Differential Calculus Concepts & Problems 5/23/2019 rd

Matrix Properties (An)-1 = (A-1)n (inverse (expt-matrix matrix n)) (expt-matrix (inverse matrix) n)) (AT )-1 = (A-1)T (AB)T = BTAT (AB)-1 = B-1A-1 det A = 1 / det A-1 Determinants can be calculated using co-factors and minors. Only square matrices have determinants. 5/23/2019 rd

Inverse Matrix AA-1 = I, the Identity matrix. Find the inverse matrix of #2A( (1 2) (3 8) ) 1a + 3b = 1 2a + 8b = 0 1c + 3d = 0 2c + 8d = 1 (solve '((1 3 1)(2 8 0)))  (4 -1) (solve '((1 3 0) (2 8 1)))  (-3/2 1/2) (4 -1) (-3/2 1/2) 5/23/2019 rd

Determinants |A| = |AT| 1 2 4 3 has determinant -5 1 4 has determinant -5 2 3 1 2 3 1 2 3 4 6 7 = 3 4 6 7 15 24 27 5 8 9 5/23/2019 rd

Co-factors Evaluate 1 2 3 -7 3 -4 4 6 7 0 0 5 0 1 0 4 -1 -2 5 -1(7 – 12) = 5 (det #2A ((1 2 3)(4 6 7)(0 1 0)))  5 Find the inverse, 5/23/2019 rd

Eigenvalues Find eigenvalues for matrix [-5 9; -6 10] (eigenvalues #2A((-5 9)(-6 10)))  (1, 4) t + 5 -9 6 t – 10 = t2 -5t + 4 = 0 => (t – 1)(t – 4) = 0 Eigenvectors are 6 6 or (1, 1), (9, -6) or (3, -2). Note orthogonality of the two vectors 1 3 -1 -5 9 1 3 4 0 1 2 -6 10 1 2 = 0 1 (M* (inverse #2A((1 3)(1 2))) #2A((-5 9)(-6 10)) #2A((1 3)(1 2))) #2A((4 0)(0 1)) 5/23/2019 rd

Cryptography A = 1 2 1 1 Message: SEND HELP 18 5 14 4 0 8 5 12 16 0 1 2 18 5 14 4 0 = 34 15 38 36 0 GOKI0ATG0 1 1 8 5 12 16 0 26 10 26 20 0 -1 2 34 15 38 36 0 = 18 5 14 36 0 recovered message 1 -1 26 10 26 20 0 8 5 12 16 0 5/23/2019 rd

Getting There From Here Limits and Continuity Getting There From Here 5/23/2019 rd

Limit of a Sequence 1 3/2 5/3 7/4 9/5 11/6 … 2 – 1/n, … limit  2 but the sequence does not contain 2. On the other hand the sequence 1 ½ 1 ¾ 1 5/6 1 6/7 …has 1 as a limit and contains 1 5/23/2019 rd

Limit of a Function 1 3/2 5/3 7/4 9/5 11/6 … 2 – 1/n  2 As x  2, f(x) = x2  4 1 9/4 25/9 49/16 81/25 121/36  4 Find limits of 1 ½ 1/3 ¼ 1/5 … 1/n 5 4 11/3 7/2 17/5 … 3 + 2/n ½ ¼ 1/8 1/16 1/32 … 5/23/2019 rd

/2 Some famous limits 5/23/2019 rd

Limit Find the limits 5/23/2019 rd

Limit of a Series 1 + ½ + ¼ + 1/8 + 1/16 + … sum of terms of a sequence (+ 1 1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256 1/512 1/1024)  1.999 5/23/2019 rd

Finding a Limit Find 5/23/2019 rd

L'Hospital's Rule Applies to Indeterminate Forms 5/23/2019 rd

e Evaluate Use L'Hospital's Rule on [ln(1 + x)]/x 5/23/2019 rd

Continuity f is continuous at x = a if 5/23/2019 rd

Where Was the Farmer? When he jumped off the roof? On the roof. That's where he was before he jumped. Well then in the air. That's where he was after he jumped. Where was he when he jumped? 5/23/2019 rd

