation: y=12(4)−432−10y = 12(\sqrt{4}) - 4^{\frac{3}{2}} - 10
    which simplifies to:
    y=12(2)−8−10=24−8−10=6y = 12(2) - 8 - 10 = 24 - 8 - 10 = 6.

Thus, the turning point is ((4, 6)).

Step 2

Find $$ \frac{d^2 y}{dx^2} $$.

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Answer

To find the second derivative, we differentiate the first derivative:

  1. The first derivative is: dydx=6x−12−32x12\frac{dy}{dx} = 6x^{-\frac{1}{2}} - \frac{3}{2} x^{\frac{1}{2}}.

  2. Differentiate again: d2ydx2=−32x−32+34x−12\frac{d^2y}{dx^2} = -\frac{3}{2} x^{-\frac{3}{2}} + \frac{3}{4} x^{-\frac{1}{2}}, which simplifies to
    d2ydx2=−32x32+34x\frac{d^2y}{dx^2} = -\frac{3}{2 x^{\frac{3}{2}}} + \frac{3}{4 \sqrt{x}}.

Step 3

State the nature of the turning point.

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Answer

To determine the nature of the turning point at (x = 4), we evaluate the second derivative:

Substituting (x = 4) into the second derivative: d2ydx2=−32(4)32+344\frac{d^2y}{dx^2} = -\frac{3}{2 (4)^{\frac{3}{2}}} + \frac{3}{4 \sqrt{4}}.

Calculating gives:
=−32⋅8+34⋅2=−316+38=−316+616=316>0= -\frac{3}{2 \cdot 8} + \frac{3}{4 \cdot 2} = -\frac{3}{16} + \frac{3}{8} = -\frac{3}{16} + \frac{6}{16} = \frac{3}{16} > 0.

Since the second derivative is positive, the turning point at ((4, 6)) is a local maximum.

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