Differentiability examples

# Multivariable calculus and vector analysis

## A set of on-line readings

### Non-Math ML version

### Differentiability examples

#### Example 1

Let f(x,y) = x2 + y2. Find Df(1, 2) and the equation for the tangent plane at (x,y) = (1, 2). Find the linear approximation to f(x,y) at (x,y) = (1, 2).

Solution:

(x,y) | = 2x |  |
(1, 2) | = 2 |  |
(x,y) | = 2y |  |
(1, 2) | = 4 |  |

So Df(1, 2) = .

Since f(1, 2) = 12 + 22 = 5, the equation for the tangent plane is

z | = f(1, 2) + (1, 2)(x - 1) + (1, 2)(y - 2) |  |
| = 5 + 2(x - 1) + 4(y - 2) |  |

For a scalar-valued function of two variables, as f(x,y) is, the tangent plane is the linear approximation. We can write the linear approximation as

L(x,y) = 5 + 2(x - 1) + 4(y - 2). |  |

#### Example 1’

If looked at the point (2, 3), what changes?

Solution:

(2, 3) | = 4 |  |
(2, 3) | = 6 |  |
Df(2, 3) | = |  |

z | = L(x,y) = f(2, 3) + (2, 3)(x - 2) + (2, 3)(y - 3) |  |
| = 13 + 4(x - 2) + 6(y - 3) |  |

#### Example 2

Find the derivative of

f(x,y,z) = (x2y2z,y + sin z) |  |

at the point (1, 2, 0).

Solution: f : R3 → R2, so the derivative is the 2 × 3 matrix of partial derivatives.

| = 2xy2z |  |
| = 2x2yz |  |
| = x2y2 |  |
| = 0 |  |
| = 1 |  |
| = cos z |  |

For any point (x,y,z) = (a,b,c), the derivative is

Df(a,b,c) = . |  |

At (1, 2, 0), the derivative is

Df(1, 2, 0) = . |  |

#### Example 3

For the function of example 2, calculate the linear approximation to f at the point (1, 2, 0).

Solution: We’ve already calculated almost everything we need. We also need the value of the function at (1,2,0):

f(1, 2, 0) = | = ((12)(22)0, 2 + sin 0) |  |
| = (0, 2) |  |

Then, the linear approximation to f at (1,2,0) is A linear approximation to f is

L(x,y,z) | = f(1, 2, 0) + Df(1, 2, 0)(x - 1,y - 2,z) |  |
| = + |  |
| = + |  |
| = (4z,y + z) |  |

#### Example 4

Use the linear approximation of f(x,y,z) from Example 3 to approximate the value of f at the point (1.1, 1.9, 0.1).

Solution:

The above linear approximation at (x,y,z) = (1.1, 1.9, 0.1) is

L(1.1, 1.9, 0.1) | = (4(0.1), 1.9 + 0.1) |  |
| = (0.4, 2.0) |  |

Note that (1.1, 1.9, 0.1) is very close to (1, 2, 0), which is the point around which we computed the linear approximation. So, we expect this linear approximation to be close to the true value of f at (1.1, 1.9, 0.1). Let’s compare the above answer to the actual value of f at (1.1, 1.9, 0.1):

f(1.1, 1.9, 0.1) | = ((1.1)2(1.9)2(0.1), 1.9 + sin(0.1)) |  |
| ≈ (0.4368, 1.9998). |  |

In this case, the approximation is quite close.

## Contents by lecture

### Part 1

### Part 2

### Part 3

### Part 4

### Part 5

### Part 6

- Differentiability in higher dimensions*
- Quiz 6: Differentiability quiz*
- More on differentiability
- The linear approximation
- Differentiability examples

### Part 7

### Part 8

### Part 9

### Part 10

### Part 11

### Part 12

### Part 13

### Part 14

### Part 15

### Part 16

### Part 17

### Part 18

### Part 19

### Part 20

### Part 21

### Part 22

### Part 23

### Part 24

### Part 25

### Part 26

### Part 27