A-Level AQA Maths Pure Implicit Differentiation: A curve is defined by the parame

A curve is defined by the parametric equations $$x = 4t^2 + 3$$ $$y = 3t^2 - 5$$ 5 (a) Show that $$\frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}$$ 5 (b) Find the Cartesian equation of the curve in the form $$xy + ax + by = c$$, where $a$, $b$, and $c$ are integers. - AQA - A-Level Maths Pure - Question 5 - 2018 - Paper 1

Question 5

$A-curve-is-defined-by-the-parametric-equations--$$x-=-4t^2-+-3$$--$$y-=-3t^2---5$$--5-(a)-Show-that-$$\frac{dy}{dx}-=--\frac{3}{4}-\times-2^{2t}$$--5-(b)-Find-the-Cartesian-equation-of-the-curve-in-the-form-$$xy-+-ax-+-by-=-c$$,-where-$a$,-$b$,-and-$c$-are-integers.-AQA-A-Level Maths Pure-Question 5-2018-Paper 1.png$

A curve is defined by the parametric equations $$x = 4t^2 + 3$$ $$y = 3t^2 - 5$$ 5 (a) Show that $$\frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}$$ 5 (b) Find the Ca... show full transcript

Worked Solution & Example Answer:A curve is defined by the parametric equations $$x = 4t^2 + 3$$ $$y = 3t^2 - 5$$ 5 (a) Show that $$\frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}$$ 5 (b) Find the Cartesian equation of the curve in the form $$xy + ax + by = c$$, where $a$, $b$, and $c$ are integers. - AQA - A-Level Maths Pure - Question 5 - 2018 - Paper 1

Step 1

Show that $$\frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}$$

96%

114 rated

Answer

To find $\frac{dy}{dx}$ , we need to differentiate the equations with respect to $t$ .

Differentiate $y = 3t^2 - 5$ : $\frac{dy}{dt} = 6t$
Differentiate $x = 4t^2 + 3$ : $\frac{dx}{dt} = 8t$
Now, we can find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t}{8t} = \frac{3}{4}$
Therefore: $\frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}$ .

Step 2

Find the Cartesian equation of the curve in the form $$xy + ax + by = c$$

99%

104 rated

Answer

Starting with the parametric equations:

From $x = 4t^2 + 3$ , we rearrange to express $t^2$ in terms of $x$ : $t^2 = \frac{x - 3}{4}$
Substitute $t^2$ in the equation for $y$ : $y = 3t^2 - 5 = 3\left(\frac{x - 3}{4}\right) - 5$ $= \frac{3(x - 3)}{4} - 5\ = \frac{3x - 9 - 20}{4}\ = \frac{3x - 29}{4}$
Rearranging gives: $4y = 3x - 29$ or $3x - 4y - 29 = 0$ .

Thus, we can express it in the required form as: $xy + 0x + 3y = 29$ , where $a = 0, b = 3, c = 29$ .

Show that $$\frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}$$

Find the Cartesian equation of the curve in the form $$xy + ax + by = c$$

Join the A-Level students using SimpleStudy...