# mathematical notation and terminology

MATH 1330X-MIDTERM # 1 Instructor: Hai Yan Liu June 10th, 2021 Last Name: First Name: ID# Instructions. You must sign below to confirm that you have read, understand, and will follow them. • You have 80-minute (70 minutes to finish your writing and 10 minutes to upload on brightspace)closedbook to complete this exam. One (1) single-sided, hand-written, (non-mechanically reproduced), 8.5×11 (Letter-size) sheet. Calculators are permitted. • The exam consists of 6 questions on 8 pages. Total value of the exam is 28 points. • All questions are long-answer. To receive full marks, your solution must be correct, complete, and show all relevant details. • Be sure to read carefully and follow the instructions for the individual problems. • For rough work or additional work space, you may use the back pages. Do not use scrap paper of your own. • Use proper mathematical notation and terminology. • You may ask for clarification. 1. (3 points) Find all values of x in the interval [0, 2π], such that √ 2 sin(x) = − 3. √ sin(x) = − x = 3 2 4π , 3 1 x= 5π 3 2. (6 points) Consider the following discrete-time dynamical system: Mt+1 = 0.6Mt + 8. (a) Write down the updating function of the dynamical system. y = f (x) = 0.6x + 8 (b) Find all equilibrium points of the dynamical system, if any. x∗ = d 8 8 = = = 20 1−r 1 − 0.6 0.4 (c) Give the solution of the dynamical system with M0 = 0. Mt = rt (M0 − M ∗ ) + M ∗ = −20 × 0.6t + 20 (d) Draw the cobweb diagram of the dynamical system with M0 = 0. (e) Determine the stability of the equilibrium point using the cobweb diagram, or stability theorem for DTDS with linear updating function. Since |r| = 0.6 < 1, so the equilibrium x∗ = 20 is stable. 2 Question 2 continue… 3 3. (8 points) Find the following limits, if they exists. |x − 4| x→4 x2 − 16 (a) lim lim− x→4 −(x − 4) 1 |x − 4| = lim− =− 2 x − 16 x→4 (x − 4)(x + 4) 8 lim+ x→4 |x − 4| (x − 4) 1 = lim+ = 2 x − 16 x→4 (x − 4)(x + 4) 8 |x − 4| DNE( Does not exist) x→4 x2 − 16 so lim 3 + 2e−4x (b) lim x→−∞ 1 − 16e−4x 3e4x + 2 2 1 3 + 2e−4x e−4x (3e4x + 2) = lim = = − lim = lim x→−∞ 1e4x − 16 x→−∞ 1 − 16e−4x x→−∞ e−4x (1e4x − 16) −16 8 x (c) x→∞ lim ln(x) − ln(2x − 1) x→∞ lim ln(x) − ln(2x − 1) = x→∞ lim ln 2x − 1 x = lim ln = ln(1/2) x→∞ x(2 − 1/x) (d) x→∞ lim lim x→∞ √ 9×2 + x − 3x √ 9×2 + x − 3x √ √ 9×2 + x + 3x 2 = lim ( 9x + x − 3x) √ 2 x→∞ 9x + x + 3x 2 2 9x + x − (3x) 1 x q = lim √ 2 = lim = x→∞ x→∞ 6 9x + x + 3x 3x( 1 + 1/(9x) + 1) 4 4. (5 points) Consider the following function g(x) = a (sin(x))2 +1 if x < π/2 kx+1 if x ≥ π/2 x+1 (a). What is the condition on a and k such that g(x) is continuous at π/2. (b). Find a and k so that g is continuous and has the horizontal asymptote y = 2 as x → ∞. If f (x) continues at x = π/2, then we have : lim− g(x) = lim+ g(x) x→ π2 lim− g(x) = x→ π2 We have kπ + 2 a = ; 2 π+2 x→ π2 k π2 + 1 a = = lim+ g(x) π (sin(π/2))2 + 1 +1 x→ π2 2 we have a = 2(kπ + 2) π+2 For (b), we have the horizontal asymptote y = 2 as x → ∞. kx + 1 lim = k so k = 2 and a = 2(2π+2) = 4(π+1) . π+2 π+2 x→∞ x + 1 5 From x→∞ lim g(x) = 5 (4 points) (a).Using only the definition of derivative as a limit, calculate f 0 (x) where f (x) = 2 1 − 3x . f (x + h) − f (x) = lim f (x) = lim h→0 h→0 h 0 = lim h→0 2[(1−3x−(1−3(x+h)) (1−3(x+h))(1−3x) h 2 1−3(x+h) − 2 1−3x h 2(3h) 6 = h→0 h(1 − 3(x + h))(1 − 3x) (1 − 3x)2 = lim (b). Find the tangent line at x = 1. Since tangent line is : y − f (x0 ) = f 0 (x0 )(x − x0 ) f (1) = −1, and f 0 (1) = 6/4 = 3/2. So tangent line is 5 3 y − (−1) = 3/2(x − 1) or y = x − 2 2 . 