Applications of Differentiation (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Finding a Tangent to $y = 2\sin x$

Find the equation of the tangent to $y = 2\sin x$ at the point where $x = \frac{\pi}{6}$.

Step 1: Find the coordinates of the point.

When $x = \frac{\pi}{6}$:

$y = 2\sin\frac{\pi}{6} = 2 \times \frac{1}{2} = 1$

The point is $P\left(\frac{\pi}{6}, 1\right)$.

Step 2: Find the gradient by differentiating.

$\frac{dy}{dx} = 2\cos x$

Step 3: Evaluate the gradient at the point.

When $x = \frac{\pi}{6}$:

$\frac{dy}{dx} = 2\cos\frac{\pi}{6} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$

Step 4: Apply the point-gradient formula.

$y - 1 = \sqrt{3}\left(x - \frac{\pi}{6}\right)$

$y = x\sqrt{3} + 1 - \frac{\pi\sqrt{3}}{6}$

Worked Example: Geometric Configuration with Tangents and Normals

Consider the curve $y = \cos x$ at the points $A\left(-\frac{\pi}{2}, 0\right)$ and $B\left(\frac{\pi}{2}, 0\right)$. Show that the tangents and normals at these points form a square.

Finding the gradients:

The derivative is:

$y' = -\sin x$

At point $A\left(-\frac{\pi}{2}, 0\right)$:

Gradient of tangent $= -\sin\left(-\frac{\pi}{2}\right) = 1$

Gradient of normal $= -1$

At point $B\left(\frac{\pi}{2}, 0\right)$:

Gradient of tangent $= -\sin\frac{\pi}{2} = -1$

Gradient of normal $= 1$

Finding the equations:

Tangent at $A$:

$y - 0 = 1 \times \left(x + \frac{\pi}{2}\right)$

$y = x + \frac{\pi}{2}$

Normal at $A$:

$y - 0 = -1 \times \left(x + \frac{\pi}{2}\right)$

$y = -x - \frac{\pi}{2}$

Tangent at $B$:

$y - 0 = -1 \times \left(x - \frac{\pi}{2}\right)$

$y = -x + \frac{\pi}{2}$

Normal at $B$:

$y - 0 = 1 \times \left(x - \frac{\pi}{2}\right)$

$y = x - \frac{\pi}{2}$

Geometric properties:

The two tangents meet on the $y$-axis at $T\left(0, \frac{\pi}{2}\right)$.

The two normals meet on the $y$-axis at $N\left(0, -\frac{\pi}{2}\right)$.

Since adjacent sides are perpendicular, $ANBT$ is a rectangle. Since the diagonals are also perpendicular, it is a rhombus. Therefore, quadrilateral $ANBT$ is a square.

Worked Example: Tangent Line and Triangle Area

Consider the tangent to $y = \tan 2x$ at the point where $x = \frac{\pi}{8}$. Find the intercepts and calculate the area of the triangle formed by the tangent and the coordinate axes.

Step 1: Find the point and gradient.

The function is $y = \tan 2x$.

Differentiating using the chain rule:

$y' = 2\sec^2 2x$

When $x = \frac{\pi}{8}$:

$y = \tan\frac{\pi}{4} = 1$

$y' = 2\sec^2\frac{\pi}{4} = 2 \times \left(\sqrt{2}\right)^2 = 4$

Step 2: Find the tangent equation.

$y - 1 = 4\left(x - \frac{\pi}{8}\right)$

$y = 4x - \frac{\pi}{2} + 1$

Step 3: Find the intercepts.

For the $y$-intercept, set $x = 0$:

$y = 1 - \frac{\pi}{2} = \frac{2 - \pi}{2}$

For the $x$-intercept, set $y = 0$:

$0 = 4x - \frac{\pi}{2} + 1$

$4x = \frac{\pi}{2} - 1 = \frac{\pi - 2}{2}$

$x = \frac{\pi - 2}{8}$

Step 4: Calculate the triangle area.

The tangent line and the coordinate axes form a triangle. Using the formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

$= \frac{1}{2} \times \frac{\pi - 2}{8} \times \frac{2 - \pi}{2}$

Note that $\frac{2 - \pi}{2} = -\frac{\pi - 2}{2}$, so:

$= \frac{1}{2} \times \frac{\pi - 2}{8} \times \frac{\pi - 2}{2} = \frac{(\pi - 2)^2}{32} \text{ square units}$

Worked Example: Sketching $y = \sin x + \cos x$

Sketch the curve $y = \sin x + \cos x$ for $0 \leq x \leq 2\pi$.

