Differentiability examples

### Multivariable calculus and vector analysis

### 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.