The Slope of a line passing through a curve 5/23/2019 rd

The Slope of a line tangent at a point on the curve 5/23/2019 rd

What is the Derivative? The derivative is one of the two central concepts of calculus. The derivative gives the slope of the line tangent to a function at a point. In this way, derivatives can be used to determine many geometrical properties of the function such as concavity or convexity. The derivative provides a mathematical formulation of the instantaneous rate of change; it measures the rate at which the function's value changes as the function's argument changes. 5/23/2019 rd

Let’s compute a derivative! 5/23/2019 rd

The Notation All of the following are equivalent when y = f(x): 5/23/2019 rd

Three Rules to live by 5/23/2019 rd

Example Find the equation of the line tangent to y = f(x) = 2x2 + 2x + 3 at the point (1,7) f’(x) = 4x + 2 slope function f’(1) = 6 Using the point-slope form of a straight line: y – 7 = 6(x – 1) or y = 6x + 1 5/23/2019 rd

The Chain Rule y=f[u(x)] 5/23/2019 rd

Alternate approach 5/23/2019 rd

The top 5 Table of Derivatives 5/23/2019 rd

Table of Derivatives – the next 5 5/23/2019 rd

Another Example 5/23/2019 rd

and some more… 5/23/2019 rd

x2, 2x and xx y = x2 => y' = 2x y = 2x => ln y = x ln 2 => y'/y = ln 2 => y' = ln 2 * 2x y = xx => ln y = x ln x => y'/y = x * 1/x + ln x = 1 + ln x y' = xx(1 + ln x) Lim (2 + h)5 – 32 h  0 h 5/23/2019 rd

Our First Application of Derivatives Analyzing Functions 5/23/2019 rd

Some Well Known Facts Given f(x) is continuous and differentiable in the interval (a,b), then If f’(x) = 0, x  (a, b), then f(x) is constant If f’(x) > 0, x  (a, b), then f(x) is strictly increasing If f’(x) < 0, x  (a, b), then f(x) is strictly decreasing + + - 5/23/2019 rd

Convex and Concave Functions If f’(x) is increasing, x  (a, b), f(x) is convex in the interval (a, b) If f’(x) is decreasing, x  (a, b),f(x) is concave in the interval (a, b) 5/23/2019 rd

Example of Some Well Known Facts f(x) = x3 – 2x + 1 and f’(x) = 3x2 – 2 5/23/2019 rd

Maxima and Minima Graph y = 3x2 – 12x - 2. Find where increasing. Graph y'. y' = 6x – 12  for x > 2 => increasing 1 2 2 5/23/2019 rd

Concavity y = 3x4 – 7x2 – 2; y' = 12x3 – 14x; y'' = 36x2 - 14 5/23/2019 rd

Our Second Application of the Derivative Finding Roots… 5/23/2019 rd

Newton-Raphson Method Problem: solve for x: f(x) = 0 when f(x) is not linear or quadratic. Assume f(x) is continuous on the interval [a, b] and f(a) and f(b) have opposite sign, then the equation f(x) = 0 has at least one real root between a and b. 5/23/2019 rd

Newton-Rhapson Method Newton Iteration 5/23/2019 rd

Newton-Raphson Method -2 f(x1) x2 x1 5/23/2019 rd

Newton-Raphson Method - 3 To find the solution to f(x) = 0 Recursion formula Repeat the above in order to obtain a better approximation 5/23/2019 rd

Newton-Raphson Method - 4 Find the root of f(x) = x4 – 4x + 1 = 0 in the interval (0,1) f(0) = 1 and f(1) = -2 (a sign change, therefore at least one root) f’(x) = 4x3 - 4 (quartic 1 0 0 -4 1)  (1.493359d0 0.250992d0 #C(-0.872175d0 1.381032d0) #C(-0.872175d0 -1.381032d0)) 5/23/2019 rd

