A-Level Mathematics May/June 2024 Q5: The equation of a curve is y = e^(sinx) / cos²x for 0 ≤ x ≤ 2π. Find dy/dx and hence fi…

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  The equation of the curve is given by $y = \frac{e^{\sin x}}{\cos^2 x}$ for $0 \le x \le 2\pi$.

  To find $\frac{dy}{dx}$, we use the quotient rule, $\frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$. Let $u = e^{\sin x}$ and $v = \cos^2 x$.

  Using the chain rule: $\frac{du}{dx} = e^{\sin x} \cdot \frac{d}{dx}(\sin x) = e^{\sin x} \cos x$. $\frac{dv}{dx} = 2(\cos x) \cdot \frac{d}{dx}(\cos x) = 2\cos x (-\sin x) = -2\sin x \cos x$.

  Now, applying the quotient rule: $\frac{dy}{dx} = \frac{(\cos^2 x)(e^{\sin x} \cos x) - (e^{\sin x})(-2\sin x \cos x)}{(\cos^2 x)^2}$ $\frac{dy}{dx} = \frac{e^{\sin x} \cos^3 x + 2e^{\sin x} \sin x \cos x}{\cos^4 x}$

  To find the stationary points, we set $\frac{dy}{dx} = 0$. $\frac{e^{\sin x} \cos^3 x + 2e^{\sin x} \sin x \cos x}{\cos^4 x} = 0$

  For the fraction to be zero, the numerator must be zero: $e^{\sin x} \cos^3 x + 2e^{\sin x} \sin x \cos x = 0$

  Factor out the common terms $e^{\sin x} \cos x$: $e^{\sin x} \cos x (\cos^2 x + 2\sin x) = 0$

  This gives three possibilities: $e^{\sin x} = 0$, $\cos x = 0$, or $\cos^2 x + 2\sin x = 0$.

  1. $e^{\sin x} = 0$: This has no solution as $e^z > 0$ for all real $z$.
  2. $\cos x = 0$: If $\cos x = 0$, the original function $y$ is undefined, so these are not stationary points on the curve.
  3. $\cos^2 x + 2\sin x = 0$: We must solve this equation.

  Use the identity $\cos^2 x = 1 - \sin^2 x$ to get an equation solely in terms of $\sin x$. $(1 - \sin^2 x) + 2\sin x = 0$ $\sin^2 x - 2\sin x - 1 = 0$

  This is a quadratic equation in $\sin x$. Let $s = \sin x$, so $s^2 - 2s - 1 = 0$. Using the quadratic formula, $s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $\sin x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$ $\sin x = \frac{2 \pm \sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$.

  We have two cases: Case 1: $\sin x = 1 + \sqrt{2} \approx 2.414$. This has no solution, as $-1 \le \sin x \le 1$. Case 2: $\sin x = 1 - \sqrt{2} \approx -0.4142$.

  Let the basic angle be $\alpha = \arcsin(\sqrt{2}-1) \approx \arcsin(0.4142) \approx 0.4271$ radians. Since $\sin x$ is negative, the solutions for $x$ in the interval $0 \le x \le 2\pi$ are in the third and fourth quadrants.

  Third quadrant solution: $x = \pi + \alpha = \pi + 0.4271... \approx 3.5687...$ Fourth quadrant solution: $x = 2\pi - \alpha = 2\pi - 0.4271... \approx 5.8560...$

  Rounding to 3 significant figures, the x-coordinates of the stationary points are: $x = 3.57$ and $x = 5.86$.

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  How the marks are awarded

  • M1 — Correctly applying the structure of the quotient rule to the function, with an attempt to differentiate the numerator and denominator.
  • A1 — Obtaining the correct, unsimplified derivative: $\frac{dy}{dx} = \frac{(\cos^2 x)(e^{\sin x} \cos x) - (e^{\sin x})(-2\sin x \cos x)}{(\cos^2 x)^2}$. The derivatives of both numerator and denominator must be correct.
  • DM1 — Setting the numerator of the calculated derivative equal to zero to find the stationary points. This mark is dependent on having earned the first M1 mark.
  • DM1 — Using the identity $\cos^2 x = 1 - \sin^2 x$ to transform the equation into a quadratic equation in terms of a single trigonometric function, $\sin x$.
  • M1 — Demonstrating a complete method to solve the three-term quadratic equation for $\sin x$, typically by using the quadratic formula.
  • A1 — Obtaining one correct solution for $x$ from the valid value of $\sin x$. For example, $x = 3.57$ (or $x=5.86$).
  • A1FT — Obtaining the second correct solution for $x$ and ensuring no other incorrect solutions are given within the specified interval. This mark can be awarded on a follow-through basis from a slightly incorrect quadratic.

  Common mistakes

  • Incorrectly applying the chain rule when differentiating $e^{\sin x}$ (e.g., getting just $e^{\sin x}$) or $\cos^2 x$ (e.g., getting $-2\sin x$).
  • Making an algebraic error when simplifying the derivative, such as incorrectly cancelling terms from the numerator and denominator.
  • After setting the numerator to zero, failing to factorise correctly or incorrectly concluding that $\cos x = 0$ gives stationary points, without checking that the original function is undefined there.
  • When solving for $x$ from $\sin x = 1 - \sqrt{2}$, finding only the principal value or finding solutions in the incorrect quadrants for the given interval $0 \le x \le 2\pi$.

  Examiner tip: This question demonstrates that complex problems are often a sequence of familiar steps; master the chain rule, quotient rule, and solving trigonometric quadratics individually so you can confidently combine them.

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