IB DP Math AA Topic: AHL 5.14 : Implicit differentiation: IB Style Questions HL Paper 2

Detailed Solution

 (a) Show that \( \frac{dy}{dx} = \frac{2x – e^{x+y}}{e^{x+y} – 2y} \)

To find the derivative \( \frac{dy}{dx} \), we use implicit differentiation on the given equation \( e^{x+y} = x^2 + y^2 \).

1. Differentiate both sides with respect to \( x \):
Left side: \( e^{x+y} \)
Using the chain rule, \( \frac{d}{dx} e^{x+y} = e^{x+y} \cdot \frac{d}{dx}(x + y) \).
Since \( y \) is a function of \( x \), \( \frac{d}{dx}(x + y) = 1 + \frac{dy}{dx} \).
So, the derivative of the left side is \( e^{x+y} (1 + \frac{dy}{dx}) \).
Right side: \( x^2 + y^2 \)
\( \frac{d}{dx}(x^2) = 2x \).
\( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \) (using the chain rule).
So, the derivative of the right side is \( 2x + 2y \frac{dy}{dx} \).

2. Set the derivatives equal:
\[
e^{x+y} \left(1 + \frac{dy}{dx}\right) = 2x + 2y \frac{dy}{dx}
\]

3. Solve for \( \frac{dy}{dx} \):
Distribute on the left: \( e^{x+y} + e^{x+y} \frac{dy}{dx} = 2x + 2y \frac{dy}{dx} \).
Bring all terms involving \( \frac{dy}{dx} \) to one side:
\[
e^{x+y} \frac{dy}{dx} – 2y \frac{dy}{dx} = 2x – e^{x+y}
\]
 Factor out \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} (e^{x+y} – 2y) = 2x – e^{x+y}
\]
 Solve for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{2x – e^{x+y}}{e^{x+y} – 2y}
\]

This matches the given expression, so part (a) is complete.

(b) (i) Show that the x-coordinates of points P and Q satisfy the equation \( 2x^2 + (\ln(2x))^2 – 2x \ln(2x) – 2x = 0 \)

Points \( P \) and \( Q \) have horizontal tangents, meaning \( \frac{dy}{dx} = 0 \).

1. Set the derivative to zero:
\[
\frac{2x – e^{x+y}}{e^{x+y} – 2y} = 0
\]
 For the fraction to be zero, the numerator must be zero (assuming the denominator is non-zero):
\[
2x – e^{x+y} = 0 \implies e^{x+y} = 2x
\]

2. Since $e^{x+y} = 2x$ and given two points on the curve C where the tangent is horizontal are P and Q. ,

we have $x+y = \ln(2x)$. Then $y = \ln(2x) – x$.

Substitute $y = \ln(2x) – x$ into $y^2 = 2x – x^2$.
$(\ln(2x) – x)^2 = 2x – x^2$
$(\ln(2x))^2 – 2x\ln(2x) + x^2 = 2x – x^2$
$2x^2 + (\ln(2x))^2 – 2x\ln(2x) – 2x = 0$

(b) (ii) Hence, find the coordinates of \( P \) and \( Q \).

Solve $2x^2 + (\ln(2x))^2 – 2x\ln(2x) – 2x = 0$ numerically. Approximate solutions are 

x = 0.331 or 1.84

Find the corresponding $y$ values using $y = \ln(2x) – x$.

y = -0.743332…
y = -0.538335…

Coordinates of P and Q are approximately (0.331,-0.743) and (1.84,-0.538)

(c) Using the line of symmetry,  the coordinates of the points on the curve \( C \) where the tangent is vertical.

Since the curve is symmetric about $y=x$, the points where the tangent is vertical will swap and are given as (-0.743,0.331) and (-0.538,1.84)

(d)

 For a gradient of -1, $\frac{2x – e^{x+y}}{e^{x+y} – 2y} = -1$. This simplifies to $2x – e^{x+y} = -e^{x+y} + 2y$, so $2x = 2y$, thus $x = y$.

 Substitute $x = y$ into $e^{x+y} = x^2 + y^2$.
$e^{2x} = 2x^2$
Solving numerically gives $x \approx 0.96$. Since $x=y$, the coordinates are approximately 

$x = -0.451$
$y = -0.451$

……………..Markscheme………………….

Solution: –

(a) attempt to use implicit differentiation

$(1+\frac{dy}{dx})e^{x+y}=2x+2y\frac{dy}{dx}$

attempt to expand brackets and collect $\frac{dy}{dx}$ terms on the same side
$e^{x+y}\frac{dy}{dx}-2y\frac{dy}{dx}=2x-e^{x+y}$ OR $e^{x+y}\frac{dy}{dx}-2y\frac{dy}{dx}+e^{x+y}=2x$ OR $(e^{x+y}-2y)\frac{dy}{dx}=2x-e^{x+y}$
$\frac{dy}{dx}=\frac{2x-e^{x+y}}{e^{x+y}-2y}$

(b) (i) recognition that $2x-e^{x+y}=0$ at P and Q
$\Rightarrow x+y=ln(2x)$

recognition that $e^{x+y}=x^{2}+y^{2}$ at P and Q
$\Rightarrow 2x=x^{2}+(ln(2x)-x)^{2}$ or equivalent
$2x^{2}+(ln(2x))^{2}-2x~ln(2x)-2x=0$

(b) (ii) x = 0.331077… or 1.84273…
x = 0.331 or 1.84

attempt to use x+y=ln(2x) OR $e^{x+y}=x^{2}+y^{2}$ to find y-coordinates
y = -0.743332…
y = -0.538335…

coordinates (0.331,-0.743) and (1.84,-0.538)

(c) coordinates (-0.743,0.331) and (-0.538,1.84)

(d) setting $\frac{2x-e^{x+y}}{e^{x+y}-2y}=-1$  OR  recognition that the point lies on line of symmetry
$2x-e^{x+y}=2y-e^{x+y}$ 
$y=x$

attempt to substitute $y=x$ into $e^{x+y}=x^{2}+y^{2}$
$e^{2x}=2x^{2}$ OR $e^{2y}=2y^{2}$

$x = -0.451$
$y = -0.451$

coordinates $(-0.451,-0.451)$