3.7 Implicit Differentiation

Transcription

1 1 3.7 Implicit Differentiation In some cases we are not given an explicit formula like y = cos(x) but an implicit relationship between x and y, like x 3 + y 3 = 6xy In some cases we can solve explicitly for y and get one or more functions x 2 + y 2 = 1 f(x) = 1 x 2 ; g(x) = 1 x 2 and decide which to use for which part of the curve. Sometime this may be very complex or impossible but we still have a well-defined (graphical) tangent line, and we want to find the corresponding slope y at the point (x 0, y 0 ) of interest. We assume y = y(x) is a function of x (at least over small intervals) and differentiate using the chain rule x 2 + y 2 = 1 2x + 2y dy dx = 0 dy dx = x y 2 which we can evaluate at a pint (x 0, y 0 ) to say dy dx = x 0. Notice (x0,y 0 ) y 0 that we need both the x and y coordinates to evaluate the derivative (as more than one y-value may correspond to an x-value in a relationship). Also, the derivative is undefined when y = 0 which we would expect from the vertical tangents to the circle as it cuts the x-axis. We can proceed as usual to find the equation of the tangent to a curve given implicitly by the point-slope method. x 3 + y 3 = 6xy 3x 2 + 3y 2 dy dx = 6y + 6x dy dx dy dx = 2y x2 y 2 2x and we find that the slope at (3, 3) is 1 and the corresponding tangent has equation y 3 = ( 1)(x 3) or y = x + 6. As before we can find horizontal tangents by setting dy dx = 0, which in this case yields 2y x 2 = 0 or y = 1 2 x2. We need to find which points (x, y) on the curve also have y = 1 2 x2, so we substitute into the original equation x 3 + y 3 = 6xy to get x = 0 or x 3 = 16. Now x = 0 y = 0 and

2 3 x = 3 16 y = Another example involving trigonometric functions is sin(x + y) = y 2 cosx where chain {}}{ dy = 2y dx cos x + y2 ( sinx) cos(x + y) (1 + dy dx ) }{{} chain rule which after collecting terms in dy dx 3.8 Higher Derivatives } {{ } product rule and rearranging yields dy dx = y2 sinx + cos(x + y) 2y cosx cos(x + y) Recall that if f is differentiable then we defined a new function f (x). Since f (x) is a function in its own right, we can ask if it is differentiable, and form a new function by differentiating again. We call f ( (x) ) = (f ) (x) the second derivative of x. In the other notation this is d dy dx dx = d2 y. dx 2 4 Say f(x) = xsinx, then f (x) = sinx + xcosx and f (x) = cosx + cosx xsinx = 2 cosx xsinx If our original function represents position as a function of time, then we have seen that the derivative corresponds to velocity. The second derivative is the rate of change of the velocity (first derivative) and is called acceleration. If s(t) = t 3 6t 2 + 9t then s (t) = 3t 2 12t + 9 and s (t) = 6t 12. Note that to tell when the particle is speeding up or slowing down we need to compare the signs of the velocity and acceleration (same sign: speeds up; opposite signs: slows down). We can continue this differentiation procedure, regarding the second derivative as a new function in its own right and finding the third derivative f (x) = (f ) (x); and so on to form the n-th derivative f (n) ( not to be confused with f n (x) = [f(x)] n ) from f by differentiating n times. We also write dn y dx n.

3 5 If f is a polynomial then eventually f (n) (x) = 0: y = 3x 3 +5x 2 +x 7 y = 9x 2 +10x 1 y = 18x+10 y = 18 y (4) = 0 and so y (n) = 0, n 4. If f is rational the derivative may never vanish y = 1 x y = 1 x 2 y = 2 x 3 y = 6 x 4 y (4) = 24 x 5 and this pattern continues to give y (n) = ( n)( n + 1)( n + 2) ( 3)( 2)( 1)x (n+1) = ( 1)n n! where we x n+1 have used the symbol n! (n factorial) to mean n(n 1)(n 2) We can also find higher derivatives from relationships via implicit differentiation. If x 4 + y 4 = 9 then 4x 3 + 4y 3 y = 0 y = x3 y 3 which we can differentiate using the quotient rule y = 3x2 y 3 3x 3 y 2 y y 6 = 3x2 y 3 3x 3 y 2 x3 y 3 y 6 = 3x2 (y 4 + x 4 ) y 7 6 but x and y must still satisfy the original relationship so y 4 + x 4 = 9 and we get y = 27 x2 y 7. The higher derivatives of sinx and cosx can be calculated quite easily once we realise that the derivatives repeat. f(x) = cosx; f (x) = sinx; f (x) = cos x; f (x) = sinx; f (4) (x) = cosx The derivatives for sinx have a similar cycle of length 4 sinx cos x cos x sinx so higher derivatives are easy to calculate: D 27 sin x = D 3 sinx = cos x