Higher Order Derivatives First derivative y’ f’(x) Second derivative y’’ f’’(x) Third derivative y’’’ f’’’(x) Fourth derivative Y(4) f(4) 5/23/2019 rd

An Example f(x) = 6x3 – 12x2 + 6x – 2 f’(x) = 18x2 – 24x + 6 f(4) = f(5) = … = 0 5/23/2019 rd

Proceeding from a single variable function to a 2 variable function. Partial Derivatives Proceeding from a single variable function to a 2 variable function. 5/23/2019 rd

Partial Derivatives To find fx(x,y), treat y as a constant and differentiate f with respect to x in the usual way. To find fy(x,y) treat, x as a constant and differentiate f with respect to y 5/23/2019 rd

Example Problems 5/23/2019 rd

More Notation 5/23/2019 rd

Partials 1st partials 2nd partials 5/23/2019 rd

Partial Differentiation y = f(x, y, z, t, …) is differentiated with respect to one of these variables, the other variables are held constant. Let w = f(x, z) = x2 + xz2. Then Wx = Wz = Consider right circular cylinder V(d, h) = d2h/4; Vd = dh/2 and Vh = d2/4; the area 5/23/2019 rd

Unconstrained Optimization Find the area bounded by the curve y = f(x) = x3 - 6x2 + 9x + 1 and the x-axis between the curve's x-extrema. y'(x) = 3x2 –12x + 9 = 0 when x = 1 or x = 3 pt of inflection y''(x) = 6x – 12; x = 2 is Pt-Inflection y''(1) < 0 => relative max at(1,5); y''(3) > 0 => relative min at (3,1) 1 2 3 = 6 (U-poly-eval '(1 9 -6 1) 1) syms x; f = x^3 - 6*x^2 + 9*x + 1; fx = diff(f); roots([3 -12 9]);ans = 3.00 1.00; fxx1 = -6.00 fxx2 = subs(diff(fx),3) =fxx2 = 6.00; subs(f,1) = 5.00; subs(f,3) = 1.00 5/23/2019 rd

Cost to Construct A box has base length 3 times the width. The cost to construct is $10/ ft2 for the top and bottom and $6/ft2 for the sides. The volume must be 50 ft3. Find the dimensions to minimize the cost to construct. C = 2(10)(3w2) + 2(6)(hw) + 2(6)(3hw) = 60w2 + 48hw; 50 = 3w2h => h = 50/3w2 C = 60w2 + 800/w C' = 120w - 800/w2 = 0 when 12w3 = 80 => w = 1.88 C" = 120 + 1600/w > 0 => relative minimum. h l=3w w 5/23/2019 rd

Z = f(x, y) = x3 + y3 + 3xy Zx = 3x2 + 3y = 0; Zy = 3y2 + 3x = 0; (0, 0); (-1 -1) x2 + y2 + x + y = 0; or (x + ½)2 + (y + ½)2 = ½ Zxx = 6x; Zxy = 3 = Zyx Circle centered at (½, ½) with r = 1/21/2 Rules: D & Zxx > 0 => rel min; D > 0 and Zxx < 0 rel max D < 0 => saddle point and D = 0 => no info D = At (-1 -1), D = 25 and Zxx < 0 => rel max; Z(-1, -1) = 1 At (0, 0), D = -9 => saddle point at (0, 0, 0) 5/23/2019 rd

f(x, y) = x2 + y2 fx = 2x; fy = 2y (0, 0) fxx = 2; fyy = 2 D = 2 0 0 2 = 4 > 0 and fxx and fyy > 0 => relative minimum 5/23/2019 rd

