Calculus II Fall PDF Free Download

Calculus II Fall 2014 Lecture 3 Partial Derivatives Eitan Angel University of Colorado Monday, December 1, 2014 E. Angel (CU) Calculus II 1 Dec 1 / 13

Introduction Much of the calculus of several variables is basically single-variable calculus applied to several variables one at a time. E. Angel (CU) Calculus II 1 Dec 2 / 13

Introduction Much of the calculus of several variables is basically single-variable calculus applied to several variables one at a time. If we fix all but one of the independent variables to a constants, and differentiate that variable, we get a partial derivative. E. Angel (CU) Calculus II 1 Dec 2 / 13

Mortgage Payments Suppose that you are interested in buying a house. You will certainly want to know what your monthly mortgage payment would be at various interest rates. E. Angel (CU) Calculus II 1 Dec 3 / 13

Mortgage Payments Suppose that you are interested in buying a house. You will certainly want to know what your monthly mortgage payment would be at various interest rates.looking at a table for a 30-year mortgage you find the following Monthly payments per $1000 (in $) rate (r) 2.00% 2.50% 3.00% 3.50% 4.00% 4.50% 5.00% 3.69 3.95 4.21 4.49 4.77 5.06 5.36 How do monthly payments change with respect to interest rates? We can estimate the derivative at, say r = 3.0%: P P (3.0 + h) P (3.0) (3.0) = lim P h 0 h r 0.56. = r=3.0 P (3.5) P (3.0) 3.5 3.0 E. Angel (CU) Calculus II 1 Dec 3 / 13

Mortgage Payments Now suppose we think we may be able to pay off our mortgage faster than 30 years. E. Angel (CU) Calculus II 1 Dec 4 / 13

Mortgage Payments Now suppose we think we may be able to pay off our mortgage faster than 30 years. Considering various mortgage terms as well, we find the following table describing P (r, t). Monthly payments per $1000 (in $) 10 9.20 9.43 9.66 9.89 10.12 10.36 10.61 15 6.43 6.66 6.90 7.14 7.39 7.64 7.90 20 5.06 5.30 5.55 5.80 6.06 6.33 6.60 25 4.24 4.49 4.74 5.01 5.28 5.56 5.85 30 3.69 3.95 4.21 4.49 4.77 5.06 5.36 t / r 2.00% 2.50% 3.00% 3.50% 4.00% 4.50% 5.00% E. Angel (CU) Calculus II 1 Dec 4 / 13

Mortgage Payments Monthly payments per $1000 (in $) 10 9.20 9.43 9.66 9.89 10.12 10.36 10.61 15 6.43 6.66 6.90 7.14 7.39 7.64 7.90 20 5.06 5.30 5.55 5.80 6.06 6.33 6.60 25 4.24 4.49 4.74 5.01 5.28 5.56 5.85 30 3.69 3.95 4.21 4.49 4.77 5.06 5.36 t / r 2.00% 2.50% 3.00% 3.50% 4.00% 4.50% 5.00% P t (4.5, 15) := P (4.5, t) t=15 = lim h 0 P (4.5, 15 + h) P (4.5, 15) h E. Angel (CU) Calculus II 1 Dec 5 / 13

Mortgage Payments Monthly payments per $1000 (in $) 10 9.20 9.43 9.66 9.89 10.12 10.36 10.61 15 6.43 6.66 6.90 7.14 7.39 7.64 7.90 20 5.06 5.30 5.55 5.80 6.06 6.33 6.60 25 4.24 4.49 4.74 5.01 5.28 5.56 5.85 30 3.69 3.95 4.21 4.49 4.77 5.06 5.36 t / r 2.00% 2.50% 3.00% 3.50% 4.00% 4.50% 5.00% P t (4.5, 15) := P P (4.5, 15 + h) P (4.5, 15) (4.5, t) t=15 = lim h 0 h P P (4.5, 20) P (4.5, 15) t = (4.5,15) 20 15 E. Angel (CU) Calculus II 1 Dec 5 / 13

Mortgage Payments Monthly payments per $1000 (in $) 10 9.20 9.43 9.66 9.89 10.12 10.36 10.61 15 6.43 6.66 6.90 7.14 7.39 7.64 7.90 20 5.06 5.30 5.55 5.80 6.06 6.33 6.60 25 4.24 4.49 4.74 5.01 5.28 5.56 5.85 30 3.69 3.95 4.21 4.49 4.77 5.06 5.36 t / r 2.00% 2.50% 3.00% 3.50% 4.00% 4.50% 5.00% P t (4.5, 15) := P P (4.5, 15 + h) P (4.5, 15) (4.5, t) t=15 = lim h 0 h P P (4.5, 20) P (4.5, 15) t = (4.5,15) 20 15 = 6.33 7.64 5 = 0.262 E. Angel (CU) Calculus II 1 Dec 5 / 13

Partial Derivatives of Two Variable Functions Let (x 0, y 0 ) be a point in the domain of a function f(x, y). E. Angel (CU) Calculus II 1 Dec 6 / 13

Partial Derivatives of Two Variable Functions Let (x 0, y 0 ) be a point in the domain of a function f(x, y). The vertical plane y = y 0 will cut the surface z = f(x, y) in the curve z = f(x, y 0 ). y is held constant at value y 0, so y is not a variable. We define the partial derivative of f with respect to x at the (x 0, y 0 ) as the ordinary derivative of f(x, y 0 ) with respect to x at the point x = x 0. E. Angel (CU) Calculus II 1 Dec 6 / 13