4 7 3.9 Related Rates In a situation where there is a fixed relationship between quantities (for example: radius and area of a circle) when one of the quantities changes with time, so does the other. If I double the radius of a circle, the area is multiplied by four. As the radius changes, r = r(t), then the area also changes, A = A(t). A changes with time via its dependence on r A(t) = A(r(t)) = π[r(t)] 2 and so differentiating with respect to t, using the chain rule we get A (t) = 2πr(t)r (t) so the rate of change in area depends on the current radius as well as the rate of change of radius (this should make physical sense..) In a related rates problem start by identifying the given information the desired unknown quantity 8 You may need to differentiate implicitly, How fast is a car moving northwards on a circular track of radius 1km when it is southeast of the center moving east at 20km/h? Let x be distance east of center and y north of center; then the car travels according to x 2 + y 2 = 1 and the car is at (1/ 2, 1/ 2). Differentiating 2x(t)x (t) + 2y(t)y (t) = 0 y (t) = (x/y)x (t) = 20km/h or use constraints to turn a function of more than one variable into a function of one variable A water tank has the shape of an inverted circular cone with radius 2m and height 4m. Find the rate at which the height is decreasing if the water in the tank is currently 1m deep and leaking out through a hole at 1m 3 /hr. So we have V = π 3 r2 h so V (t) depends on both r(t) and h(t). But r and h cannot vary independently, the geometry of the

5 9 problem means that the radius of the surface of the water and the height must bear the same relationship as the radius and the height of the full cone. So r(t) : h(t) as 2 : 4 or r = h/2. Then we can write V = π 3 (h/2)2 h = π 12 h3 V (t) = π 4 [h(t)]2 h (t) Now we know V (t) = 1m 3 /hr and h = 1 so 1 = π 4 [1]2 h (t) and h (t) = 4 π. The water level is decreasing at a rate of 4 π m/hr Make sure that you have differentiated and have a relationship between all the known and the unknown quantity before you substitute and solve for the unknown quantity; or solve (algebraically) for the unknown quantity and then substitute As a general strategy, remember to Read the problem, make note of what is being asked the unknown and what is given 2. Draw a diagram showing the relationship between the quantities in the problem (if possible) 3. Assign symbols for the quantities that vary with time (for example A(t), r(t) for area and radius) 4. Identify the unknown and the given quantities in your notation 5. Write down an equation relating the various quantities, using the constraints (geometry) in the problem to eliminate variables by substitution 6. Differentiate both sides of the equation with respect to t using the Chain Rule 7. Substitute the given information into the resulting equation and solve for the unknown.

6 Linear Approximation and Differentials Recall that the tangent line to a curve lies close to the curve near the point they touch each other. This means that the function values for the tangent line and the function are close as long as we stay near the point of tangency. We have seen that the equation of the tangent line to a differentiable function f at a is y = f(a) + f (a)(x a) so we can approximate the value of f near x = a with the linear approximation or tangent line approximation f(x) f(a)+f (a)(x a) or using h = x a f(a+h) f(a)+f (a)h We call L(x) = f(a) + f (a)(x a) the linearization of f at a. This approximation is useful if f(x) is hard to calculate in general, but easy for particular values of x. To linearize a function at a we need to find the equation of the tangent line through (a, f(a)). 12 If we want to find a linear approximation to f(x) = x + 3 at a = 1, we find f (x) = 1 2 x+3 and so f (1) = 1/4. The linear approximation is then L(x) = x+7 4 and we can approximate x + 3 near x = 1 using this linearization = ( )/4 = [See concavity later to check if this approximation over- or under-estimates the exact value] We can also approach approximations by talking about differentials: If y = f(x) is differentiable, then we regard the differential dx as an independent variable; and define the differential dy via dy = f (x)dx The actual change in f between x and x + x is given by So if we take x = dx we see that y = f(x + x) f(x) y = f(x + dx) f(x) f (x)dx = dy

