Announcements. Related Rates (last week), Linear approximations (today) l Hôpital s Rule (today) Newton s Method Curve sketching Optimization problems

Transcription

1 Announcements Assignment 4 is now posted. Midterm results should be available by the end of the week (assuming the scantron results are back in time). Today: Continuation of applications of derivatives: Related Rates (last week), Linear approximations (today) l Hôpital s Rule (today) Newton s Method Curve sketching Optimization problems AMAT 217 (University of Calgary) Fall / 18

2 An observation Draw the tangent line to f (x) at x = a: For values of x close to a: f (x) and the tangent line are very close to each other. For values of x far from a: f (x) and the tangent line are far from each other. AMAT 217 (University of Calgary) Fall / 18

3 We also use the terminology tangent line approximation and linear approximation. AMAT 217 (University of Calgary) Fall / 18 Tangent lines and linear approximations Geometric Interpretation of Derivative f (x) equals is the slope of the tangent line to the graph at the point (x, f (x)). Recall the point-slope formula for a straight line: y y 1 = m(x x 1 ). Using m = f (a) and (x, y) = (a, f (a)) we can write the equation of tangent line: Equation of Tangent Line at x=a y = f (a) + f (a)(x a). The previous observation is the the motivation behind the concept of linearizations. In particular, we call the tangent line formula a linearization of the function f (x) at x = a: Linearization of f(x) at x=a L(x) = f (a) + f (a)(x a). By the observation, for values of x close to a, the linearization provides a good approximation for the function. Thus, we have: For x near a: f (x) L(x).

4 Example Determine the linear approximation for f (x) = 3 x at x = 8. Use it to approximate Solution: We simply find the equation of a tangent line (and give it the name L(x)). The slope of the tangent line to f (x) at x = 8 is f (8). f (x) = 1 3 x 2/3 = 1 3x 2/3 Therefore, f (8) = 1 3(8 2/3 ) = 1 3( 3 8) = L(x) = f (8) + f (8)(x 8) = (x 8). 12 To approximate we use the fact that when x is close to 8, the tangent line gives a good approximation to the function: = f (8.1) L(8.1) = 2 + (8.1 8) = = In other words, we plug 8.1 into the equation of the tangent line to obtain the approximation for f (8.1) = AMAT 217 (University of Calgary) Fall / 18

5 Question Is the approximation an overestimate or underestimate? 120 Answer with and without using a calculator. Solution: First, by using a calculator we see that the approximation gives while the actual value is equal to , = Thus, the approximation is an overestimate of the actual value. AMAT 217 (University of Calgary) Fall / 18

6 Without a calculator: We plot the function f (x) = 3 x and the tangent line at x = 8: Based on the shape of the graph, we can conclude that the approximation for is larger than the exact value, since the tangent line lies above the function for values of x close to 8. Hence, it is an overestimate. AMAT 217 (University of Calgary) Fall / 18

7 Example Determine the linear approximation for f (x) = sin x at x = 0. Solution: We must compute the equation of the tangent line to sin x at x = 0. First we compute f (0) to get: f (0) = sin 0 = 0 Now taking the derivative of f (x) = sin x gives: f (x) = cos x Hence, m = f (a) gives: f (0) = cos 0 = 1 Therefore, the linear approximation is: L(x) = f (0) + f (0)(x 0) = 0 + (1)(x 0) = x Hence, for values of x close to 0, we have sin x x. Note: The approximation sin x x (for x close to 0) is a very important one. It is frequently used in optics to simplify formulas, and helps describe the motion of pendulums and vibrations in springs. AMAT 217 (University of Calgary) Fall / 18

8 Better approximations (Not covered in this course) We can obtain better approximations to f (x) by using higher degree polynomials. We call these approximations Taylor polynomials, and denote by T n(x) (degree n). f (x) = e x T 0 (x) = 1 T 1 (x) = 1 + x T 2 (x) = 1 + x x 2 T 3 (x) = 1 + x x x 3 T 4 (x) = 1 + x x x x 4 AMAT 217 (University of Calgary) Fall / 18

9 L Hôpital s Rule Uses derivative to compute its. Discovered by Bernoulli (but purchased by l Hôpital). Also spelled l Hospital but pronounced low-pea-tahl not la-hospital. L Hôpital s Rule Suppose we have: where a is any real number (or ± ). Then, f (x) x a g(x) = 0 ± or 0 ±, f (x) x a g(x) f (x) x a g (x). That is, if it is type 0/0 or ± / ± then all we need to do is differentiate the numerator and differentiate the denominator, then take the it. We call a it of type 0/0 or ± / ± an indeterminate form. Caution 1: Don t use the quotient rule on the function f (x)/g(x). Simply differentiate the top, then differentiate the bottom. Caution 2: Can only use for types 0 0 and ± ±. AMAT 217 (University of Calgary) Fall / 18

