Parametric gradients and tangent directions | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

\begin{array}{c} \frac{d y}{d x} = \frac{d y / d t}{d x / d t} \\ \frac{d x}{d t} \neq 0 \end{array}

Evaluate both rates at the same parameter. A nonzero vector derivative gives the tangent direction even when no finite gradient exists.

Find the point and the two rates

A parametric curve gives both coordinates in terms of $t$ . Find the point by substituting into $x (t), y (t)$ , then find the gradient by dividing the $y$ rate by the $x$ rate.

The chain-rule ratio is valid only when $d x / d t \neq 0$ . Use the same parameter value in the coordinates and in both derivatives before writing the tangent equation.

\begin{array}{c} y - y_{0} = m (x - x_{0}) \\ m = {\frac{y^{'} (t)}{x^{'} (t)} |}_{t = t_{0}} \end{array}

Check zero rates before dividing

If $x^{'} = 0$ and $y^{'} \neq 0$ , the tangent is vertical: $x = x_{0}$ . If $y^{'} = 0$ and $x^{'} \neq 0$ , the tangent is horizontal: $y = y_{0}$ .

If both derivatives vanish, the ratio is indeterminate. Inspect a nearby derivative ratio and its limit, or the Cartesian curve. The zero vector alone does not decide whether a tangent exists.

\begin{array}{c} r^{'} (t_{0}) = (0, b), b \neq 0 \\ ⟹ x = x (t_{0}) \end{array}

Use a direction vector in three dimensions

When $r^{'} (t_{0}) \neq 0$ , it points along the tangent line. In three dimensions one scalar gradient cannot describe the line; keep the full direction vector.

Use a fresh parameter $λ$ for the line. Its points lie along the local tangent, while the original $t$ traces the curve.

\begin{array}{c} x = r (t_{0}) + λ r^{'} (t_{0}) \\ λ \in ℝ \end{array}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find a tangent with a finite gradient

Find the tangent line to $x = t^{2}$ , $y = t^{3}$ at $t = 1$ .

Worked solution

1
Find the point on the curve.
$\begin{aligned} (x (1), y (1)) & = (1^{2}, 1^{3}) \\ = (1, 1) \end{aligned}$
The tangent must pass through the point at the specified parameter.
2
Differentiate both coordinates and form their ratio.
$\begin{aligned} \frac{d x}{d t} & = 2 t \\ \frac{d y}{d t} & = 3 t^{2} \\ m & = {\frac{3 t^{2}}{2 t} |}_{t = 1} \\ = \frac{3}{2} \end{aligned}$
At $t = 1$ , the denominator is $2 \neq 0$ , so the ratio is valid.
3
Use point-gradient form.
$y - 1 = \frac{3}{2} (x - 1)$
The line has the required gradient and contains $(1, 1)$ .

Answer

$y - 1 = \frac{3}{2} (x - 1)$ .

EXAMPLE 02

Read a vertical tangent

For $x = t^{2}$ , $y = t$ , find the tangent at $t = 0$ . Explain why a zero $x$ rate does not remove the tangent.

The right-opening curve x=t squared, y=t, with increasing-parameter arrows and the vertical tangent x=0 at the origin.

Worked solution

1
Find the point and the derivative vector.
$\begin{aligned} r (0) & = (0, 0) \\ r^{'} (t) & = (2 t, 1) \\ r^{'} (0) & = (0, 1) \end{aligned}$
The point is the origin. The nonzero derivative vector points straight upward.
2
Check the gradient denominator.
$\begin{aligned} {\frac{d x}{d t} |}_{t = 0} & = 0 \\ {\frac{d y}{d t} |}_{t = 0} & = 1 \end{aligned}$
Dividing $1$ by $0$ gives no finite gradient. These rates identify a vertical tangent.
3
Write the vertical line through the point.
$x = 0$
The graph turns to the right of the origin, while its tangent at the origin is the vertical axis.

Answer

The tangent is $x = 0$ . Its direction vector $(0, 1)$ is nonzero, although the tangent has no finite gradient.

EXAMPLE 03

Point a probe along a 3D route

An inspection probe follows $r (t) = (t, t^{2}, t^{3})$ metres. At $t = 1$ second, its sensor must point along the forward tangent. Give a suitable direction vector and a vector equation for the tangent line.

Worked solution

1
Locate the probe and differentiate the route.
$\begin{aligned} r (1) & = (1, 1, 1) \\ r^{'} (t) & = (1, 2 t, 3 t^{2}) \end{aligned}$
The position supplies a point on the line, and the derivative supplies its direction.
2
Evaluate the direction at the same time.
$r^{'} (1) = (1, 2, 3)$
This vector is nonzero and points in the direction of increasing $t$ . Normalisation is unnecessary because only direction is requested.
3
Use a new parameter to describe the tangent line.
$\begin{array}{c} x = (\begin{array}{cccccccccc} 1 \\ 1 \\ 1 \end{array}) + λ (\begin{array}{cccccccccc} 1 \\ 2 \\ 3 \end{array}) \\ λ \in ℝ \end{array}$
At $λ = 0$ the line passes through the probe. Positive $λ$ follows the forward tangent direction.