Maxima and Minima f(x, y) = 5x2 + 10y2 + 12xy – 4x – 6y + 1 fx = 10x + 12y – 4 = 0; fy = 20y + 12x – 6 = 0 (solve '((10 12 4)(12 20 6)))  (1/7 3/14) f(1/7, 3/14) = -1/14 fxx = 10 fxy = 12 fyy = 20 => (1/7, 3/14, -1/14) is a relative minimum. 5/23/2019 rd

f(x, y) = 3x2y + x2 - 6x – 3y - 2 fx = 6xy + 2x – 6; fy = 3x2 - 3 fxy = 6x; fyx = 6x fxx = 6y + 2; fyy = 0 Substitute x = 6/(6y + 2) to get 36/(6y + 2)2 = 1 or 36 = 36y2 + 24y + 4 or (quadratic 36 24 -32) (2/3 -4/3) -8 4 D = -16 => saddle points; no rel max or min. 4 0 5/23/2019 rd

Cross Product Let vector u = 2i – 3j + k and v = i + j – 2k. Then Note u  (u x v) = 10 -15 + 5 = 0 and v  (u x v) = 5 + 5 -10 = 0 => to (u x v) u x v = |u||v| sin  i, j and k form a right-handed triple u  v = |u||v| cos  in that i x i = j x j = k x k = 0 i x j = k; j x k = i, k x i = j 5/23/2019 rd

Applied Problems In Extrema 1. Divide 120 into 2 parts such that the product of one times the square of the other is a maximum. f(x) = x2(120 – x) = on [0, 120]. f'(x) = 240x – 3x2 = 0 when x = 0, x = 80 f"(x) = 240 – 6x f"(0) = 240 > 0 => relative minimum f"(80) = -240 < 0 => relative maximum f(0) = 0, f(120) = 0 => f(80) = 256,000 is max point. 5/23/2019 rd

2. Show that the curve y = x3 – 8 has no maximum or minimum. f '(x) = 3x2 = 0 at x = 0 When x is less or greater than zero, the slope is positive and thus constantly increasing. f "(0) = 0 => point of inflection. 5/23/2019 rd

(dD2/dt)' = 650 => relative minimum 3. Ship B is 65 miles due east of ship A and is sailing due west at 10 mph while A is sailing due south at 15 mph. Find their closest distance to each other if they continue. 65 – 10t 30, 45 54.08 A B 15*131/2 15t D2 = (15t)2 + (65 – 10t)2 = 225t2 + 4225 – 1300t + 100t2 dD2/dt = 650t – 1300 = 0 when t = 2 (dD2/dt)' = 650 => relative minimum 5/23/2019 rd

Optimization Given a length of L units to create a square and a circle, find the dimensions to maximize the area. Of all planar areas the circle has minimum perimeter. x L - x x/4 C = 2r = L – x C2 = 42r2 f(x) = 0 x L f ‘(x) = f(0) = => all for the circle; absolute maximum f(L) = => all for the square f ‘(x) = f() = ; absolute minimum f(x) = x2/16 + (L – x)2/4, 0  x  L; L2/4; L2 /16 f'(x) = x/8 + (L – x)/2 = 0 at x = 4L/(+ 4) 5/23/2019 rd

Z = f(x, y) = x3 + y3 + 3xy Zx = 3x2 + 3y = 0; Zy = 3y2 + 3x = 0; Test: (0, 0); (-1 -1) x2 + y2 + x + y = 0; or (x + ½)2 + (y + ½)2 = ½ Zxx = 6x; Zxy = 3 = Zyx; Zyy = 6y D = At (-1 -1), D = 36 – 9 = 25; Zxx < 0 => rel max; Z(-1, -1) = 1 At (0, 0), D = -9 => saddle point at (0, 0, 0) 5/23/2019 rd

Exercises The sum of 2 positive numbers is 20. Find if product is a maximum. (10, 10) Find if sum of their squares is a minimum. (10, 10) Find if product of the square of one and the cube other the other is a maximum. (12, 8) 5/23/2019 rd

Exercises The product of two positive numbers is 16. a) Find the numbers if the sum is least. (4, 4) b) If the sum of one and square of other is least. (8, 2) Find equation of line through the point (3, 4) which cuts the first quadrant a triangle of minimum area. 4x + 3y – 24 = 0 (3, 4) 5/23/2019 rd