The Definition Definition The partial derivative of f(x, y) with respect to x at the point (x 0, y 0 ) is f f(x 0 + h, y 0 ) f(x 0, y 0 ) x = lim, (x0,y 0 ) h 0 h provided the limit exists. E. Angel (CU) Calculus II 1 Dec 7 / 13

The Definition Definition The partial derivative of f(x, y) with respect to x at the point (x 0, y 0 ) is f f(x 0 + h, y 0 ) f(x 0, y 0 ) x = lim, (x0,y 0 ) h 0 h provided the limit exists. Notice that to distinguish partial derivatives from ordinary derivatives, we use the symbol rather than d. An equivalent expression is d dx f(x, y 0) Some other notations are f x (x 0, y 0 ) f x (x 0, y 0 ) x=x0 z x (x0,y 0 ) E. Angel (CU) Calculus II 1 Dec 7 / 13

Partial Derivative wrt y Let (x 0, y 0 ) be a point in the domain of a function f(x, y). E. Angel (CU) Calculus II 1 Dec 8 / 13

Partial Derivative wrt y Let (x 0, y 0 ) be a point in the domain of a function f(x, y). The vertical plane x = x 0 will cut the surface z = f(x, y) in the curve z = f(x 0, y). E. Angel (CU) Calculus II 1 Dec 8 / 13

Partial Derivative wrt y Let (x 0, y 0 ) be a point in the domain of a function f(x, y). The vertical plane x = x 0 will cut the surface z = f(x, y) in the curve z = f(x 0, y). x is held constant at value x 0, so x is not a variable. We define the partial derivative of f with respect to y at the (x 0, y 0 ) as the ordinary derivative of f(x 0, y) with respect to y at the point y = y 0. E. Angel (CU) Calculus II 1 Dec 8 / 13

The Definition Definition The partial derivative of f(x, y) with respect to y at the point (x 0, y 0 ) is f f(x 0, y 0 + h) f(x 0, y 0 ) y = lim, (x0,y 0 ) h 0 h provided the limit exists. E. Angel (CU) Calculus II 1 Dec 9 / 13

The Definition Definition The partial derivative of f(x, y) with respect to y at the point (x 0, y 0 ) is f f(x 0, y 0 + h) f(x 0, y 0 ) y = lim, (x0,y 0 ) h 0 h provided the limit exists. An equivalent expression is d dy f(x 0, y) y=y0 Some other notations are f y (x 0, y 0 ) f y (x 0, y 0 ) z y (x0,y 0 ) E. Angel (CU) Calculus II 1 Dec 9 / 13

Picture of f x and f y at (x 0, y 0 ) The two previous definitions give two tangent lines associated with the surface z = f(x, y) at the point P (x 0, y 0, f(x 0, y 0 )). What kind of shape is determined by two intersecting lines? E. Angel (CU) Calculus II 1 Dec 10 / 13

Partial Derivatives at a Point Example Find the values of f/ x and f/ y at the point (4, 5) if f(x, y) = x 2 + 3xy + y 1 E. Angel (CU) Calculus II 1 Dec 11 / 13

Partial Derivatives at a Point Example Find the values of f/ x and f/ y at the point (4, 5) if f(x, y) = x 2 + 3xy + y 1 To find f/ x, we treat y as a constant and differentiate with respect to x: f x = x (x2 + 3xy + y 1) = 2x + 3 1 y + 0 0 = 2x + 3y. So f/ x at (4, 5) is 2(4) + 3( 5) = 7. To find f/ y, we treat x as a constant and differentiate with respect to y: f y = y (x2 + 3xy + y 1) E. Angel (CU) Calculus II 1 Dec 11 / 13

Partial Derivative as a Function Example Find the function f/ y if f(x, y) = y sin xy. E. Angel (CU) Calculus II 1 Dec 12 / 13

Partial Derivative as a Function Example Find the function f/ y if f(x, y) = y sin xy. and f as a product of y and sin xy: We treat x as a constant f y = (y sin xy) y = y y sin xy + (sin xy) y (y) E. Angel (CU) Calculus II 1 Dec 12 / 13

Partial Derivative as a Function Example Find the function f/ y if f(x, y) = y sin xy. and f as a product of y and sin xy: We treat x as a constant f y = (y sin xy) y = y y sin xy + (sin xy) y (y) = (y cos xy) (xy) + sin xy y E. Angel (CU) Calculus II 1 Dec 12 / 13

Slope of a Surface in the y-direction Example The plane x = 1 intersects the paraboloid z = x 2 + y 2 in a parabola. Find the slope of the tangent line to the parabola at (1, 2, 5). E. Angel (CU) Calculus II 1 Dec 13 / 13

Slope of a Surface in the y-direction Example The plane x = 1 intersects the paraboloid z = x 2 + y 2 in a parabola. Find the slope of the tangent line to the parabola at (1, 2, 5). The slope is the value of the partial derivative z/ y at (1, 2): z y = (1,2) y (x2 + y 2 ) = 2y (1,2) = 4 (1,2) Alternatively, we can think of the parabola as the graph of a single-variable function z = 1 + y 2 in the plane x = 1, and find the slope at y = 2. Now we can calculate the ordinary derivative: dz dy = d y=2 dy (1 + y2 ) = 2y y=2 = 4. y=2 E. Angel (CU) Calculus II 1 Dec 13 / 13