Answer

A forward direction is $(1, 2, 3)$ . The tangent is $x = (1, 1, 1) + λ (1, 2, 3)$ , $λ \in ℝ$ .

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Find a point and a gradient

Find the tangent to $x = t$ , $y = t^{2}$ at $t = 3$ .

Show answer & reasoning

$y - 9 = 6 (x - 3)$ .

The point is $(3, 9)$ . Since $x^{'} = 1$ and $y^{'} = 2 t$ , the gradient at $3$ is $6$ . Substitute the point and gradient into point-gradient form.

CHECK 02

Interpret a zero derivative vector

For $x = t^{3}$ , $y = t^{2}$ , both coordinate derivatives vanish at $t = 0$ . Use $d x / d y$ for nearby nonzero $t$ to find the tangent there.

Show answer & reasoning

The tangent is $x = 0$ , even though $r^{'} (0) = 0$ .

For $t \neq 0$ , $d x / d y = (3 t^{2}) / (2 t) = 3 t / 2 \to 0$ as $t \to 0$ . The nearby tangent direction therefore approaches the vertical direction. The curve passes through the origin, so the limiting tangent is $x = 0$ .

CHECK 03

Check a cart’s level heading

A cart follows $x = 2 t$ , $y = t^{2} - 4 t$ metres for $0 \leq t \leq 4$ seconds. At its lowest point, a straight inspection guide must match its tangent. Find the guide line and say whether the cart heads left or right.

Show answer & reasoning

The guide is $y = - 4$ m. The cart heads right through $(4, - 4)$ m at $t = 2$ s.

The lowest point occurs at $y^{'} = 2 t - 4 = 0$ , giving $t = 2$ ; $y = {(t - 2)}^{2} - 4$ confirms it is the minimum. Here $x^{'} = 2 > 0$ , so the tangent is horizontal and the cart moves right. Substitution gives $(x, y) = (4, - 4)$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find $\frac{d y}{d x}$ for $x = t^{2} + 1, y = 3 t$ , $t$ nonzero.
2 marks
Find the tangent at $2$ for $x = t, y = t^{3}$ .
2 marks
Find the tangent at $0$ for $(cos t, sin t)$ .
2 marks
Use a limit to find the tangent at $0$ for $x = t^{3}, y = t^{2}$ .
2 marks
Give a vector tangent equation at $1$ for $(t, t^{2}, t^{3})$ .
2 marks
Find the horizontal tangent of $x = 2 t, y = t^{2} - 4 t$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $\frac{d y}{d x}$ for $x = t^{2} + 1, y = 3 t$ , $t$ nonzero.
2 marks
Derivatives are $2 t$ and $3$ .
Divide $y$ rate by $x$ rate.
$\frac{3}{2 t}$
Find the tangent at $2$ for $x = t, y = t^{3}$ .
2 marks
The point is $(2, 8)$ .
Gradient is $3 t^{2} = 12$ .
$y - 8 = 12 (x - 2)$ .
Find the tangent at $0$ for $(cos t, sin t)$ .
2 marks
The point is $(1, 0)$ .
The derivative is $(0, 1)$ , giving a vertical line.
$x = 1$ .
Use a limit to find the tangent at $0$ for $x = t^{3}, y = t^{2}$ .
2 marks
For nonzero $t$ , $\frac{d x}{d y} = \frac{3 t}{2}$ .
This tends to $0$ , giving a vertical tangent.
$x = 0$ .
Give a vector tangent equation at $1$ for $(t, t^{2}, t^{3})$ .
2 marks
The point is $(1, 1, 1)$ .
The derivative there is $(1, 2, 3)$ .
$x = (1, 1, 1) + s (1, 2, 3)$ .
Find the horizontal tangent of $x = 2 t, y = t^{2} - 4 t$ .
2 marks
$\frac{d y}{d t} = 2 t - 4 = 0$ gives $2$ ; $\frac{d x}{d t} = 2$ .
The corresponding point is $(4, - 4)$ .
$y = - 4$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics school Vector calculus: polynomial and trigonometric parameterisations, related coordinate rates, Bézier paths, componentwise differentiation and integration, motion, tangent directions, circular motion and arc length. Reasoning and proofs use these school topics; vector-field calculus is outside this chapter.

Official curriculum source ↗

In this lesson

Understand the method.

Find the point and the two rates

Check zero rates before dividing

Use a direction vector in three dimensions

See the method in action.

Find a tangent with a finite gradient

Find the point on the curve.

Differentiate both coordinates and form their ratio.

Use point-gradient form.

Read a vertical tangent

Find the point and the derivative vector.

Check the gradient denominator.

Write the vertical line through the point.

Point a probe along a 3D route

Locate the probe and differentiate the route.

Evaluate the direction at the same time.

Use a new parameter to describe the tangent line.

Try it yourself.

Find a point and a gradient

Interpret a zero derivative vector

Check a cart’s level heading

Keep practising