7 13 so we approximate the actual change in f by the change in the tangent line. If we in increase the radius of a sphere from 10cm to 12cm, estimate how much we have increased the volume. We use V (r) = 4π 3 r3 so and V (r) = 4πr 2 V (10 + 2) V (10) V (10) 2 = 800π cm Applications of Differentiation 4.1 Maximum and Minimum Values A function f has a local (or relative) maximum at c if f(x) f(c) for all x near c (all x in some open interval containing c). Similarly, f has a local minimum at c if f(x) f(c) for all x near c. If f is continuous on [a, b] then f attains both an absolute maximum and absolute minimum on [a, b] A critical number of a function f is a c in the domain of f such that f (c) = 0 or f (c) does not exist. If f is a local extremum (maximum or minimum) at c, and f (c) exists, then f (c) = 0 (c is a critical number). So to find local extremes look at critical points: All local extremes occur at critical points BUT not all critical points are local extremes.

8 15 To find the absolute extremes on an interval, find the local extremes (which must occur at critical points) and compare these to the endpoint values. 4.2 Mean Value Theorem If f is continuous on [a, b] and differentiable on (a, b) then there is a c (a, b) such that f (c) = f(b) f(a) b a or f(b) f(a) = f (c)(b a) If f (x) = g (x) for all x in an interval, then f g is constant on that interval, that is f(x) = g(x) + C, for some constant C. 4.3 Derivatives and the shape of a graph If f is differentiable on an interval I, then if for all x I, f (x) = 0 then f is constant on I f (x) > 0 then f is increasing on I 16 f (x) < 0 then f is decreasing on I If f is differentiable on disjoint intervals no comparisons can be made between intervals: Think f(x) = 1/x, which has negative derivative on (, 0) and (0, ) so is decreasing on both those intervals, but f(positive) > f(negative). If f lies above (below) all its tangents on an interval I then it is concave upward (concave downward). If f (x) > 0 for all x in I, then the graph of f is concave upward on I If f (x) < 0 for all x in I, then the graph of f is concave downward on I A point P on a curve y = f(x) is called an inflection point if f is continuous there and the curve changes concavity at P Note that a point of inflection need not be a critical point (if the slope is non-zero there) but will be if the slope there is zero or undefined.

9 First Derivative Test If c is a critical number of a continuous function f and 1. f changes from positive to negative at c, then f has a local maximum at c (+0 ) 2. f changes from negative to positive at c, then f has a local minimum at c ( 0+ ) 3. f does not change sign at c then f has no local max. or min. at c Second Derivative Test Suppose f is continuous near c with f (c) = 0 and 1. f (c) > 0 then f has a local minimum at c 2. f (c) < 0 then f has a local maximum at c Asymptotes For f defined on [a, ) we say lim f(x) = L if f(x) can be made x arbitrarily close to L by taking x sufficiently large [similarly lim f(x) = L]. x The line y = L is called a horizontal asymptote of y = f(x) if either lim f(x) = L or lim f(x) = L x x The line x = a is called a vertical asymptote of y = f(x) if (at least) one of the following is true lim x a x a f(x) = ± or lim f(x) = ± + The line y = mx + c is called a slant asymptote of y = f(x) if lim f(x) (mx + c) = 0 or lim x f(x) (mx + c) = 0 x

