Home / IB Mathematics AHL 5.15 Slope fields and their diagrams AI HL Paper 1

IB Mathematics AHL 5.15 Slope fields and their diagrams AI HL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AI HL Paper 1- All Topics

Question

The slope field illustrated below represents the differential equation:
$$\frac{dy}{dx} = x^4 + 4x^2 y + 4y^2$$
Slope Field
(a) On the provided slope field, sketch the specific solution curve that passes through the point $(0, 1)$.
Certain solutions possess points where the gradient is zero. All such points are located on a curve defined by $y = ax^2$, where $a$ is a constant.
(b) Sketch this parabolic curve on the slope field.
(c) Determine the value of $a$ by analyzing the differential equation.

Most-appropriate topic codes (IB Mathematics: Applications and Interpretation HL):

• AHL 5.15: Slope fields and their diagrams — parts (a), (b)
• AHL 5.14: Setting up a model/differential equation from a context— part (c)
▶️ Answer/Explanation

(a)
Draw a smooth curve starting at the point $(0, 1)$.
Ensure the curve follows the direction of the slope segments at every point.
The resulting curve decreases as $x$ increases, crossing the $x$-axis at approximately $x \approx -0.9$.

Solution Curve

(b)
Sketch the parabola $y = ax^2$ where $a$ is negative.
The curve passes through the origin and opens downwards.
This curve identifies the locations where the slope segments in the field are perfectly horizontal.

Parabola of Zero Gradients

(c)
To find where the gradient is zero, set $\frac{dy}{dx} = 0$:
$0 = x^4 + 4x^2 y + 4y^2$
Recognize the expression as a perfect square trinomial:
$0 = (x^2 + 2y)^2$
Take the square root of both sides:
$0 = x^2 + 2y$
Isolate $y$ to find the equation of the curve:
$2y = -x^2$
$y = -\frac{1}{2} x^2$
Comparing this result to the given form $y = ax^2$, we find:
$\boxed{a = -\frac{1}{2}}$

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