Row Your Boat You are 5 miles from shore and wish to land 6 miles along the shore in minimum time. You can row at 2 mph and walk at 4 mph. Where should you land? P 5 (25 + x2)1/2 C A x 6 - x 5/23/2019 rd

Tangents and Normals Problems 5/23/2019 rd

2x – xy' - y + 2yy' = 0 => y' = (y – 2x)/(2y – x) 1. Find the points of tangency of horizontal and vertical tangents to x2 –xy + y2 = 27. 2x – xy' - y + 2yy' = 0 => y' = (y – 2x)/(2y – x) Horizontal: y – 2x = 0 or y = 2x and x2 –xy + y2 = 27 => x2 – x(2x) + 4x2 = 27 => 3x2 = 27 and x2 = 9, Points (3, 6) and (-3, -6) Vertical: 2y – x = 0 or x = 2y and x2 –xy + y2 = 27 4y2 -2y2 + y2 = 27 => 3y2 = 27; (6, 3) and (-6, -3) 5/23/2019 rd

2. Find the equation of the tangent and the normal to x2 + 3xy + y2 = 5 at (1, 1). 2x + 3xy' + 3y + 2yy' = 0 => y' = -(2x + 3y)/(3x + 2y) The slope of the tangent at (1, 1) is -1. Tangent: (y – 1) = -1(x – 1) or x + y = 2 Normal: (y – 1) = 1(x – 1) or x – y = 0 5/23/2019 rd

2yy' = 4p or y' = 2p/y. Let(xo, yo) be the point of tangency. Then 3. Show that the equation of the tangent of slope m  0 to the parabola y2 = 4px is y = mx + p/m. 2yy' = 4p or y' = 2p/y. Let(xo, yo) be the point of tangency. Then y02 = 4px0 and m = y'(y0) = 2p/y0 m = 2p/y => y0 = 2p/m and x0 = (4p2/m2)/4p = p/m2 (y – 2p/m = m(x – p/m2) y = mx + p/m 5/23/2019 rd

4. Find extrema of y = (x – 2)2/3. y' = 2(x – 2)-1/3 /3 = 2/3(x – 2)1/3. Critical value at x = 2 Use first derivative test As x  2+ positive slope; As x  2- negative slope indicating relative minimum 2 5/23/2019 rd

Related Rates Air is escaping a spherical balloon at the rate of 2 ft3/min/ How fast is the surface area shrinking when the radius is 12 ft? Vsphere = 4r3/3 and Ssphere = 4r2 dV/dt = 4r2 * dr/dt; dS/dt = 8r * dr/dt dS/dt /dV/dt = dS/dV = 2/r = 2/12 = 1/6 ft2/ft3 dS/dV * dV/dt = dS/dt = 1/6 * -2 = -1/3 ft2/min ft2/ft3 * ft3/min = ft2 / min 5/23/2019 rd

Exponential Extrema Find extrema of f(x) = x2ex. f'(x) = x2ex + 2xe = xex(x + 2) = 0 for x = 0, -2 (0, 0) is relative min and (-2, 4e-2) is relative max. f"(x) = 2ex + 4xex + x2e = ex(2 + 4x + x2) (quadratic 1 4 2)  (-0.585786, -3.414214) (-2  21/2) are points of inflection. 5/23/2019 rd

Rate of Change Show that the rate of change of the area of a circle with respect to its circumference is its radius. A = R2 = 2R2/2 = CR/2 = C2/4 => dA/dC = 2C/4 = R C = 2R dA/dC = dA/dR * dR/dC = 2R * 1/2 = R 5/23/2019 rd

Motion A ball thrown vertically upwards has position given by s = 218t – 16t2 . Find the times when the velocity is 64 ft/sec. s' = ds/dt = v = 218 – 32t = 64 when t = 154/32 = 4.8125 sec. 5/23/2019 rd

Rate of Change The volume of a ball is V = d3/6. Find the rate of change of the volume with respect to the diameter when the diameter is 2 ft. dV/dd = 3d2/6 = ½ *  * 4 = 2. 5/23/2019 rd

