PARTIAL DERIVATIVES MATH 200 GOALS Figure out how to take - PowerPoint PPT Presentation

MATH 200 WEEK 4 - MONDAY PARTIAL DERIVATIVES

MATH 200 GOALS ▸ Figure out how to take derivatives of functions of multiple variables and what those derivatives mean. ▸ Be able to compute first-order and second-order partial derivatives . ▸ Be able to perform implicit partial differentiation . ▸ Be able to solve various word problems involving rates of change, which use partial derivatives.

MATH 200 REVIEW ▸ Calculus I: ▸ dy/dx = f’(x) ▸ f’(a) = “slope of the tangent line to f at x = a” ▸ e.g. f ( x ) = x 2 − 1 f (1) = (1) 2 − 1 = 0 THE POINT (1,0) IS ON THE GRAPH OF f f ′ ( x ) = 2 x f ′ (1) = 2(1) = 2 THE SLOPE OF THE LINE TANGENT TO F AT (1,0) IS 2

MATH 200 CONSIDER A 3D EXAMPLE ▸ Let f(x,y) = x 2 + y 2 ▸ Consider the trace of f on the plane y=1 ▸ f(x,1) = x 2 + 1

MATH 200 ▸ We can certainly find slope of the line tangent to z = x 2 +1 at any point on the xz-plane… ▸ Differentiating with respect to x, we get dz/dx = 2x ▸ The slope of the line at x=1 is 2

MATH 200 ▸ So, can we write parametric equations or a vector-valued function for that same line in 3-Space…? ▸ Need: (1) a point on the line and (2) a direction vector parallel to the line ▸ We were at x=1 on the plane y=1. Since f(1,1) = 2, the point of tangency is (1,1,2) ▸ The slope of 2 is telling us [change in z]/[change in x]  ▸ We need a direction vector x = 1 + t  for which z/x = 2 and y=0…  L : y = 1 ▸ <1,0,2> works! (There are infinitely many other choices  z = 2 + 2 t  of course.

MATH 200 ▸ What if we repeat the same process for the trace of f on the plane y=2? ▸ We’d be looking at the trace f(x,2) = x 2 + 4 ▸ Here it is on the xz-plane with its tangent line at x=1

MATH 200 ▸ We can find slope of the line tangent to z = x 2 +4 at any point on the xz-plane… ▸ Differentiating with respect to x, we get dz/dx = 2x ▸ It’s the same! ▸ The slope of the line at x=1 is 2 ▸ Since [change in z]/[change in x] = 2, we can use the direction vector <1,0,2> again, with starting point (1,2,5)  x = 1 + t   L : y = 2  z = 5 + 2 t 

MATH 200 WHAT JUST HAPPENED…? ▸ We just our first partial derivative! ▸ Notice that in both cases (whether we set y=1 or y=2) we got dz/dx = 2x ▸ This would’ve been the case with any choice of constant value for y ▸ We could have done the same work on any plane of the form x=constant ▸ In that case, we’d find [change in z]/[change in y]

MATH 200 DEFINITIONS ▸ The partial derivative of f with respect to x is what you get when you… WE USE THIS PARTIAL SYMBOL INSTEAD OF ▸ treat y as a constant … JUST d TO INDICATE THAT THERE IS MORE ▸ and differentiate with respect to x THAN ONE INDEPENDENT VARIABLE ▸ We write any of the following: ∂ ∂ f ∂ z f x ( x, y ) ∂ x, ∂ x,

MATH 200 DEFINITIONS ▸ The partial derivative of f with respect to y is what you get when you… NOTICE: WE’RE NOT USING “PRIME” NOTATION ▸ treat x as a constant … ANYMORE… ▸ and differentiate with respect to y IF I WRITE f’(x,y), YOU DON’T KNOW WHICH ▸ We write any of the following: VARIABLE I’M HOLDING CONSTANT ∂ f ∂ z f y ( x, y ) ∂ y , ∂ y ,

MATH 200 A LITTLE PRACTICE ▸ Compute both first-order partial derivatives (fancy way of saying first derivatives) for the following functions ▸ i.e. compute the partial derivative with respect to x and the partial derivative with respect to y for each function 1. f ( x, y ) = 3 x 2 − 2 y + 1 2. g ( x, y ) = xy 2 − ln y 3. h ( x, y ) = e x 2 + y 3 4. z = e x 2 y 3

MATH 200 WITH RESPECT TO X, THESE POWER RULE EXAMPLE 1 ARE CONSTANT TERMS f ( x, y ) = 3 x 2 − 2 y + 1 f x ( x, y ) = 3(2 x 1 ) − 0 + 0 WITH RESPECT TO Y, THESE = 6 x ARE CONSTANT TERMS f ( x, y ) = 3 x 2 − 2 y + 1 POWER RULE f y ( x, y ) = 0 − 2(1) + 0 = − 2

MATH 200 WITH RESPECT TO X, EXAMPLE 2 THIS WHOLE TERM IS g ( x, y ) = xy 2 − ln y CONSTANT g x ( x, y ) = (1) y 2 − 0 y 2 IS TREATED AS CONSTANT HERE, SO IT’S LIKE DIFFERENTIATING 5x = y 2 WITH RESPECT TO x x IS TREATED AS CONSTANT HERE, SO IT’S LIKE g ( x, y ) = xy 2 − ln y DIFFERENTIATING 5y 2 WITH RESPECT TO y g y ( x, y ) = x (2 y ) − 1 y = 2 xy − 1 y