10 Example sin(nx) Using l Hôpital s Rule show that = 1, where n is a positive constant. x 0 nx Solution: Plugging in x = 0 gives a it of the type 0/0, since sin(0) = 0. Applying l Hôpital s Rule: sin(nx) x 0 nx H x 0 n cos(nx), Since (sin(nx)) = n cos(nx) and (nx) = n, n = n cos(0), Plug in x = 0, n = 1, Since cos(0) = 1. Notation We use the symbol H = to denote we are using l Hôpital s Rule in that step. AMAT 217 (University of Calgary) Fall / 18

11 Example Using l Hôpital s Rule, evaluate x 1 x 2 1 x 1. Solution: Plugging in x = 1 gives x 2 1 x 1 x 1 = = 0 = indeterminate form! 0 Applying l Hôpital s Rule x 2 1 H 2x x 1 x 1 x 1 1 = 2 1 = 2 Alternatively, using just algebraic simplification x 2 1 x 1 x 1 (x + 1)(x 1) x + 1 = 2 x 1 x 1 x 1 AMAT 217 (University of Calgary) Fall / 18

12 Example Evaluate Solution: e x x x 2. Taking x to gives a it of the type /. Therefore, l Hôpital s Rule applies: e x H e x x x 2 x 2x. The new it is also of type /, thus, we can apply l Hôpital s Rule again: e x x 2x H e x x 2. The last it gives a type of /2 =. Therefore, the value of the original it is + : e x x x 2 =. AMAT 217 (University of Calgary) Fall / 18

13 Other it types Sometimes when plugging in x = a we get a it of the type: 0 (± ) On these it problems, we can not apply l Hôpital s Rule directly. However, we can always turn a product of 0 and ± into one of the required forms by taking one of the functions to the bottom as follows: f (x) g(x) = g(x) 1/f (x) or f (x) g(x) = f (x) 1/g(x) Tip: Try both (one will be easier than the other). AMAT 217 (University of Calgary) Fall / 18

14 Example Evaluate x ln x. x 0 + Solution: Taking x to the bottom (as a reciprocal) gives: x ln x ln x x 0 + x 0 + 1/x. Now our it is of the form ( )/( ) and l Hôpital s Rule can be applied. Note that if we brought ln x to the bottom instead, our it would be of the form 0/0 and l Hôpital s Rule would still apply. ln x ln x) x 0 +(x x 0 + 1/x H x 0 + 1/x 1/x 2 x 2 x 0 + x x 0 +( x) Rewrite the it Derivative of top and bottom Simplifying Cancelling = 0 Taking the it AMAT 217 (University of Calgary) Fall / 18

15 In the next problem we see how to deal with indeterminate forms of the types: 0 0, 0 and 1 Side note: A common misconception is that 1 is 1, but this is incorrect. Infinity is not a real number and can be thought of as shorthand for a it. The misconception comes from the fact that 1 x = 1 for any finite power x. When dealing with its, we must allow both the base and exponent to vary as follows: 1 a 1,x ax. Now, it becomes clearer that we can get any value we want by making x and a vary simultaneously at different rates. For example, if a was approaching 1 from the right (i.e., a > 1) at a very slow rate, and x was approaching at a very fast rate, you might expect the answer to be. On the other hand, if a was approaching 1 from the left (i.e., a < 1) at a very slow rate, and x was approaching at a fast rate, you might expect the answer to be 0. The next example illustrates how a it of the type 1 can result in something unexpected. AMAT 217 (University of Calgary) Fall / 18

16 Example Show that x 1/(x 1) = e. x 1 + Solution: Plugging in x = 1 (from the right) gives a it of the type 1. To deal with this type of it we will use logarithms. Let L denote our it: L x 1/(x 1). x 1 + Now, take the natural log of both sides: ( ln L ln x 1/(x 1)). x 1 + Using log properties we have: ln x ln L x 1 + x 1. The right side it is now of the type 0/0, therefore, we can apply l Hôpital s Rule: ln x ln L x 1 + x 1 H 1/x x = 1 Thus, ln L = 1 and hence, our original it (denoted by L) is: L = e 1 = e. AMAT 217 (University of Calgary) Fall / 18

17 Example Compute Solution: x 4 + e x x x 2 + 2e x. x 4 + e x H 4x 3 + e x x x 2 + 2e x x 2x + 2e x Since / type (can take derivatives) H 12x 2 + e x x 2 + 2e x Since / type (can take derivatives) H 24x + e x x 2e x H 24 + e x x 2e x H e x x 2e x = 1 2 Since / type (can take derivatives) Since / type (can take derivatives) Since / type (can take derivatives) Cancelling and taking the it AMAT 217 (University of Calgary) Fall / 18

18 Example Compute ln(cos x) x 0 ln(cos 3x). Solution: This is a 0/0 it type, thus: ln(cos x) x 0 ln(cos 3x) H x 0 sin x cos x 3 sin(3x) cos(3x) tan x x 0 3 tan(3x) H sec 2 x x 0 9 sec 2 (3x) = 1 9 Taking derivatives of top and bottom Simplifying giving 0/0 type Taking derivatives Since sec(0) = 1 AMAT 217 (University of Calgary) Fall / 18