Parametric gradients and tangent directions | Year 12

01 / LEARN

The idea, step by step

Use the chain-rule ratio

Where $d x / d t \neq 0$ , tangent gradient is $(d y / d t) / (d x / d t)$ . Evaluate both derivatives at the same parameter, and find that parameter first if a point rather than t is given.

A nonzero derivative vector gives tangent direction. In three dimensions use $x = r (t_{0}) + s r^{'} (t_{0})$ with a fresh parameter s for the tangent line.

Vertical and indeterminate cases

If dx/dt=0 but dy/dt is nonzero, the tangent is vertical. Its equation is x equal to the current x coordinate. If only dy/dt is zero, the tangent is horizontal.

When both derivatives vanish, the ratio is indeterminate. Simplify a derivative ratio away from the point and examine a limit, or inspect the local Cartesian relation. A zero derivative vector does not automatically mean there is no tangent.

A common trap

A zero denominator proves there is no tangent.

Instead: A vertical tangent has no finite gradient but is still a valid line.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

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

Derivatives are 2 t and $3 t^{2}$ .
At 1 the point is (1, 1) and gradient 3/2.
Use point-gradient form.

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

Example 2

Worked example

Find the tangent at t=0 for $x = t^{2}, y = t$ .

Coordinate derivatives at zero are 0 and 1.
The direction vector is (0, 1).
The tangent is vertical through the origin.

x = 0

Check your understanding

Find the gradient at 3 for x=t, y= $t^{2}$ .

Show answer

The derivative ratio is 2 t/1.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find dy/dx 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 dy/dx for $x = t^{2} + 1, y = 3 t$ , t nonzero.
2 marks
Derivatives are 2 t and 3.
Divide y rate by x rate.
$3 / (2 t)$
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, $d x / d y = 3 t / 2$ .
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
$d y / d t = 2 t - 4 = 0$ gives 2; dx/dt=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 topic 6.3–6.5: polynomial and trigonometric parameterisations, related coordinate rates, motion, tangent directions and arc length. This is school Vector calculus, not vector-field calculus.

Official curriculum source ↗

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