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The curve C has equation $$3^{x} + xy - y^{2} + 5 = 0$$ Show that $ rac{dy}{dx}$ at the point (1, 3) on the curve C can be written in the form $ rac{1}{ ho} ext{ln}( u e^{3})$, where $ ho$ and $ u$ are integers to be found. - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 1

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The-curve-C-has-equation--$$3^{x}-+-xy---y^{2}-+-5-=-0$$--Show-that-$-rac{dy}{dx}$-at-the-point-(1,-3)-on-the-curve-C-can-be-written-in-the-form-$-rac{1}{-ho}--ext{ln}(-u-e^{3})$,-where-$-ho$-and-$-u$-are-integers-to-be-found.-Edexcel-A-Level Maths Pure-Question 6-2013-Paper 1.png

The curve C has equation $$3^{x} + xy - y^{2} + 5 = 0$$ Show that $ rac{dy}{dx}$ at the point (1, 3) on the curve C can be written in the form $ rac{1}{ ho} ext{l... show full transcript

Worked Solution & Example Answer:The curve C has equation $$3^{x} + xy - y^{2} + 5 = 0$$ Show that $ rac{dy}{dx}$ at the point (1, 3) on the curve C can be written in the form $ rac{1}{ ho} ext{ln}( u e^{3})$, where $ ho$ and $ u$ are integers to be found. - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 1

Step 1

Differentiate implicitly

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Answer

To find rac{dy}{dx}, we differentiate the given equation implicitly:

rac{d}{dx}(3^{x}) + rac{d}{dx}(xy) - rac{d}{dx}(y^{2}) + rac{d}{dx}(5) = 0

Using chain rule, the derivatives are:

  1. For 3x3^{x}, the derivative is 3xextln(3)3^{x} ext{ln}(3).
  2. For xyxy, we use the product rule: x rac{dy}{dx} + y.
  3. For y2y^{2}, the derivative is 2y rac{dy}{dx}.

Thus, the differentiated equation becomes:

3^{x} ext{ln}(3) + rac{dy}{dx}(x + 2y) = 0

Step 2

Evaluate at the point (1, 3)

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Answer

Substituting (x,y)=(1,3)(x, y) = (1, 3) into the differentiated equation:

3^{1} ext{ln}(3) + rac{dy}{dx}(1 + 2(3)) = 0

This simplifies to:

3 ext{ln}(3) + rac{dy}{dx}(7) = 0

Solving for rac{dy}{dx} gives:

rac{dy}{dx} = - rac{3 ext{ln}(3)}{7}

Step 3

Express in the required form

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Answer

We need to express rac{dy}{dx} in the form rac{1}{ ho} ext{ln}( u e^{3}).

We can rewrite our expression as:

- rac{3 ext{ln}(3)}{7} = - rac{1}{7} ext{ln}(3^{3}) \ = - rac{1}{7} ext{ln}(3 imes e^{3})

Thus, we identify:

  • ho=7 ho = 7
  • u=3 u = 3

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