A-Level Mathematics May/June 2024 Q4: The equation of a curve is ye^(2x) + y²e^x = 6. Find the gradient of the curve at the p…

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  The equation of the curve is given by $ye^{2x} + y^2e^x = 6$.

  Step 1: Find the value of $e^x$ at the point where $y=1$.

  Substitute $y=1$ into the equation: $(1)e^{2x} + (1)^2e^x = 6$ $e^{2x} + e^x - 6 = 0$

  This is a quadratic equation in terms of $e^x$. Let $u = e^x$. The equation becomes: $u^2 + u - 6 = 0$ Factorising the quadratic: $(u+3)(u-2) = 0$ So, $u = -3$ or $u = 2$.

  Substituting back $u = e^x$: $e^x = -3$ or $e^x = 2$. Since $e^x > 0$ for all real $x$, the solution $e^x = -3$ is not possible. Therefore, we must have $e^x = 2$.

  Step 2: Differentiate the equation implicitly with respect to $x$.

  We differentiate $ye^{2x} + y^2e^x = 6$ term by term.

  For the first term, $ye^{2x}$, we use the product rule: $\frac{d}{dx}(ye^{2x}) = y \cdot \frac{d}{dx}(e^{2x}) + e^{2x} \cdot \frac{d}{dx}(y) = y(2e^{2x}) + e^{2x} \frac{dy}{dx} = 2ye^{2x} + e^{2x} \frac{dy}{dx}$

  For the second term, $y^2e^x$, we use the product rule: $\frac{d}{dx}(y^2e^x) = y^2 \cdot \frac{d}{dx}(e^x) + e^x \cdot \frac{d}{dx}(y^2) = y^2e^x + e^x (2y \frac{dy}{dx}) = y^2e^x + 2ye^x \frac{dy}{dx}$

  The derivative of the right-hand side is $\frac{d}{dx}(6) = 0$.

  Combining these gives the full differentiated equation: $(2ye^{2x} + e^{2x} \frac{dy}{dx}) + (y^2e^x + 2ye^x \frac{dy}{dx}) = 0$

  Step 3: Substitute known values and solve for the gradient, $\frac{dy}{dx}$.

  At the point in question, we have $y=1$ and $e^x=2$. We also need $e^{2x} = (e^x)^2 = 2^2 = 4$.

  Substitute these values into the differentiated equation: $2(1)(4) + (4)\frac{dy}{dx} + (1)^2(2) + 2(1)(2)\frac{dy}{dx} = 0$ $8 + 4\frac{dy}{dx} + 2 + 4\frac{dy}{dx} = 0$

  Now, collect terms: $10 + 8\frac{dy}{dx} = 0$ $8\frac{dy}{dx} = -10$ $\frac{dy}{dx} = -\frac{10}{8}$ $\frac{dy}{dx} = -\frac{5}{4}$

  The gradient of the curve at the point where $y=1$ is $-1.25$.

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  How the marks are awarded

  • M1 — Substituting y=1 into the curve's equation to form a quadratic equation in e^x, which is e^(2x) + e^x - 6 = 0, and attempting to solve it.
  • A1 — Correctly solving the quadratic to find the only valid solution, e^x = 2, and rejecting the negative solution.
  • B1 — Correctly differentiating the first term, ye^(2x), using the product rule to get 2ye^(2x) + e^(2x) dy/dx.
  • B1 — Correctly differentiating the second term, y²e^x, using the product rule and chain rule to get y²e^x + 2ye^x dy/dx.
  • M1 — Substituting the known values (y=1, e^x=2, and e^(2x)=4) into the full differentiated equation (which must be equated to zero) and attempting to make dy/dx the subject.
  • A1 — Obtaining the final correct gradient, dy/dx = -5/4 or -1.25, from correct preceding work.

  Common mistakes

  • Errors in applying the product rule, for example differentiating ye^(2x) to (dy/dx) * (2e^(2x)) or forgetting to differentiate one of the factors.
  • Incorrectly applying the chain rule when differentiating y², resulting in 2y instead of the correct 2y(dy/dx).
  • Failing to reject the e^x = -3 solution after solving the quadratic, as e^x must be positive.
  • Making algebraic errors when isolating dy/dx after substitution, such as sign errors or incorrect grouping of terms.

  Examiner tip: When finding a numerical gradient on an implicitly defined curve, always find the coordinates of the point first before differentiating, then substitute these values immediately after differentiating to simplify the algebra.

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