10 19 If r > 0 is rational then lim lim x x 1 x r = 0 and, if xr is defined for all x, 1 x r = 0 Asymptotes are liable to occur in rational functions (ratio of polynomials) vertical asymptotes where the denominator has a zero [unmatched by one in the numerator] horizontal asymptotes occur in the form y = 0 if the denominator has higher degree than the numerator y = C if they have the same degree slant asymptotes occur where the numerator has degree one higher than the denominator. Then polynomial long division alows us to write lower degree remainder term that 0 f(x) = mx+c+ = mx+c+ denominator as x 20 A formula for slant asymptotes: Write down the coefficients of the leading two terms of the numerator above those in the denominator (remember zero coefficients!) in a 2-by-2 square 3x 3 4x 2 + 7x 2 2x 2 + 5x Now divide everything by the leading power of the denominator, namely / /2 Now the slant asymptote will be mx+b where m is the value, 3/2, in the top left corner (the ratio of the leading coefficients 3 and 2) To find b take the top right entry, 2, and subtract the product of the two diagonal entries 3/2 and 5/2, that is 15/4, to get 2 (15/4) = 23/4

11 Curve Sketching Domain Determine where f is (un)defined Intercepts y-intercept f(0) and x-intercepts found by solving f(x) = 0 Symmetry 3 kinds even f( x) = f(x) symmetric about the y-axis odd f( x) = f(x) symmetric through the origin [rotate graph for x 0 by 180 degrees]. periodic f(x + p) = f(x) for some positive constant p, the smallest possible p is called the period of f and we need only determine what the graph is on one interval of length p and translate. Asymptotes Horizontal, Vertical, Slant Intervals of Increase/Decrease Compute f (x) and find intervals where positive (f increasing) or negative (f decreasing) Local max/min Critical numbers, First derivative Test, if necessary the 22 Second Derivative Test [unless it is obviously preferable] Concavity and points of inflection Compute f (x) and find intervals of concave upwards/downwards and the points (of inflection) where concavity changes. Sketch 4.7 Optimization Understand Wanted? Unknown? Given? Conditions (constraints)? Diagram Usually useful Notation Chose a symbol (such as A for area) for the quantity to be optimized, and variables (a, b,..., x, y, z) for the other quantities Relationships between quantities Express quantity to optimize in terms of the other variables (area is length times breadth: A = xy) Reduce quantity to optimize to function of one variable (if necessary) using the conditions/constraints.

12 23 Find absolute max/min value of this function (as the problem requires) remember to justify the (appropriate) kind of extremum. Formulate answer Check your answer is reasonable (e.g. lengths,areas positive etc.). Write a sentence answering the question remember you invented all the notation, the answer is not x = 4 it is something like The breadth of a rectangle with minimum area is 4 feet Some questions like those on the homework you may want to revise: If p = 20 q 2 +5 find dq dp Write q = 20/p and then 2q dq dp = 20/p2 or dq dp = 10 p 2 q Find the equation of the tangent to the curve (x 2 + y 2 x) 2 = 1 2 (x2 + y 2 ) at (1, 1) 2(x 2 + y 2 x)(2x + 2yy 1) = x + yy and now substitute x = 1, y = 1 to get 2( )(2 + 2y 1) = 1 + y and so 2 + 4y = 1 + y and y = 1/3. This gives the equation y = ( 1/3)(x 1) + 1 = x/3 + 4/3 24 Find the speed and direction of the x-coordinate of a point moving on 4x 3 + 3y 3 = xy at the point P = ( 1 7, 1 7 ) when the y-coordinate increases at 3 units per second Differentiating w.r.t. t we get 12x 2 dx dt dy + 9y2 dt = dx dt + x dy dt dy dt = 5 dx 49 Which after substituting for X and y is 2 49 dt and finally dy dx dt = +3 and so dt = 6/5 so x is decreasing at 6/5 = 1.2 units per second. If you are standing 6m from the release point of a weather balloon rising at 50m/min how fast is your distance to the balloon increasing when the balloon is 8m high. With s(t) =distance and h(t) =height we have s 2 (t) = h 2 (t) + 36 and so 2s ds dt dh dh = 2h dt. We have h and dt but we need s to find ds dt. Using Pythagoras s2 = (8) 2 + (6) 2 = 100 so s = 10 [or use triangle]. So finally

