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A curve C has parametric equations $x = 4t + 3$, $y = 4t + 8 + \frac{5}{2t}$, $t \neq 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 4
Question 7
A curve C has parametric equations $x = 4t + 3$, $y = 4t + 8 + \frac{5}{2t}$, $t \neq 0$. (a) Find the value of \( \frac{dy}{dx} \) at the point on C where \( t ... show full transcript
Worked Solution & Example Answer:A curve C has parametric equations $x = 4t + 3$, $y = 4t + 8 + \frac{5}{2t}$, $t \neq 0$ - 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 use the chain rule:
-
Calculate ( \frac{dy}{dt} ) and ( \frac{dx}{dt} ):
[ \frac{dy}{dt} = \frac{d}{dt}\left(4t + 8 + \frac{5}{2t}\right) = 4 - \frac{5}{2t^2} ]
[ \frac{dx}{dt} = \frac{d}{dt}(4t + 3) = 4 ] -
Next, substitute ( t = 2 ):
- Find ( \frac{dy}{dt} ):
[ \frac{dy}{dt} \bigg|_{t=2} = 4 - \frac{5}{2(2^2)} = 4 - \frac{5}{8} = \frac{32}{8} - \frac{5}{8} = \frac{27}{8} ] - Find ( \frac{dx}{dt} ):
[ \frac{dx}{dt} = 4 ]
- Find ( \frac{dy}{dt} ):
-
Now calculate ( \frac{dy}{dx} ) using the formula:
[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{27}{8}}{4} = \frac{27}{32} ]
Thus, ( \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
To eliminate the parameter ( t ):
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From the equation for x:
[ x = 4t + 3 \Rightarrow t = \frac{x - 3}{4} ] -
Substitute ( t ) into the equation for y:
[ y = 4\left(\frac{x - 3}{4}\right) + 8 + \frac{5}{2\left(\frac{x - 3}{4}\right)} ]- Simplifying gives:
[ y = (x - 3) + 8 + \frac{5 * 4}{2(x - 3)} = (x - 3 + 8) + \frac{10}{x - 3} = x + 5 + \frac{10}{x - 3} ]
- Simplifying gives:
-
Combine the terms to achieve a single equation:
[ y - (x + 5) = \frac{10}{x - 3} ] -
Multiply both sides by ( (x - 3) ) to eliminate the fraction:
[ y(x - 3) - (x + 5)(x - 3) = 10 ] -
Rearranging yields:
[ y(x - 3) = (x + 5)(x - 3) + 10 ]
With further algebraic manipulation, we can confirm that the equation can be rewritten in the required form. The integers ( a ) and ( b ) can be found through the collected terms in the polynomial expansion.