Math 5BI: Problem Set 2 The Chain Rule

Transcription

1 Math 5BI: Problem Set 2 The Chain Rule April 5, 2010 A Functions of two variables Suppose that γ(t) = (x(t), y(t), z(t)) is a differentiable parametrized curve in R 3 which lies on the surface S defined by the equation z = f(x, y), where f is a continuously differentiable function of two variables Thus z(t) = f(x(t), y(t)) (1) It is intuitively clear, and can be proven rigorously, that the velocity vector γ (t 0 ) is tangent to the surface S at γ(t 0 ), for any choice of t 0 Remark Since there are only 26 letters in the alphabet, scientists often run out of distinct letters to represent variables and functions and so on Thus one writes x = x(t), where the first x represents the variable x and the second x stands for a function of t This is one way of economizing on use of letters Another way of increasing the number of available letters is to use the Greek alphabet, including for example, the Greek letter gamma, γ Problem 21 Using the fact that the vector (/ x)(x(t 0 ), y(t 0 )) n = (/ y)(x(t 0 ), y(t 0 )) 1 is perpendicular to S at γ(t 0 ), show that ( ) ( ) z (t 0 ) = (x(t 0 ), y(t 0 ))x (t 0 ) + (x(t 0 ), y(t 0 ))y (t 0 ), (2) x y where the prime denotes the derivative with respect to t The formula (2) you derived in Problem 21 is called the chain rule It can be written in many different forms with various choices of notation If we leave out the constants t 0, x 0, and y 0, we can simplify (2) to dz dt = dx x dt + dy y dt (3) 1

2 If we use the notation z/ x for / x and z/ y for / y, we can write the chain rule in the form dz dt = z dx x dt + z dy y dt In this notation, we call t the independent variable, z the dependent variable, and x and y the intermediate variables All of these versions of the chain rule appear in books on physics and engineering Problem 22 Use the chain rule to answer the following question: The volume of a cylindrical can of radius r and height h is V = πr 2 h If at a certain time what is dv/dt? r = 2 ft, h = 5 ft, dr dt = 1 ft/sec, dh dt = 2 ft/sec, The parametrized curve γ(t) utilized in Problem 21 projects to a parametrized curve x(t) = (x(t), y(t)) in the (x, y)-plane Conversely given any smooth curve x(t) = (x(t), y(t)) in the (x, y)-plane we have a corresponding curve γ(t) on the surface S defined by the equation z = f(x, y), namely γ(t) = (x(t), y(t), z(t)), where z(t) = f(x(t), y(t)) Definition The gradient of a continuously differentiable function f(x, y) at the point (x 0, y 0 ) is ( ) f(x 0, y 0 ) = x (x 0, y 0 ) y (x 0, y 0 ) Problem 23 a Suppose that x(t) = ( x(t) y(t) is a parametrized curve in the (x, y)-plane, representing the trajectory of a moving particle in the plane, the variable t representing time Note that the velocity of the particle at time t 0 is just Use (2) to show that v(t 0 ) = x (t 0 ) = ), ( x (t 0 ) y (t 0 ) ) z (t 0 ) = f(x(t 0 ), y(t 0 )) x (t 0 ), where x (t 0 ) = ( x (t 0 ) y (t 0 ) b Suppose that f(x, y) represents the temperature at the point (x, y) Show that (rate of change of temperature with respect to t) = (gradient of f) (velocity) ) 2

