MATH 251 Lecture 11

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Test 1 in 1 week

Multidimensional Limits

We only have a way to show that limits do not exist.

"Functions are continuous except when they are not continuous."

Chain Rule

Recall from Calc 2 that

${\frac {\mathrm {d} }{\mathrm {d} x}}\left(f\left(g\left(x\right)\right)\right)=f'\left(g\left(x\right)\right)\,g'\left(x\right)$

The multivariable version:

${\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} t}}\left(f\left(x\left(t\right),y\left(t\right)\right)\right)&=f_{x}\left(x\left(t\right),\,y\left(t\right)\right)\,x'\left(t\right)+f_{y}\left(x\left(t\right),\,y\left(t\right)\right)\,y'\left(t\right)\\&={\frac {\partial f}{\partial x}}\,{\frac {\mathrm {d} x}{\mathrm {d} t}}+{\frac {\partial f}{\partial y}}\,{\frac {\mathrm {d} y}{\mathrm {d} t}}\end{aligned}}$

Example

${\begin{aligned}f(x,y)&=e^{x}\cos {y}\\x(t)&=t\\y(t)&=-2t^{2}\end{aligned}}$

Calculate ${\frac {\mathrm {d} f}{\mathrm {d} t}}$

${\begin{aligned}{\frac {\mathrm {d} f}{\mathrm {d} t}}&=f_{x}\cdot 1+f_{y}\cdot \left(-4t\right)\\f_{x}&=e^{x}\cos {y}\\f_{y}&=-e^{x}\sin {y}\\{\frac {\mathrm {d} f}{\mathrm {d} t}}&=e^{t}\left[\cos {2t^{2}}-4t\sin {2t^{2}}\right]\end{aligned}}$

Evil Example

${\begin{aligned}f&=f(x,y,t)\\x&=x(y,t)\\y&=y(t)\\\therefore &f(x(y(t),t),y(t),t)\end{aligned}}$

Create a Tree: f |-- x | |-- y | | `--t | `-- t |-- y | `-- t `-- t

${\frac {\mathrm {d} f}{\mathrm {d} t}}={\frac {\partial f}{\partial x}}\,\left({\frac {\partial x}{\partial y}}\,y'+{\frac {\partial x}{\partial t}}\right)+{\frac {\partial f}{\partial y}}\,y'+{\frac {\partial f}{\partial t}}$