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A curve C has parametric equations x = 4t + 3, y = 4t + 8 + \frac{5}{2t}, \ t ≠ 0 - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 4
Question 7
A curve C has parametric equations x = 4t + 3, y = 4t + 8 + \frac{5}{2t}, \ t ≠ 0. (a) Find the value of \( \frac{dy}{dx} \) at the point on C where \( t = 2 \), gi... show full transcript
Worked Solution & Example Answer:A curve C has parametric equations x = 4t + 3, y = 4t + 8 + \frac{5}{2t}, \ t ≠ 0 - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 4
Step 1
Find the value of \( \frac{dy}{dx} \) at the point on C where \( t = 2 \)
Answer
To find ( \frac{dy}{dx} ), we will use the chain rule.
- Calculate ( \frac{dy}{dt} ) and ( \frac{dx}{dt} ):
- ( y = 4t + 8 + \frac{5}{2t} ) \n - Differentiate to find ( \frac{dy}{dt} = 4 - \frac{5}{2t^2} )
- ( x = 4t + 3 ) \n - Differentiate to find ( \frac{dx}{dt} = 4 )\n
- Now, substitute ( t = 2 ) into the derivatives:
- ( \frac{dy}{dt} = 4 - \frac{5}{2(2^2)} = 4 - \frac{5}{8} = \frac{32}{8} - \frac{5}{8} = \frac{27}{8} )
- ( \frac{dx}{dt} = 4 )\n
- Now compute ( \frac{dy}{dx} ): ( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{27/8}{4} = \frac{27}{32} )
Thus, the value of ( \frac{dy}{dx} ) at the point where ( t = 2 ) is ( \frac{27}{32} ).
Step 2
Show that the Cartesian equation of the curve C can be written in the form \( y = \frac{x^2 + ax + b}{x - 3} \)
Answer
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From the parametric equations, we have:
- ( x = 4t + 3 )
- Rearranging gives ( t = \frac{x - 3}{4} ).
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Substitute ( t ) in ( y ):
- ( y = 4\left(\frac{x - 3}{4}\right) + 8 + \frac{5}{2\left(\frac{x - 3}{4}\right)} )
- Simplifying this gives:
- ( y = (x - 3) + 8 + \frac{10}{x - 3} )
- Further simplifying:
- ( y = x + 5 + \frac{10}{x - 3} )
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To express it as required:
- Multiply everything by ( (x - 3) )
- This gives ( y(x - 3) = x^2 + 5x - 15 + 10 )
- Therefore, rearranging leads to the form ( y = \frac{x^2 + 5x - 5}{x - 3} ), identifying that ( a = 5 ) and ( b = -5 ).
Thus confirmed, the equation can be expressed in the required form.