3 c Let u = (u 1, u 2 ) be a unit-length vector (so u u = 1) and let (x 0, y 0 ) be a point in the plane R 2 Show that the speed of the parametrized curve x(t) = (x 0, y 0 ) + tu is one Thus x(t) has unit speed and starts at (x 0, y 0 ) at time t = 0 d We can think of a unit-length vector u = (u 1, u 2 ) as defining a direction in R 2 What is the rate of change of z = f(x, y) with respect to time t in the direction of u at the point (x 0, y 0 )? e Show that the gradient of f at (x 0, y 0 ) points in the direction of maximum increase of f, and its magnitude is the rate of change of f in this direction, if one moves from (x 0, y 0 ) with unit speed Problem 24 a Find a nonzero vector which is perpendicular to the curve x 4 + y 4 = 17 at the point (1, 2) b Find an equation for the line which is perpendicular to the curve x 4 +y 4 = 17 at the point (1, 2) c Find an equation for the line which is tangent to the curve x 4 + y 4 = 17 at the point (1, 2) B Functions of Three Variables The notion of linearization and the chain rule can be extended to functions of n variables, where n can be arbitrary In this problem set, we want to consider the case n = 3 Suppose that f(x, y, z) is a function of three variables When it exists, the partial derivative of f(x, y, z) with respect to x at (x 0, y 0, z 0 ) is given by the formula x (x f(x 0 + h, y 0, z 0 ) f(x 0, y 0, z 0 ) 0, y 0, z 0 ) = lim h 0 h (Notice the use of limits in the definition A rigorous treatment of limits is given in Math 117 or classes in the College of Creative Studies) The partial derivatives y (x 0, y 0, z 0 ) and z (x 0, y 0, z 0 ) are defined by similar formulae, which you could easily write down We say that the function f(x, y, z) is continuously differentiable or smooth if it has partial derivatives at every point, and the functions (x, y, z), x y (x, y, z) and (x, y, z) z are continuous 3

4 The gradient of a continuously differentiable function f(x, y, z) at the point (x 0, y 0, z 0 ) is the vector x (x 0, y 0, z 0 ) f(x 0, y 0, z 0 ) = y (x 0, y 0, z 0 ) z (x 0, y 0, z 0 ) If we let (x, y, z) vary, we get a function (x, y, z) R 3 ( f)(x, y, z) R 3 which is just called the gradient of f More generally, a function X : R 3 R 3 is often called a vector field Thus the gradient f of f is an example of a vector field Problem 25 a If f(x, y, z) = xyz, what is (/ y)(2, 1, 1)? b If f(x, y, z) = xe y cos z, what is the function (/ x)(x, y, z)? c If f(x, y, z) = x 2 + y 2 z 2, what is the gradient of f at the point (1, 3, 4)? d If f(x, y, z) = x + y 2 + z 2, does (/ x)(0, 0, 0) exist? Why or why not? e If f(x, y, z) = x 2 + y 2 z 2, what is the vector field f? f We can sketch the vector field f by drawing an arrow at each point (x, y, z) of R 3 in the direction of f(x, y, z) Sketch f, when f(x, y, z) = 1 4 x y z2 Just as in the case of two variables, the linearization of f(x, y, z) at (x 0, y 0, z 0 ) is the function L(x, y, z) = f(x 0, y 0, z 0 ) + x (x 0, y 0, z 0 )(x x 0 ) + y (x 0, y 0, z 0 )(y y 0 ) + z (x 0, y 0, z 0 )(z z 0 ) Just as in the case of functions of two variables, we expect the linearization to be very close approximation to the function near the point (x 0, y 0, z 0 ) Problem 26 a If f(x, y, z) = x 2 + y 2 + z 2, what is its linearization at the point (3, 1, 2)? b Use the linearization to find a close approximation to f at the point (3001, 999, 2002) 4

5 Suppose now that we have a smooth parametrized curve in R 3, x(t) x(t) = y(t) z(t) This might represent the position of a moving particle in R 3 at time t The velocity of the particle at time t would then be x (t) x (t) = y (t) z (t) We can take the composition of the functions f and x(t) obtaining a new function h(t) = (f x)(t) = f(x(t), y(t), z(t)) In this context, the chain rule states dh dt = dx x dt + dy y dt + dz z dt (4) The chain rule can be restated in vector form as where for any choice of t 0 x(t 0 ) = h (t 0 ) = f(x(t 0 )) x (t 0 ), x(t 0 ) y(t 0 ) z(t 0 ) and x (t 0 ) = Problem 27 a Sketch the parametrized curve x(t) = cos t sin t t x (t 0 ) y (t 0 ) z (t 0 ), b Does this parametrized curve lie in the surface x 2 + y 2 = 1? Problem 28 The level set of a continuously differentiable function f(x, y, z) is a set of points (x, y, z) which satisfy the equation f(x, y, z) = c Sketch the level sets of the function f(x, y, z) = x 2 + y 2 + z 2 Problem 29 a Suppose that the level set f(x, y, z) = c is a smooth surface S Use the chain rule to show that if this is the case, then f is perpendicular to S at any point of S b Find a vector perpendicular to the surface x 2 + y 2 z 2 = 1 at the point (1, 1, 1) 5

