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 .
MATH 200 WEEK 4 - MONDAY
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) = x2 − 1 f ′(x) = 2x f ′(1) = 2(1) = 2 f(1) = (1)2 − 1 = 0 THE POINT (1,0) IS ON THE GRAPH OF f THE SLOPE OF THE LINE TANGENT TO F AT (1,0) IS 2
MATH 200
CONSIDER A 3D EXAMPLE
▸ Let f(x,y) = x2 + y2 ▸ Consider the trace of f on the plane y=1 ▸ f(x,1) = x2 + 1
MATH 200
▸ We can certainly find slope of the line tangent to z = x2+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
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
for which z/x = 2 and y=0…
▸ <1,0,2> works! (There are
infinitely many other choices
L : x = 1 + t y = 1 z = 2 + 2t
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) = x2 + 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 = x2+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)
L : x = 1 + t y = 2 z = 5 + 2t
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…
▸ treat y as a constant… ▸ and differentiate with respect to x ▸ We write any of the following:
∂f ∂x, ∂z ∂x, fx(x, y)
WE USE THIS PARTIAL SYMBOL INSTEAD OF JUST d TO INDICATE THAT THERE IS MORE THAN ONE INDEPENDENT VARIABLE
MATH 200
DEFINITIONS
▸ The partial derivative of f with respect to y is what you get
when you…
▸ treat x as a constant… ▸ and differentiate with respect to y ▸ We write any of the following:
NOTICE: WE’RE NOT USING “PRIME” NOTATION ANYMORE… IF I WRITE f’(x,y), YOU DON’T KNOW WHICH VARIABLE I’M HOLDING CONSTANT
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
MATH 200
EXAMPLE 1
f(x, y) = 3x2 − 2y + 1 fx(x, y) = 3(2x1) − 0 + 0 = 6x
f(x, y) = 3x2 − 2y + 1 fy(x, y) = 0 − 2(1) + 0 = −2
WITH RESPECT TO X, THESE ARE CONSTANT TERMS POWER RULE WITH RESPECT TO Y, THESE ARE CONSTANT TERMS POWER RULE
MATH 200
EXAMPLE 2
g(x, y) = xy2 − ln y gx(x, y) = (1)y2 − 0 = y2
g(x, y) = xy2 − ln y gy(x, y) = x(2y) − 1 y = 2xy − 1 y
WITH RESPECT TO X, THIS WHOLE TERM IS CONSTANT y2 IS TREATED AS CONSTANT HERE, SO IT’S LIKE DIFFERENTIATING 5x WITH RESPECT TO x x IS TREATED AS CONSTANT HERE, SO IT’S LIKE DIFFERENTIATING 5y2 WITH RESPECT TO y
MATH 200
EXAMPLE 3
h(x, y) = ex2+y2 hx(x, y) = 2xex2+y2 h(x, y) = ex2+y2 hy(x, y) = 2yex2+y2
THE DERIVATIVE OF eu(x) IS u’(x)eu(x), SO IN TERMS OF PARTIAL DERIVATIVES, WE SHOULD WRITE ux(x,y)eu(x,y)
THE DERIVATIVE OF eu(x) IS u’(x)eu(x), SO IN TERMS OF PARTIAL DERIVATIVES, WE SHOULD WRITE uy(x,y)eu(x,y)
MATH 200
EXAMPLE 4
z = ex2y3 ∂z ∂x = 2xy3ex2y3
z = ex2y3 ∂z ∂y = 3x2y2ex2y3
∂ ∂x(x2y3) = 2xy3
WE CAN WRITE “THE PARTIAL DERIVATIVE WITH RESPECT TO x OF x2y3” LIKE THIS:
MATH 200
WHAT PARTIAL DERIVATIVES GIVE US
▸ Let’s look at f(x,y) = 3x2 - 2y + 1 from Example 2 at the
point (1,2)
▸ We found that fx(x,y) = 6x ▸ Evaluating the x partial at (1,2) we get fx(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 ∂x
= 6 = ⇒ ⟨1, 0, 6⟩
L : x = 1 + t y = 2 z = 6t
MATH 200
HIGHER ORDER PARTIAL DERIVATIVES
▸ Consider the function z = 3x2 - x3y4 ▸ Let’s find the two first order partial derivatives:
∂z ∂x = 6x − 3x2y4; ∂z ∂y = 4x3y3
▸ 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 ∂ ∂x ∂z ∂x
∂x2 ; ∂ ∂y ∂z ∂x
∂y∂x; ∂ ∂x ∂z ∂y
∂x∂y ∂ ∂y ∂z ∂y
∂y2 ;
MATH 200
∂2z ∂x2 = 6 − 6xy4
∂2z ∂y∂x = 12x2y3 ∂2z ∂x∂y = 12x2y3 ∂2z ∂y2 = 12x3y2
NOTICE THAT THE SECOND ORDER MIXED PARTIALS ARE THE SAME! THIS WILL BE TRUE FOR ANY TIME THEY ARE BOTH CONTINUOUS
MATH 200
IMPLICIT DIFFERENTIATION
▸ Recall from calc 1:
x4 + y4 = xy d dx(x4 + y4) = d dx(xy) 4x3 + 4y3 = y + xdy dx dy dx = 4x3 + 4y3 − y x
TREAT Y AS AN IMPLICIT FUNCTION OF X
MATH 200
IMPLICIT DIFFERENTIATION WITH THREE VARIABLES
Find ∂z ∂x xey+z − 2z2 = 3y + 1 ∂ ∂x(xey+z − 2z2) = ∂ ∂x(3y + 1) ey+z + x ∂z ∂xey+z − 4z ∂z ∂x = 0
∂z ∂x = −ey+z xey+z − 4z
TREAT Z AS A FUNCTION OF X AND Y AS A CONSTANT
MATH 200
ONE MORE EXAMPLE
▸ Compute dy/dz for 3xy2 − zey = 4z3
∂ ∂z (3xy2 − zey) = ∂ ∂z (4z3) 3x
∂z
∂z
6xy ∂y ∂z − ey − zey ∂y ∂z = 12z2 ∂y ∂z (6xy − zey) = 12z2 + ey ∂y ∂z = 12z2 + ey 6xy − zey
DIFFERENTIATE BOTH SIDES WITH RESPECT TO Z, TREATING Y AS A FUNCTION OF Z AND X AS A CONSTANT PRODUCT RULE GET ALL OF THE DY/DZ TERMS ON ONE SIDE AND FACTOR SOLVE
MATH 200
RECAP
▸ 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
▸ E.g. f(x,y) = x
3+y 3
▸ fx(x,y) = 3x
2
▸ When we evaluate partial derivative
we get the slope of the line tangent to the trace of f on the plane y = y0
(x0,y0) Trace on y=y0