Math 241: Multivariable calculus, Lecture 11 Implicit Differentiation, Directional Derivative and Gradient. Section 14.5, 14.6 go.illinois.edu/math241fa17 Friday, Sept 22nd, 2017 Exam I Monday Evening, No class on Wednesday go.illinois.edu/math241fa17. • y = s + t, 2 2 • z = es +t . Math 241: Problem of the day Use the chain rule to calculate @w (1; 0); @t 2 y where w = x + z2 , • x = x(s; t) = s cos(t), go.illinois.edu/math241fa17. 2 2 • z = es +t . Math 241: Problem of the day Use the chain rule to calculate @w (1; 0); @t 2 y where w = x + z2 , • x = x(s; t) = s cos(t), • y = s + t, go.illinois.edu/math241fa17. Math 241: Problem of the day Use the chain rule to calculate @w (1; 0); @t 2 y where w = x + z2 , • x = x(s; t) = s cos(t), • y = s + t, 2 2 • z = es +t . go.illinois.edu/math241fa17. n @f @ X @f @xi = f (x1(t1;:::; tk );:::; xn(t1;:::; tk )) = : @tj @tj @xi @tj i=1 Chain rule For a function f (x1;:::; xn) and functions x1 = x1(t1;:::; tk );:::; xn = xn(t1;:::; tk ) the chain rule says: go.illinois.edu/math241fa17. Chain rule For a function f (x1;:::; xn) and functions x1 = x1(t1;:::; tk );:::; xn = xn(t1;:::; tk ) the chain rule says: n @f @ X @f @xi = f (x1(t1;:::; tk );:::; xn(t1;:::; tk )) = : @tj @tj @xi @tj i=1 go.illinois.edu/math241fa17. Example 0.1 Suppose that x; y are related by x3 − y 3 − 6xy = 0: dy Assume y = y(x) and find dx . Usually you can solve for y, so that y = y(x) satisfying F (x; y(x)) = 0: Now take the derivative wrt x on both sides and use the chain rule: @F dx @F dy dy @F + = 0 =) = − @x : @x dx @y dx dx @F @y Implicit Differentiation. Suppose x; y are related by an equation F (x; y) = 0 go.illinois.edu/math241fa17. Example 0.1 Suppose that x; y are related by x3 − y 3 − 6xy = 0: dy Assume y = y(x) and find dx . Now take the derivative wrt x on both sides and use the chain rule: @F dx @F dy dy @F + = 0 =) = − @x : @x dx @y dx dx @F @y Implicit Differentiation. Suppose x; y are related by an equation F (x; y) = 0 Usually you can solve for y, so that y = y(x) satisfying F (x; y(x)) = 0: go.illinois.edu/math241fa17. Example 0.1 Suppose that x; y are related by x3 − y 3 − 6xy = 0: dy Assume y = y(x) and find dx . Implicit Differentiation. Suppose x; y are related by an equation F (x; y) = 0 Usually you can solve for y, so that y = y(x) satisfying F (x; y(x)) = 0: Now take the derivative wrt x on both sides and use the chain rule: @F dx @F dy dy @F + = 0 =) = − @x : @x dx @y dx dx @F @y go.illinois.edu/math241fa17. Implicit Differentiation. Suppose x; y are related by an equation F (x; y) = 0 Usually you can solve for y, so that y = y(x) satisfying F (x; y(x)) = 0: Now take the derivative wrt x on both sides and use the chain rule: @F dx @F dy dy @F + = 0 =) = − @x : @x dx @y dx dx @F @y Example 0.1 Suppose that x; y are related by x3 − y 3 − 6xy = 0: dy Assume y = y(x) and find dx . go.illinois.edu/math241fa17. Usually you can solve z in terms of x; y, giving a function z = z(x; y). So z has partial derivatives with respect to x; y. Take partial derivative first with respect to x and use the chain rule: @F @x @F @y @F @z + + = 0 @x @x @y @x @z @x @x @y Now @x = 1; @x = 0, so that we find @F @z @F @z = − @x ; same way: = − @y @x @F @y @F @z @z Implicit Differentiation for more variables Now assume that x; y; z are related by F (x; y; z) = 0: go.illinois.edu/math241fa17. So z has partial derivatives with respect to x; y. Take partial derivative first with respect to x and use the chain rule: @F @x @F @y @F @z + + = 0 @x @x @y @x @z @x @x @y Now @x = 1; @x = 0, so that we find @F @z @F @z = − @x ; same way: = − @y @x @F @y @F @z @z Implicit Differentiation for more variables Now assume that x; y; z are related by F (x; y; z) = 0: Usually you can solve z in terms of x; y, giving a function z = z(x; y). go.illinois.edu/math241fa17. Take