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

$\frac{\mathrm{d}}{\mathrm{d}x}\left(f\left(g\left(x\right)\right)\right)=f^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)$

The multivariable version:

$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\left(f(x\left(t\right),y\left(t\right))\right)& =f_x(x\left(t\right),y\left(t\right))x^{\prime }\left(t\right)+f_y(x\left(t\right),y\left(t\right))y^{\prime }\left(t\right)\\ & =\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}\end{array}$

Example

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

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

$\begin{array}{rl}\frac{\mathrm{d}f}{\mathrm{d}t}& =f_x\cdot 1+f_y\cdot (-4t)\\ f_x& =e^x\cos y\\ f_y& =-e^x\sin y\\ \frac{\mathrm{d}f}{\mathrm{d}t}& =e^t[\cos 2t^2-4t\sin 2t^2]\end{array}$

Evil Example

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

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

$\frac{\mathrm{d}f}{\mathrm{d}t}=\frac{\partial f}{\partial x}(\frac{\partial x}{\partial y}{y}^{\prime }+\frac{\partial x}{\partial t})+\frac{\partial f}{\partial y}{y}^{\prime }+\frac{\partial f}{\partial t}$