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 $t = 2$, giving your answer as a fraction in its simplest form. (b) Show that the cartesian equation of the curve C can be written in the form $y = \frac{x^2 + ax + b}{x - 3}, \: x \neq 3$ where $a$ and $b$ are integers to be determined.

A-Level Edexcel Maths Pure Exponential & Logarithms: A curve C has parametric equatio

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

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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 $t = 2$, ... 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 $t = 2$

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Answer

To find $\frac{dy}{dx}$ , we will first calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$ .

Calculate $\frac{dx}{dt}$ : $\frac{dx}{dt} = \frac{d}{dt}(4t + 3) = 4.$
Calculate $\frac{dy}{dt}$ : $\frac{dy}{dt} = \frac{d}{dt}\left(4t + 8 + \frac{5}{2t}\right) = 4 - \frac{5}{2t^2}.$
Substituting $t = 2$ gives: $\frac{dy}{dt}\bigg|_{t=2} = 4 - \frac{5}{2(2^2)} = 4 - \frac{5}{8} = \frac{32}{8} - \frac{5}{8} = \frac{27}{8}.$
Now, substitute these into the formula for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{27}{8}}{4} = \frac{27}{32}.$

Thus, the value of $\frac{dy}{dx}$ at the point where $t = 2$ is $\frac{27}{32}$ .

Step 2

Show that the cartesian equation can be written in the given form

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Answer

We start with the parametric equations:

$x = 4t + 3$ ,

From $x = 4t + 3$ , we can express $t$ in terms of $x$ :

$t = \frac{x - 3}{4}.$

Now, substitute $t$ into the equation for $y$ :

$y = 4t + 8 + \frac{5}{2t} = 4\left(\frac{x - 3}{4}\right) + 8 + \frac{5}{2 \left(\frac{x - 3}{4}\right)}.$

This simplifies to:

$y = (x - 3) + 8 + \frac{5 \cdot 4}{2(x - 3)} = x + 5 + \frac{10}{x - 3}.$

Thus, rearranging gives:

$y - 5 = x + \frac{10}{x - 3} \\ y - 5 = \frac{x(x - 3) + 10}{x - 3} = \frac{x^2 - 3x + 10}{x - 3}.$

Therefore, we can write:

$y = \frac{x^2 - 3x + 10}{x - 3}.$

Comparing this with the form $y = \frac{x^2 + ax + b}{x - 3}$ , we find that:

$a = -3$
$b = 10$ .

Therefore, $a$ and $b$ are integers, and we have verified the required equation.

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

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

Find the value of $\frac{dy}{dx}$ at the point on C where $t = 2$

Show that the cartesian equation can be written in the given form

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