Tangent and normal equations and gradient conditions
| English | Español |
|---|---|
| normal line | normal line |
A road’s tangent gives its local direction; a cross-road follows the normal. Why must both lines use the same curve point?
- A road’s tangent gives its local direction; a cross-road follows the normal. Why must both lines use the same curve point?
- This lesson studies normal line 法线: A line through the curve point perpendicular to its local tangent.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines normal line?
A line through the curve point perpendicular to its local tangent.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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.
f′=3x²−3, giving 9 at x=2.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
A normal to a horizontal tangent has finite gradient zero. This claim is false. Explain which definition or assumption it violates.
Find the normal gradient there.
For the finite nonzero tangent gradient 9, the normal gradient is −1/9.
A normal to a horizontal tangent has finite gradient zero.
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
How many tangents to f are parallel to y=9x+4?
3x²−3=9 has the two allowed inputs ±2.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · G. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
A line through the curve point perpendicular to its local tangent. Choose the relationship, show the method, check its assumptions and interpret the result.