Step 1: Find values at domain endpoints.

When $x = 0$:

$y = \sin 0 + \cos 0 = 0 + 1 = 1$

When $x = 2\pi$:

$y = \sin 2\pi + \cos 2\pi = 0 + 1 = 1$

Step 2: Find $x$-intercepts.

Set $y = 0$:

$\sin x + \cos x = 0$

$\sin x = -\cos x$

$\tan x = -1$ (dividing by $\cos x$)

This means $x$ is in quadrant 2 or 4, with related angle $\frac{\pi}{4}$.

Therefore: $x = \frac{3\pi}{4}$ or $x = \frac{7\pi}{4}$

Step 3: Find stationary points.

Differentiate:

$y' = \cos x - \sin x$

Set $y' = 0$:

$\cos x - \sin x = 0$

$\sin x = \cos x$

$\tan x = 1$

This means $x$ is in quadrant 1 or 3, with related angle $\frac{\pi}{4}$.

Therefore: $x = \frac{\pi}{4}$ or $x = \frac{5\pi}{4}$

Step 4: Find the $y$-values at stationary points.

When $x = \frac{\pi}{4}$:

$y = \sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

When $x = \frac{5\pi}{4}$:

$y = \sin\frac{5\pi}{4} + \cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$

Step 5: Determine the nature of stationary points using the second derivative.

$y'' = -\sin x - \cos x$

When $x = \frac{\pi}{4}$:

$y'' = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} < 0$

This is a maximum turning point at $\left(\frac{\pi}{4}, \sqrt{2}\right)$.

When $x = \frac{5\pi}{4}$:

$y'' = -\left(-\frac{\sqrt{2}}{2}\right) - \left(-\frac{\sqrt{2}}{2}\right) = \sqrt{2} > 0$

This is a minimum turning point at $\left(\frac{5\pi}{4}, -\sqrt{2}\right)$.

Step 6: Find points of inflection.

The second derivative $y'' = -\sin x - \cos x$ equals zero when:

$-\sin x - \cos x = 0$

These are the same values as the $x$-intercepts: $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.

We can verify the second derivative changes sign at these points by checking values:

The $x$-intercepts $\left(\frac{3\pi}{4}, 0\right)$ and $\left(\frac{7\pi}{4}, 0\right)$ are also inflection points.

Worked Example: Sketching $y = x - \sin x$

Sketch the curve $f(x) = x - \sin x$.

Step 1: Determine the domain.

The domain of $f(x) = x - \sin x$ is all real numbers.

Step 2: Test for symmetry.

Both $\sin x$ and $x$ are odd functions.

Therefore, $f(x) = x - \sin x$ is odd, meaning $f(-x) = -f(x)$.

The graph has rotational symmetry about the origin.

Step 3: Find zeroes and examine the sign.

The function equals zero at $x = 0$ and nowhere else.

This is because:

• For $x > 0$: $\sin x < x$, so $x - \sin x > 0$
• For $x < 0$: $\sin x > x$, so $x - \sin x < 0$

Step 4: Examine behaviour as $x \to \pm\infty$.

Since $\sin x$ always remains between $-1$ and $1$:

As $x \to \infty$: $f(x) = x - \sin x \to \infty$

As $x \to -\infty$: $f(x) = x - \sin x \to -\infty$

Step 5: Find stationary points.

Differentiate:

$f'(x) = 1 - \cos x$

Set $f'(x) = 0$:

$\cos x = 1$

This occurs at $x = ..., -2\pi, 0, 2\pi, 4\pi, ...$

Step 6: Determine the nature of stationary points.

Note that $f'(x) = 1 - \cos x$ is never negative because $\cos x \leq 1$ always.

This means the curve is always increasing except at the stationary points.

At each stationary point:

• $x = ..., -2\pi, 0, 2\pi, 4\pi, ...$
• $y = x$ at these points

These are stationary inflection points: $..., (-2\pi, -2\pi), (0, 0), (2\pi, 2\pi), (4\pi, 4\pi), ...$

Step 7: Find other points of inflection.

Differentiate again:

$f''(x) = \sin x$

This equals zero at $x = ..., -\pi, 0, \pi, 2\pi, 3\pi, ...$

Since $\sin x$ changes sign around each of these points, the points $..., (-\pi, -\pi), (\pi, \pi), (3\pi, 3\pi), ...$ are also inflection points.

At these inflection points:

$f'(\pi) = 1 - \cos\pi = 1 - (-1) = 2$

The gradient at these additional inflection points is $2$.