PARTIAL DERIVATIVES TANGENT PLANES Suppose a surface S has equation z = f(x, y), 15.4 where f has continuous first partial derivatives. Tangent Planes and Linear Approximations Let P(x0, y0, z0) be a point on S. In this section, we will learn how to: Approximate functions using tangent planes and linear functions. TANGENT PLANES Equation 2 TANGENT PLANES Example 1 Suppose f has continuous partial derivatives. Find the tangent plane to the elliptic paraboloid z = 2 x2 + y2 at the point (1, 1, 3). An equation of the tangent plane to the Let f(x, y) = 2 x2 + y2. surface z = f(x, y) at the point P(x0, y0, z0) is: Then, fx(x, y ) = 4 x fy(x, y ) = 2 y z – z0 = fx(x0, y0)( x – x0) + fy(x0, y0)( y – y0) fx(1, 1) = 4 fy(1, 1) = 2 TANGENT PLANES Example 1 TANGENT PLANES So, Equation 2 gives the equation The figure shows the elliptic paraboloid of the tangent plane at (1, 1, 3) as: and its tangent plane at (1, 1, 3) that we z – 3 = 4( x – 1) + 2( y – 1) found in Example 1. or z = 4 x + 2 y – 3 1 LINEAR APPROXIMATIONS LINEAR APPROXIMATIONS In Example 1, we found that an equation of Thus, in view of the visual evidence in the tangent plane to the graph of the function the previous two figures, the linear function f(x, y) = 2 x2 + y2 at the point (1, 1, 3) is: of two variables L(x, y) = 4 x + 2 y – 3 z = 4 x + 2 y – 3 is a good approximation to f(x, y) when ( x, y) is near (1, 1). LINEARIZATION & LINEAR APPROXIMATION LINEAR APPROXIMATIONS The function L is called the linearization of f For instance, at the point (1.1, 0.95), the linear at (1, 1). approximation gives: f(1.1, 0.95) The approximation ≈ 4(1.1) + 2(0.95) – 3 f(x, y) ≈ 4x + 2 y – 3 = 3.3 is called the linear approximation or tangent This is quite close to the true value plane approximation of f at (1, 1). of f(1.1, 0.95) = 2(1.1) 2 + (0.95) 2 = 3.3225 LINEAR APPROXIMATIONS LINEAR APPROXIMATIONS However, if we take a point farther away In general, we know from Equation 2 that from (1, 1), such as (2, 3), we no longer get an equation of the tangent plane to the graph a good approximation. of a function f of two variables at the point (a, b, f(a, b)) is: In fact, L(2, 3) = 11, whereas f(2, 3) = 17. z = f(a, b) + fx(a, b)( x – a) + fy(a, b)( y – b) 2 LINEARIZATION Equation 3 LINEAR APPROXIMATION Equation 4 The linear function whose graph is The approximation this tangent plane, namely f(x, y) ≈ f(a, b) + fx(a, b)( x – a) + f (a, b)( y – b) L(x, y) = f(a, b) + fx(a, b)( x – a) y + fy(a, b)( y – b) is called the linear approximation or is called the linearization of f at ( a, b). the tangent plane approximation of f at ( a, b). LINEAR APPROXIMATIONS Theorem 8 LINEAR APPROXIMATIONS Example 2 xy If the partial derivatives fx and fy exist Show that f(x, y) = xe is differentiable near ( a, b) and are continuous at ( a, b), at (1, 0) and find its linearization there. then f is differentiable at ( a, b). Then, use it to approximate f(1.1, –0.1). LINEAR APPROXIMATIONS Example 2 LINEAR APPROXIMATIONS Example 2 The partial derivatives are: The linearization is: L(x, y) = f(1, 0) + fx(1, 0)( x – 1) + fy(1, 0)( y – 0) xy xy 2 xy . fx(x, y) = e + xye fy(x, y) = x e = 1 + 1( x – 1) + 1 y fx(1, 0) = 1 fy(1, 0) = 1 = x + y Both fx and fy are continuous functions. So, f is differentiable by Theorem 8. 3 LINEAR APPROXIMATIONS Example 2 DIFFERENTIALS The corresponding linear approximation For a differentiable function of one variable, is: y = f(x), we define the differential dx to be xe xy ≈ x + y an independent variable. So, That is, dx can be given the value f(1.1, – 0.1) ≈ 1.1 – 0.1 = 1 of any real number. Compare this with the actual value of f(1.1, –0.1) = 1.1 e–0.11 ≈ 0.98542 DIFFERENTIALS Equation 9 DIFFERENTIALS Then, the differential of y is defined The figure shows the as: relationship between the increment ∆y and the differential dy . dy = f’(x) dx See Section 3.10 DIFFERENTIALS DIFFERENTIALS ∆y represents the change in height of For a differentiable function of two variables, the