التفاضل والتكامل متعدد المتغيرات والتحليل الشعاعي
مُقدَّم| English | المعنى |
|---|---|
| gradient/ˈɡreɪdɪənt/ | تدرج |
| directional derivative/daɪˈrekʃənl dɪˈrɪvətɪv/ | The rate of change of a function at a point in a chosen unit direction. |
| Jacobian/dʒæˈkəʊbɪən/ | جاكوبي |
One point, many slopes
- Standing on a hillside, the slope you feel depends on the direction you walk. A function of two variables has a different rate of change in every direction.
- GRE multivariable questions test the gradient 梯度, directional derivatives 方向导数, the chain rule and the area factor in a change of coordinates.
Partials, gradient, direction
- $f_x$ varies $x$ with $y$ fixed. For $f = x^2 + 3xy$: $f_x = 2x + 3y$, $f_y = 3x$. The gradient is $\nabla f = (f_x, f_y)$.
- For differentiable $f$, the directional derivative along a unit vector $\mathbf{u}$ is $\nabla f \cdot \mathbf{u}$. It is largest along $\nabla f$, with value $|\nabla f|$.
$f = x^2 + 3y^2$, so $\nabla f = (2x, 6y)$ and $\nabla f(1,1) = (2, 6)$. Direction $(3,4)$ has length 5, so $\mathbf{u} = (\tfrac35, \tfrac45)$ and $D_{\mathbf{u}} f = \tfrac{6}{5} + \tfrac{24}{5} = 6$. Using $(3,4)$ unnormalised would give 30.
$\dfrac{\partial}{\partial x}(x^2 y)$ equals
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
Treat y as a constant: ∂/∂x(x²y) = y · 2x = 2xy. The option 2xy + x² wrongly differentiates y as well.
للدفاع $f = x^2 + y^2$, find the directional derivative at $(1, 2)$ in the direction of $(3, 4)$.
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
$\nabla f = (2, 4)$, $\mathbf{u} = (0.6, 0.8)$: $1.2 + 3.2 = 4.4$.
Differentiability and the chain rule
- Differentiable means a good linear approximation; partials existing at a point is not enough. Continuous first partials nearby are sufficient.
- Along a path: $\dfrac{d}{dt} f(x(t), y(t)) = f_x x'(t) + f_y y'(t)$.
Multiple integrals and Jacobians 雅可比行列式
- Sketch the region first; reversing the order of a non-rectangular region changes the limits.
- Polar: $dA = r\,dr\,d\theta$. A general change $(u,v) \mapsto (x,y)$ scales area by $\left|\det \dfrac{\partial(x,y)}{\partial(u,v)}\right|$, always non-negative.
- Disc of radius 2: $\displaystyle\int_0^{2\pi}\!\!\int_0^2 r\,dr\,d\theta = 4\pi$.
The area element in polar coordinates is
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
The Jacobian of the polar map is r.
تقييم $\iint_D (x^2 + y^2)\,dA$ over the unit disc $D$, to three decimal places.
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
$\int_0^{2\pi}\int_0^1 r^3\,dr\,d\theta = \pi/2$.
When is the graph a plane?
- $f_x = a$ and $f_y = b$ everywhere on $\mathbb{R}^2$ force $f = ax + by + c$.
- $f_{xy} = 0$ does not: $x^2 + y^2$. Parallel straight level lines do not: $e^x$.
Does it force a plane?
Distinguish conditions that force a linear function on the plane.
If $f_{xy} = 0$ everywhere on $\mathbb{R}^2$, the graph of $f$ is a plane.
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
$x^2 + y^2$ has zero mixed partial.
Recap
- Normalise the direction before dotting with $\nabla f$.
- Polar area element $r\,dr\,d\theta$; general changes use $|\det J|$.
- Constant first partials give a plane; weaker conditions do not.
- directional derivative: The rate of change of a function at a point in a chosen unit direction.