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Math 2433 Week 4 (part 2)

15.1 - DIFFERENTIABILITY AND

We say that f is differentiable at x if there exists a vector y such that f (x + h) − f (x) = y i h + o(h).

g()h We will say that g(h) is o(h) if lim= 0 h→0 h Example: For fxy(, )=+ 3 x y2 :

Let f be differentiable at x. The gradient of f at x is the unique vector ∇ f (x) such that f (x + h) − f (x) = ∇ f (x) i h + o(h). In other words, the gradient of a f (x, y) is ∇f (x, y) = fx i + fy j

And the gradient of a function f (x, y,z) is ∇f (x, y,z) = fx i + fy j + fz k

Continuing the previous example:

More examples:

1. Find the gradient of fxy(, )=+ 2 ex sin( x2 y )

2. Find the gradient of fxyz(, ,)= xy− z at (2, -1, 4)

Popper 4

3x 5. Find the gradient of f (x, y,z) = + ln z y a. ∇fxyz(, ,)= 3/ y− 3 xy /2 + 1/ z b. ∇fxyz(, ,)= ( 3/ y)ijk−( 3/ xy2 ) +( 1/ z) c. ∇fxyz(, ,)=+ 3 xijk( 3/ y) +( 1/ z) d. None of these

15.2 and Directional

Properties of gradients:

Directional Derivatives: f 'u gives the directional of f in the direction u. In other words, fu′ gives the rate of change of f in the direction of u.

f ' = ∇f (x) i u u

Example:

1. Find the at the point P in the direction indicated.

fxy(, )=+ x22 3 y at P(1,1) in the direction of i!–!j! ! !

22 ⎛⎞1 ⎛⎞3 2. Find the directional derivative for fxy(, )=+ x 3 y at Q⎜⎟−1, towards R ⎜⎟2, . ⎝⎠2 ⎝⎠2

Note that the directional derivative in a direction u is the component of the gradient vector in that direction.

Important:

Example: Find a in the direction in which f increases most rapidly at P and give the rate of change of f in that direction; find a unit vector in the direction in which f decreases most rapidly at P and give the rate of change of f in that direction.

−2 2x f (x, y) = y e at P(0,1)

Popper 4

x 7. What is the direction in which the function fxy(, )= yx2 − decreases most rapidly at the y2 point (-2, 1)? a. −5i b. 32ij+ c. 0 d. 5i e. 3ij+ f. None of these

15.3 - The ; Chain Rules

What was the MVT for functions of one variable?

Example: Let fxy(, )= x3 − xy and let a = (0,1) and b = (1,3). Find a point c on the line segment ab for which the mean value theorem (for several variables) is satisfied.

A nonempty U (in the plane or in three-space) is said to be connected if any two points of U can be joined by a polygonal path that lies entirely in U.

Thm – Let U be an open connected set and let f be a on U. If ∇ f (x) = 0 for all x in U, then f is constant on U.

Thm - Let U be an open connected set and let f and g be differentiable functions on U. If ∇ f (x) = ∇ g(x) for all x in U, then f and g differ by a constant on U.

Examples: d 1. Find [ ft(())r ] given f (x, y) = 6x + y, r(t) = t i + 7t j dt

2. Find the rate of change of f with respect to t along the given curve. f (x, y) = x2y, r(t) = e t i + e−t j

3. Find the rate of change of f with respect to t along the given curve.

!

Popper 4

8. Suppose that g(t) = f (r(t)) . How would you find g′(t) ? a. gt′′()= f(r( t)) b. gt′′′()= f(r ( t)) c. g′(t) = ∇f r t i r′ t ( ( )) ( ) d. g′(t) = f ′ r t i r′ t ( ( )) ( ) e. None of these

Other chain rules:

If

Then

Example: du 3. ux= 22−32 xyy+= xt ()cos()sin tyt = t. Find dt

Popper 4

9. Find du/dt.

a) b) c) d) e)

10. The directional derivative of fxyz(, ,)=+ xeyz xye z at the point P(2,1,1)− in the direction of the yz−−11 line x +=2 = is −23 a. 244eeeijk−− b. −2e 2e c. − 14 1 d. (2eeeijk−− 4 4 ) 14 e. -2 f. None of these

Math 2433 – Week 5

15.4 - The Gradient as a ; Lines and Tangent Planes

Suppose that f (x, y) is a non-constant function that is continuously differentiable. That means f is differentiable and its gradient ∇ f is continuous. We saw last week that at each point in the domain ∇ f (if ≠ 0) points in the direction of the most rapid increase of f. Also, at each point of the domain, the gradient vector ∇ f (if ≠ 0) is perpendicular to the level curve of f that passes through that point.

The vector

Is perpendicular to the gradient so it is the .

The equation of the tangent line is:

And the equation of the normal line is:

Example: Write an equation for the tangent line and an equation for the normal line at point P.

For f(x,y,z), the equation for the tangent plane is

f x (x0 , y0 , z0 )(x − x0 ) + f y (x0 , y0 , z0 )(y − y0 ) + f z (x0 , y0 , z0 )(z − z0 ) = 0 And the normal line is

Example: Find an equation for the tangent plane and scalar parametric equations for the normal line at the point P.

! ! !

More examples: The curve r(t) = 2t i + 3t −1 j − 2t2 k and the ellipsoid x2 +y2 +3z2 = 25 intersect at (2, 3,−2). What is the angle of intersection?