curve y = f(x). z = f(x, y), we define the differentials dx and dy represents the change in height of dy to be independent variables. the tangent line when x changes That is, they can be given any values. by an amount dx = ∆x. 4 TOTAL DIFFERENTIAL Equation 10 DIFFERENTIALS Then the differential dz , also called If we take dx = ∆x = x – a and dy = ∆y = y – b the total differential, is defined by: in Equation 10, then the differential of z ∂ ∂ is: = + =+z z dz fx(,) xydx f y (,) xydy dx dy dz = f (a, b)( x – a) + f (a, b)( y – b) ∂x ∂ y x y Compare with Equation 9. So, in the notation of differentials, the linear approximation in Equation 4 can be written as: Sometimes, the notation df is used in place of dz . f(x , y) ≈ f(a, b) + dz DIFFERENTIALS DIFFERENTIALS The figure is the three-dimensional It shows the geometric interpretation of counterpart of the previous figure. the differential dz and the increment ∆z. DIFFERENTIALS DIFFERENTIALS dz is the change in height of the tangent ∆z represents the change in height of plane. the surface z = f(x, y) when ( x, y) changes from ( a, b) to ( a + ∆x, b + ∆y). 5 DIFFERENTIALS Example 4 DIFFERENTIALS Example 4 a a. If z = f(x, y) = x2 + 3 xy – y2, find Definition 10 gives: the differential dz . ∂z ∂ z dz= dx + dy b. If x changes from 2 to 2.05 and y changes ∂x ∂ y from 3 to 2.96, compare ∆z and dz . =+(2x 3) ydx +− (3 x 2) ydy DIFFERENTIALS Example 4 b DIFFERENTIALS Example 4 b Putting The increment of z is: x = 2, dx = ∆x = 0.05, y = 3, dy = ∆y = –0.04, ∆z = f(2.05, 2.96) – f(2, 3) we get: = [(2.05) 2 + 3(2.05)(2.96) – (2.96) 2] – [2 2 + 3(2)(3) – 32] dz = [2(2) + 3(3)]0.05 + [3(2) – 2(3)](–0.04) = 0.6449 = 0.65 Notice that ∆z ≈ dz, but dz is easier to compute. DIFFERENTIALS DIFFERENTIALS Example 5 In Example 4, dz is close to ∆z because The base radius and height of a right circular the tangent plane is a good approximation cone are measured as 10 cm and 25 cm, to the surface z = x2 + 3 xy – y2 near (2, 3, 13). respectively, with a possible error in measurement of as much as 0.1 cm in each. Use differentials to estimate the maximum error in the calculated volume of the cone. 6 DIFFERENTIALS Example 5 DIFFERENTIALS Example 5 The volume V of a cone with base radius r Each error is at most 0.1 cm. and height h is V = πr2h/3. So, we have: So, the differential of V is: |∆r| ≤ 0.1 ∂V ∂ V2π rh π r 2 dV= dr + dh = dr + dh |∆h| ≤ 0.1 ∂r ∂ h 3 3 DIFFERENTIALS Example 5 DIFFERENTIALS Example 5 To find the largest error in the volume, That gives: we take the largest error in the measurement 500π 100 π of r and of h. dV =(0.1) + (0.1) 3 3 Therefore, we take dr = 0.1 and dh = 0.1 = 20 π along with r = 10, h = 25. So, the maximum error in the calculated volume is about 20 π cm 3 ≈ 63 cm 3. FUNCTIONS OF THREE OR MORE VARIABLES MULTIPLE VARIABLE FUNCTIONS Example 6 The differential dw is defined in terms of the The dimensions of a rectangular box are differentials dx , dy , and dz of the independent measured to be 75 cm, 60 cm, and 40 cm, variables by: and each measurement is correct to within 0.2 cm. ∂w ∂ w ∂ w dw= dx + dy + dz ∂x ∂ y ∂ z Use differentials to estimate the largest possible error when the volume of the box is calculated from these measurements. 7 MULTIPLE VARIABLE FUNCTIONS Example 6 MULTIPLE VARIABLE FUNCTIONS Example 6 If the dimensions of the box are x, y, and z, We are given that its volume is V = xyz. |∆x| ≤ 0.2, | ∆y| ≤ 0.2, | ∆z| ≤ 0.2 Thus, To find the largest error in the volume, ∂V ∂ V ∂ V dV= dx + dy + dz we use ∂ ∂ ∂ dx = 0.2, dy = 0.2, dz = 0.2 x y z together with =yz dx + xz dy + xy dz x = 75, y = 60, z = 40 MULTIPLE VARIABLE FUNCTIONS Example 6 MULTIPLE VARIABLE FUNCTIONS Example 6 Thus, So, an error of only 0.2 cm in measuring each dimension could lead to an error of as much ∆V ≈ dV as 1980 cm 3 in the calculated volume. = (60)(40)(0.2) + (75)(40)(0.2) + (75)(60)(0.2) This may seem like a large error. = 1980 However, it’s only about 1% of the volume of the box.