13 25 20 ds dt = ds dt = 40m/min Find the second derivative of f(x) = xg(x 2 ) + x 2 if g is twice-differentiable Find the third degree polynomial Q if Q(1) = 5, Q (1) = 11, Q (1) = 18, Q (1) = 24. Estimate 15 1/4 Let f(x) = x 1/4. Then we want f(15) and f(16) = 2 is easy to calculate. So f(15) = f(16 1) f(16) + f (16)( 1). We need f (x) = 1 4 x 3/4 and f (16) = = 1 32 so finally f(15) 2 1/32 = 63/32 An alternate approach is linearize f(x) = x 1/4 near 16 to get L(x) = (x 16) = x and find L(15) Find the absolute maximum of f(x) = sin 2x + cos 2x on [0, π/6] Determine if Rolle s theorem applies to x2 3x 18 x 3 on [ 3, 6] No, fails to be continuous on [ 3, 6], namely at x = 3 26 Show in two ways that somewhere the graph of f(x) = sinx has slope 2/π (1) To show the slope is somewhere 2/π, show that the derivative cosx is somewhere 2/π. Now 1 < 2/π < 1 so 2/π is possibly in the range of cosx. Now cosx is continuous everywhere, and so by the Intermediate Value Theorem it takes on every value between 1 = cos(0) and 1 = cos(π), and so is somewhere 2/π. (2) sinx is also continuous and differentiable everywhere so we can also use the Mean Value Theorem to show the slope of sinx is somewhere 2/π as long as we can find a secant with that slope. The most obvious would be a rise of 2 over a run of π which happens between sin( π/2) = 1 and sin(π/2) = 1. How many real roots does x 5 + 5x + 1 = 0 have We have a polynomial x 5 + 5x + 1 of odd degree so by the IVT it has a least one root. But it cannot have two: since otherwise by Rolle s theorem the derivative would have to

14 27 vanish somewhere, while that derivative is 5x which is never zero. Alternatively, since the derivative is positive everywhere the function x 5 + 5x + 1 is always increasing and can never revisit 0. Where is f(x) = 4 3 x x3 21x 2 8x + 7 concave down If f(x) = x cos2x is defined on π x π; where is it decreasing What is lim x( x + 5 x 3) x Find the correct coefficient A that makes y = 2x + 1 a slant asymptote for 4x3 + Ax x 2 + x 1 4 A (1) Using our trick: A/2. And so the 1 1/2 constant term in the slant asymptote is A/2 2(1/2) = A/2 1 which should be 1. Thus A = (2) We have that 4x 3 + Ax x 2 + x 1 = 2x x terms and constants 2x 2 + x 1 so multiplying by the denominator we get 4x 3 +Ax 2 +3 = (2x+1)(2x 2 +x 1)+x terms and constants Remember that two polynomials are equal if their coefficients agree. A is the coefficient of x 2 on the left, so look for the x 2 terms on the right. These only come from the first term (2x+1)(2x 2 +x 1) = 4x 3 +2x 2 2x+2x 2 +x 1 = 4x 3 +4x 2 x 1 and so A = 4. Find the asymptotes of 3x 2 3x 6 4x 2 10x + 4 Find the equation of the tangent line to f(x) = 1 3 x3 x 2 + 2x + 7 which has the minimal slope. Of all rectangles of a given area which has the smallest perimeter

15 29 How should a 10ft piece of wood be used in the corner of a room to make the triangle of biggest area Let the distances along the two walls be x and y. Then we must have that x 2 + y 2 = (10) 2 = 100 and 0 x, y 10. (1) The area is A = 1 2 xy = 1 2 x 100 x 2 and so A (x) = x 2 x 2 2 = 50 x2 100 x x which is 0 if 2 50 x 2 = 0 or x = 50. Then y = 50 and A = 25 (2) sneaky way: If x, y maximize 2xy they will maximize A = 1 2 xy. We can only choose x, y with x2 + y 2 = 100. Now 0 (x y) 2 = x 2 + y 2 2xy = 100 2xy Maximizing 2xy is the same as minimizing 100 2xy = (x y) 2 which happens when x = y. Thus x 2 = y 2 = 50 and we get A = 25 as before. If the surface area of a cone of radius r and height h is given by S = πr r 2 + h 2 what ratio of height to radius gives the maximum volume V = 1 3 πr2 h for a fixed surface area [Hard!]