partial derivative first with respect to x and use the chain rule: @F @x @F @y @F @z + + = 0 @x @x @y @x @z @x @x @y Now @x = 1; @x = 0, so that we find @F @z @F @z = − @x ; same way: = − @y @x @F @y @F @z @z Implicit Differentiation for more variables Now assume that x; y; z are related by F (x; y; z) = 0: Usually you can solve z in terms of x; y, giving a function z = z(x; y). So z has partial derivatives with respect to x; y. go.illinois.edu/math241fa17. @x @y Now @x = 1; @x = 0, so that we find @F @z @F @z = − @x ; same way: = − @y @x @F @y @F @z @z Implicit Differentiation for more variables Now assume that x; y; z are related by F (x; y; z) = 0: Usually you can solve z in terms of x; y, giving a function z = z(x; y). So z has partial derivatives with respect to x; y. Take partial derivative first with respect to x and use the chain rule: @F @x @F @y @F @z + + = 0 @x @x @y @x @z @x go.illinois.edu/math241fa17. @F @z @F @z = − @x ; same way: = − @y @x @F @y @F @z @z Implicit Differentiation for more variables Now assume that x; y; z are related by F (x; y; z) = 0: Usually you can solve z in terms of x; y, giving a function z = z(x; y). So z has partial derivatives with respect to x; y. Take partial derivative first with respect to x and use the chain rule: @F @x @F @y @F @z + + = 0 @x @x @y @x @z @x @x @y Now @x = 1; @x = 0, so that we find go.illinois.edu/math241fa17. @F @z same way: = − @y @y @F @z Implicit Differentiation for more variables Now assume that x; y; z are related by F (x; y; z) = 0: Usually you can solve z in terms of x; y, giving a function z = z(x; y). So z has partial derivatives with respect to x; y. Take partial derivative first with respect to x and use the chain rule: @F @x @F @y @F @z + + = 0 @x @x @y @x @z @x @x @y Now @x = 1; @x = 0, so that we find @z @F = − @x ; @x @F @z go.illinois.edu/math241fa17. Implicit Differentiation for more variables Now assume that x; y; z are related by F (x; y; z) = 0: Usually you can solve z in terms of x; y, giving a function z = z(x; y). So z has partial derivatives with respect to x; y. Take partial derivative first with respect to x and use the chain rule: @F @x @F @y @F @z + + = 0 @x @x @y @x @z @x @x @y Now @x = 1; @x = 0, so that we find @F @z @F @z = − @x ; same way: = − @y @x @F @y @F @z @z go.illinois.edu/math241fa17. Example 0.2 Suppose that z is given implicitly as a function z = z(x; y) by the equation: x2 + 2y 2 + 3z2 = 1: p Knowing that z(0; 0) = 1= 3, find: @z @z ; : @x @y go.illinois.edu/math241fa17. 2 f : R ! R and a UNIT vector u, how does f change in the direction u at the point (a; b)? Define the directional derivative of f in the direction u at the point (a; b) to be d Duf (a; b) = f (a + tu1; b + tu2) dt t=0 here u = hu1; u2i. (Evaluate the derivative at t = 0.) Directional derivatives go.illinois.edu/math241fa17. Define the directional derivative of f in the direction u at the point (a; b) to be d Duf (a; b) = f (a + tu1; b + tu2) dt t=0 here u = hu1; u2i. (Evaluate the derivative at t = 0.) Directional derivatives 2 f : R ! R and a UNIT vector u, how does f change in the direction u at the point (a; b)? go.illinois.edu/math241fa17. Directional derivatives 2 f : R ! R and a UNIT vector u, how does f change in the direction u at the point (a; b)? Define the directional derivative of f in the direction u at the point (a; b) to be d Duf (a; b) = f (a + tu1; b + tu2) dt t=0 here u = hu1; u2i. (Evaluate the derivative at t = 0.) go.illinois.edu/math241fa17. Example: What are Dif (a; b) and Djf (a; b)? In general, for u = hu1; u2i, Duf (a; b) = ? Duf (a; b) = fx (a; b)u1 + fy (a; b)u2 Define the gradient of f at (a; b) as the vector with partials as components: rf (a; b) = hfx (a; b); fy (a; b)i: Then the directional derivative is a dot product: Duf (a; b) = rf (a; b) · u.