MATH 200 THE DERIVATIVE OF EXAMPLE 3 e u(x) IS u’(x)e u(x) , SO IN TERMS OF PARTIAL h ( x, y ) = e x 2 + y 2 DERIVATIVES, WE SHOULD WRITE h x ( x, y ) = 2 xe x 2 + y 2 u x (x,y)e u(x,y) THE DERIVATIVE OF e u(x) IS u’(x)e u(x) , SO IN TERMS OF PARTIAL h ( x, y ) = e x 2 + y 2 DERIVATIVES, WE SHOULD WRITE h y ( x, y ) = 2 ye x 2 + y 2 u y (x,y)e u(x,y)

MATH 200 WE CAN WRITE “THE PARTIAL EXAMPLE 4 DERIVATIVE WITH RESPECT TO z = e x 2 y 3 x OF x 2 y 3 ” LIKE THIS: ∂ ∂ x ( x 2 y 3 ) = 2 xy 3 ∂ z ∂ x = 2 xy 3 e x 2 y 3 z = e x 2 y 3 ∂ z ∂ y = 3 x 2 y 2 e x 2 y 3

MATH 200 WHAT PARTIAL DERIVATIVES GIVE US ▸ Let’s look at f(x,y) = 3x 2 - 2y + 1 from Example 2 at the point (1,2) ▸ We found that f x (x,y) = 6x ▸ Evaluating the x partial at (1,2) we get f x (1,2) = 6(1) = 6 ▸ What does this 6 tell us? ▸ The rate of change of f (or z) in the x-direction at (1,2) is 6 ▸ The slope of the line tangent to the trace of f(x,y) on the plane y=2 is 6

MATH 200 � ∂ z � ⇒ ⟨ 1 , 0 , 6 ⟩ = 6 = � ∂ x � (1 , 0)  x = 1 + t   L : y = 2  z = 6 t 

MATH 200 HIGHER ORDER PARTIAL DERIVATIVES ▸ Consider the function z = 3x 2 - x 3 y 4 ▸ Let’s find the two first order partial derivatives: ∂ z ∂ z ∂ x = 6 x − 3 x 2 y 4 ; ∂ y = 4 x 3 y 3 ▸ We could now differentiate either of these with respect to x or with respect to y… ▸ …for a total of four second order partial derivatives � ∂ z � ∂ z = ∂ 2 z = ∂ 2 z � � ∂ ∂ ∂ x 2 ; ∂ y ∂ x ; ∂ x ∂ x ∂ y ∂ x = ∂ 2 z = ∂ 2 z � ∂ z � � ∂ z � ∂ ∂ ∂ y 2 ; ∂ x ∂ y ∂ x ∂ y ∂ y ∂ y

MATH 200 ∂ 2 z ∂ x 2 = 6 − 6 xy 4 NOTICE THAT THE SECOND ORDER ∂ 2 z ∂ y ∂ x = 12 x 2 y 3 MIXED PARTIALS ARE THE SAME! ∂ 2 z THIS WILL BE TRUE ∂ x ∂ y = 12 x 2 y 3 FOR ANY TIME THEY ARE BOTH CONTINUOUS ∂ 2 z ∂ y 2 = 12 x 3 y 2

MATH 200 IMPLICIT DIFFERENTIATION ▸ Recall from calc 1: x 4 + y 4 = xy TREAT Y AS AN dx ( x 4 + y 4 ) = d d dx ( xy ) IMPLICIT FUNCTION OF X 4 x 3 + 4 y 3 = y + xdy dx dx = 4 x 3 + 4 y 3 − y dy x

MATH 200 IMPLICIT DIFFERENTIATION WITH THREE VARIABLES Find ∂ z ∂ x xe y + z − 2 z 2 = 3 y + 1 TREAT Z AS A ∂ x ( xe y + z − 2 z 2 ) = ∂ ∂ FUNCTION OF X ∂ x (3 y + 1) AND Y AS A e y + z + x ∂ z ∂ xe y + z − 4 z ∂ z ∂ x = 0 CONSTANT − e y + z ∂ z ∂ x = xe y + z − 4 z

MATH 200 ONE MORE EXAMPLE ▸ Compute dy/dz for 3 xy 2 − ze y = 4 z 3 DIFFERENTIATE BOTH SIDES WITH ∂ z (3 xy 2 − ze y ) = ∂ ∂ ∂ z (4 z 3 ) RESPECT TO Z, TREATING Y AS A FUNCTION OF Z AND X AS A CONSTANT � � � � 2 y ∂ y e y + ze y ∂ y = 12 z 2 3 x PRODUCT RULE − ∂ z ∂ z 6 xy ∂ y ∂ z − e y − ze y ∂ y GET ALL OF THE DY/DZ TERMS ∂ z = 12 z 2 ON ONE SIDE AND FACTOR ∂ y ∂ z (6 xy − ze y ) = 12 z 2 + e y ∂ z = 12 z 2 + e y ∂ y SOLVE 6 xy − ze y

MATH 200 RECAP Trace on y=y 0 ▸ To compute the partial derivative of f with respect to x , we… ▸ Treat all other variables as constants ▸ Use all of the derivative rules we know from Calculus 1 3 +y 3 ▸ E.g. f(x,y) = x 2 ▸ f x (x,y) = 3x ▸ When we evaluate partial derivative of f with respect to x at a point (x 0 ,y 0 ), we get the slope of the line tangent to the trace of f on the plane y = y 0 (x 0 ,y 0 )