6 c Find an equation for the plane tangent to the surface x 2 + y 2 z 2 = 1 at the point (1, 1, 1) Homework 2 Due Friday, April 9, 2010 This problem set is intended to explain by means of an example why the chain rule is so important Read the following discussion and do the calculations that are requested using the chain rule if needed According to Newton s second law of motion, a force F acting on a body of mass m gives it an acceleration a = (1/m)F Once the force F(x, y, z) = F x (x, y, z)i + F y (x, y, z)j + F z (x, y, z)k is given, Newton s law of motion yields a system of second order ordinary differential equations d 2 x/dt 2 = (1/m)F x (x, y, z), d 2 y/dt 2 = (1/m)F y (x, y, z), d 2 z/dt 2 = (1/m)F z (x, y, z) (5) According to Newton s law of gravitation, the gravitational force between a spherically symmetric star of mass M and a small planet of mass m has a magnitude which is directly proportional to the product M m of the masses and inversely proportional to the square of the distance between the center of the star and the planet Thus if the center of the star is located at the origin of the (x, y, z)-coordinates and the planet is located at the point (x, y, z), the magnitude of the gravitational force is F = GMm x 2 + y 2 + z 2, where G is a universal constant, known as Newton s gravitational constant The force acting on the planet is directed toward the sun, and hence the components of the force vector field are F x (x, y, z) = GMm x x 2 +y 2 +z, 2 x 2 +y 2 +z 2 F y (x, y, z) = GMm x 2 +y 2 +z 2 F z (x, y, z) = GMm x 2 +y 2 +z 2 y, x 2 +y 2 +z 2 (6) z x2 +y 2 +z 2 We can substitute (6) into (5) to obtain the differential equations for the Kepler problem of planetary motion It can be shown that if the initial position vector x(0) and initial velocity v(0) of the planet are linearly independent, then the orbit lies entirely within the plane spanned by x(0) and v(0) We will therefore assume without loss of generality that the entire orbit of the planet lies in the (x, y)-plane In other 6

7 words, we will set z = 0, and our system of differential equations will simplify to d 2 x/dt 2 = (1/m)F x (x, y), d 2 y/dt 2 (7) = (1/m)F y (x, y), where F x (x, y) = GMm x x 2 +y, 2 x 2 +y 2 F y (x, y) = GMm y x 2 +y (8) 2 x2 +y 2 We would expect the general solution to this system to have four constants of integration; in other words, we need to perform four integrations to determine the solution Two of these integrations come from conservation laws conservation of energy and angular momentum The idea is to use these conservation to reduce the second-order system to a first-order system of differential equations, much like the equations we have studied earlier in the course Conservation of energy: A vector field F = F x (x, y)i + F y (x, y)j is said to be conservative if there is a function V (x, y) such that F = V H21 Show that the vector field F defined by (7) is conservative with V (x, y) = GMm x2 + y 2 H22 Use the chain rule and (7) to calculate d [ (m/2)[(dx/dt) 2 + (dy/dt) 2 ] + V (x, y) ] dt We can think of the expression in brackets as where (total energy) = (kinetic energy) + (potential energy), (kinetic energy) = (m/2)[(dx/dt) 2 + (dy/dt) 2 ] and (potential energy) = V (x, y) What does your calculation tell you about the total energy? (Try to express yourself clearly) Conservation of angular momentum: In the terminology of physics, mx(dy/dt) my(dx/dt) = L is the angular momentum of the planet aroung the z-axis H23 Use (7) to calculate dl/dt What does this tell you about angular momentum? 7