Parametric first derivatives and vertical tangents
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| parametric gradient/ˌpærəˈmetrɪk ˈɡreɪdɪənt/ | hệ số góc tham số |
How do their separate rates combine into the curve’s slope?
- Two coordinates depend on the same slider.
- This lesson studies parametric gradient 参数曲线斜率: The ratio of the y-rate to the x-rate along a parametrised curve, when the x-rate is nonzero.
The method
- For x=x(t), y=y(t), the chain rule gives dy/dt=(dy/dx)(dx/dt), so dy/dx=(dy/dt)/(dx/dt) when dx/dt≠0.
- Differentiate the two coordinates separately, then divide in y-over-x order.
- The point is obtained from the same parameter value.
- A parameter need not represent physical time.
For x=t²+1, y=t³-3t at t=2, why is dy/dx=9/4?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The y-rate divides by the x-rate - never the reverse. Both derivatives come from the same parameter value, and the point (5,2) also comes from t=2.
A worked case
For x=t²+1, y=t³−3t, dx/dt=2t and dy/dt=3t²−3. At t=2 the point is (5,2) and gradient is 9/4, so the tangent is y−2=(9/4)(x−5). At t=1, the point (2,−2) has horizontal tangent y=−2 because dy/dt=0 while dx/dt=2. At t=0, (1,0) has vertical tangent x=1: dx/dt=0 but dy/dt=−3≠0. A regular local parametrisation then changes y while x has zero instantaneous change. For x=2cos t,y=3sin t, at t=π/4 the point is (√2,3√2/2) and slope is −3/2. At t=0 the ellipse tangent is vertical, x=2.
Parametric first derivatives and vertical tangents
For x=x(t), y=y(t), the chain rule gives dy/dt=(dy/dx)(dx/dt), so dy/dx=(dy/dt)/(dx/dt) when dx/dt≠0
Identify the appropriate changing variable and validate the derivative denominator or local branch.
For the cubic parametrisation at t=2, find dy/dx.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At t=2, dy/dt=9 and dx/dt=4, so the ratio is 9/4.
Where it goes wrong
- The ratio is dy/dt divided by dx/dt, not its reciprocal.
- If both rates vanish, 0/0 is inconclusive: x=t³,y=t³ still traces y=x with slope 1 through zero.
- If dx/dt=0 alone, inspect the local curve rather than report a finite numerical gradient.
- Repeated x values can correspond to different y values or branches; keep the stated parameter value.
Not true: When both parametric coordinate rates are zero, the curve must have no tangent. When both rates are zero, dy/dx=0/0 is undetermined; examine the curve more closely, because a tangent may still exist.
For the same curve at t=1, find dy/dx.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At t=1, the numerator is zero and denominator is 2, giving zero.
When both parametric coordinate rates are zero, the curve must have no tangent.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
When both rates are zero, dy/dx=0/0 is undetermined; examine the curve more closely, because a tangent may still exist.
Applying it
- Find the coordinate pair, calculate both first rates, then test whether the ratio is valid.
- State horizontal/vertical tangent lines separately when appropriate and preserve the parameter range.
- Check by eliminating the parameter when that gives a simple local relation.
- G5 requires first derivatives only; a second parametric derivative is outside this lesson’s stated scope.
For the ellipse at t=π/4, find dy/dx.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
(3cos t)/(−2sin t) at π/4 is −3/2.
Match each parametric situation to its interpretation.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
A zero x-rate alone forbids the quotient but not the tangent; x=t³,y=t³ shows both rates zero yet the curve has slope 1.
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.
- Parametric gradient: The ratio of the y-rate to the x-rate along a parametrised curve, when the x-rate is nonzero.
- For x=x(t), y=y(t), the chain rule gives dy/dt=(dy/dx)(dx/dt), so dy/dx=(dy/dt)/(dx/dt) when dx/dt≠0.
- Watch for: the ratio is dy/dt divided by dx/dt, not its reciprocal.