536-56 Given s = 9/(2t 2 + 3), find v at t = 1. s' = v = -9(4t)/(2t 2 + 3)2 v(1) = -36/25 ft/sec 5/23/2019 rd

Logarithmic Differentiation y = (2x - 5)3 / x2(x2 + 1)1/4 ln y = 3 ln (2x – 5) – 2 Ln x – ¼ ln(x2 + 1) y'/y = 6/(2x – 5) – 2/x – ¼ (2x/x2 + 1) y' = (2x - 5)3 / x2(x2 + 1)1/4 * 6/(2x – 5) – 2/x – ¼ (2x/x2 + 1) 5/23/2019 rd

Logarithmic Differentiation y = (1 + ex) ln x ln y = (ln x) ln (1 + ex) y'/y = ln x * ex/(1 + ex) + ln (1 + ex) (1/x) y' = (1 + ex) ln x *[ ln x * ex/(1 + ex) + ln (1 + ex) (1/x)] 5/23/2019 rd

Implicit Differentiation Given ey = (y + 1)ex , find dy/dx. ey y' = ex(y + 1) + y'ex y'(ey - ex ) = ex (y + 1) y' = ex(y + 1)/(ey - ex ) = ey / (ey – ey / (y+ 1) = 1/[1 – 1/(y + 1) = (y + 1)/y 5/23/2019 rd

Implicit Differentiation Find y' at (1,1) given that x2y - xy2 + x2 + y2 = 0. 2xy + x2y' -2xyy' –y2 + 2x + 2yy' = 0 y'(x2 – 2xy + 2y) = y2 – 2xy -2x y' = -3 5/23/2019 rd

Tangency Between Curves Find the angle of intersection of curves y2 = 4x and 2x2 = 12 – 5y Pts of intersection (1, 2) and (4, -4) 2yy' = 4 => y' = 2/y; y' = -4x/5 At (1, 2), m1 = 1; m2 = -4/5 At (4,-4), m1 = -1/2; m2 = -16/5 At P1, tan  = (m1 – m2)/(1 + m1m2) (atan 9) = 1.46 radians = (1 + 4/5)/(1 – 4/5) = 9 =>  = 83 40' At P2 tan  = (-1/2 + 16/5)/(1 + 8/5) = 1.0385 =>  = 46 5' 5/23/2019 rd

Asymptotes f(x) = (x2 – 4x) / (x2 - 4x + 3) Horizontal : lim as x   is 1 Vertical: (quadratic 1 –4 3)  1, 3 5/23/2019 rd

Vector Analysis 5/23/2019 rd

Vector Derivative Geometric Interpretation r(t) = xi + yj + zk dr/ds = T, a unit vector along tangent dr/dt = velocity d2r/dt2 = acceleration 5/23/2019 rd

Gradient, Divergence, Curl Let (x,y,z) be a scalar function and A(x,y,z) a vector function 5/23/2019 rd

Gradient, Divergence, Curl Let (x, y, z) = x2yz3, A = xzi –y2j + 2x2yk Then  = 2xyz3i + x2z3j + 2x2yz2k, perpendicular to    A = z -2y  x A = = 2x2i + (x – 4xy)j 5/23/2019 rd

F(x, y) = y3 + x2 - 6xy + 3x + 6y - 7 Fx = 2x – 6y + 3 = 0; Fy = 3y2 - 6x + 6 = 0 2(3y2 + 6)/6 – 6y + 3 = 0 (quadratic 1 -6 5)  y-values (1, 5); x-values (3/2, 27/2) Fxx = 2; Fxy = -6 Fyx = -6 Fyy = 6y D = 12y – 36 For y = 1, D < 0 => saddle point at (3/2, 1) y = 27/2, D > 0 & Fxx > 0 => relative minimum 5/23/2019 rd

Conics locus of points in plane whose distance from a fixed point (focus) and a from a given straight line (directrix) have constant ratio. e = r/a e = 0 => circle 0 < e < 1 => ellipse a r e = 1 => parabola e > 1 => hyperbola a r 5/23/2019 rd