Tangent and normal equations and gradient conditions
Введено| English | Значение |
|---|---|
| normal line/ˈnɔːml laɪn/ | нормаль |
Why must both lines use the same curve point?
- A road’s tangent gives its local direction; a cross-road follows the normal.
- This lesson studies normal line 法线: A line through the curve point perpendicular to its local tangent.
The method
- At a differentiable point (a,f(a)), the tangent gradient is m=f′(a) and its equation is y−f(a)=m(x−a).
- For a finite nonzero m, the normal gradient is −1/m.
- Find points satisfying a given gradient condition by solving for f′(x), then evaluate the original f.
- Parallel lines have equal gradients; finite nonzero perpendicular gradients multiply to −1.
At x=1 on f=x³-3x the tangent is horizontal. What is the normal?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The normal is perpendicular to the tangent; a horizontal tangent (m=0) forces a vertical normal, stated as x=1. The formula -1/m applies only to finite nonzero m.
A worked case
For f=x³−3x at x=2, f(2)=2 and f′(2)=9. The tangent is y−2=9(x−2), or y=9x−16. The normal is y−2=−(1/9)(x−2). Tangents parallel to y=9x+4 satisfy 3x²−3=9, giving x=±2: points (2,2) and (−2,−2). The second tangent is y+2=9(x+2). At x=1 the tangent is horizontal y=−2 and the normal is vertical x=1; no finite negative reciprocal of zero exists. For the circle at (5,0), the tangent is vertical x=5 and the normal horizontal y=0. The secant through (−2,−2) and (2,2) has gradient 1, unlike either local tangent gradient 9.
Tangent and normal equations and gradient conditions
At a differentiable point (a,f(a)), the tangent gradient is m=f′(a) and its equation is y−f(a)=m(x−a)
Connect derivative calculations to their original point, domain and stated rate law.
For f=x³−3x, find the tangent gradient at x=2.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
f′=3x²−3, giving 9 at x=2.
Where it goes wrong
- The derivative supplies a gradient, not the y-coordinate.
- Use the original curve for the point.
- The normal’s reciprocal includes a minus sign and is unavailable for a horizontal tangent; give a vertical line instead.
- A tangent can meet the curve again elsewhere.
- A given line’s intercept does not affect a parallel-gradient condition, but both tangent and normal must pass through the stated curve point.
Not true: A normal to a horizontal tangent has finite gradient zero. The normal to a horizontal tangent is vertical, so its gradient is undefined.
Find the normal gradient there. Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
For the finite nonzero tangent gradient 9, the normal gradient is −1/9.
The normal to a horizontal tangent is vertical, so its gradient is undefined.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. The normal to a horizontal tangent is vertical, so its gradient is undefined.
Applying it
- Find the point and gradient separately, then use point-slope form.
- If a gradient condition gives several inputs, retain every allowed point and line.
- State horizontal/vertical cases without inventing an infinite numerical gradient.
- Verify the line passes through the point and that a finite nonzero tangent/normal gradient pair multiplies to −1.
How many tangents to f are parallel to y=9x+4?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
3x²−3=9 has the two allowed inputs ±2.
Match each line at (a,f(a)) to its equation or gradient.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The derivative supplies the gradient and the ORIGINAL curve supplies the point; the two must be evaluated separately at the same x.
In your course
- AQA A-level Mathematics (7357), specification section G.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Normal line: A line through the curve point perpendicular to its local tangent.
- At a differentiable point (a,f(a)), the tangent gradient is m=f′(a) and its equation is y−f(a)=m(x−a).
- Watch for: the derivative supplies a gradient, not the y-coordinate.