6 6 (4 points) Biologists have determined that a certain population of zebrafish grows according to a Beveton-Holton model, namely , the number of the zebrafish per litre satisfies the DTDS for measured in weeks: xt+1 = 31.2xt 13 + xt (a). Give the updating function for this DTDS y = f (x) = 31.2x 13 + x (b). Find all fixed points of this DTDS. Let x = f (x) and solve it. x = f (x) = 31.x 13 + x 31.2x 0 x(13 + x) = x(13 + x − 31.2) = x(x − 18.2) = 0 x∗ = 0, x∗ = 18.2 We have two fixed points x∗ = 0, x∗ = 18.2. 7 Space for additional work 8 MAT1330X CALCULUS FOR THE LIFE SCIENCES I 9. Derivatives: The Definition and Basic Rules Lec 8 mini review. continuous at a number x = a lim f (x) = f (a) x→a discontinuous at a number x = a I f (a) undefined I a new limit law: if f (x) is continuous at x = g(a), then lim f (g(x)) = f lim g(x) x→a ( May 27) I (fails 1) lim f (x) DNE (fails 2) lim f (x) 6= f (a) (fails 3) x→a x→a x→a √ Exercise 9.1. From the definition, calculate the derivative of f (x) = 8×2 + 1 at x = 0 and at x = −1. When calculating these derivatives at different points for the same function, do you notice any patterns or similarities? D EFINITION OF THE D ERIVATIVE AS A F UNCTION The derivative of f (x), denoted f 0 (x), is the function • The domain of f 0 (x) is the set of all x in the domain of f , for which the above limit exists. • The domain of f 0 can be smaller than the domain of f . • f (x) is differentiable at all x such that the limit f 0 (x) exists. • As a function of x, we can regard the value of f 0 (x) geometrically as the slope of the tangent line of f at x Example 9.2. Using the definition of the derivative, calculate the derivative of f (x) = ∗ 2 . 1 − 3x These notes are solely for the personal use of students registered in MAT1330; their creator does not give permission for this material to be uploaded or shared online. 1 Example 9.3. Using your derivative from Example 9.2, find the slope and equation of the 2 tangent line to the graph of f (x) = at x = −1. 1 − 3x 2 O THER N OTATION FOR T HE D ERIVATIVE For y = f (x), here are several equivalent notations for the derivative f 0 (x). H OW F UNCTIONS C AN FAIL T O B E D IFFERENTIABLE Discontinuities. Corners. Vertical Tangents. 3 B ASIC R ULES OF D IFFERENTIATION : P OWERS Constants Powers n=0 n=1 n=2 n=3 The Power Rule (n ∈ R) 4 B ASIC R ULES OF D IFFERENTIATION : C ONSTANT M ULTIPLES AND S UMS Constant Multiples Sums and Differences Example 9.4. Differentiate each of the following functions. Bonus: Find their 2nd derivatives. √ a. g(x) = 2×3 + − x − 5 b. f (x) = π+ √ 3 x − x8 x4/3 5 T HE P RODUCT R ULE Important Remark. d • We might wish the Product Rule was dx f g = f 0 g 0 , just like we might sometimes wish √ √ √ that a + b = a + b, but both of these wishes are actually false in general. • The (actual) Product Rule is the correct rule that describes how rates of change are related in a product. The Product Rule 6 Example 9.5. Find the derivative of f (x) = (x7 + 8×3 − 16x + 2)(15x−1 − 4x−3 + x − ln 4). Then find f 0 (1). S TUDY G UIDE Important terms and concepts: Definition of the Derivative at a: slope of tangent line f (a+h)−f (a) h h→0 f 0 (a) = lim instantaneous rate of change derivative at a point f 0 (x) = lim Definition of the Derivative as a Function: h→0 f (x+h)−f (x) h Relationship between the graph of f and the graph of f 0 f 0 (x) y 0 Other notation for derivative of y = f (x): Differentiability =⇒ Continuity dy dx df dx d f (x) dx Dx f (x) Failing to be differentiable (discontinuity, corner, vertical tangent line) Constant Multiple Rule: for any k ∈ R d , kf (x) = kf 0 (x) dx Constant Rule: for any c ∈ R d , c =0 dx Sum/Difference Rule: d f (x) ± g(x) = f 0 (x) ± g 0 (x) dx Power Rule: for any n ∈ R d n , x = nxn−1 dx Exercises: Adler & Lovrić textbook, 2nd ed. Prof. Lutscher’s Notes (posted on Virtual Campus) 7 Product Rule: d f g = f 0g + f g0 dx §4.5 # 8–28, 39–53 §5.1 #1–26 §8.1 pg. 31 # 1–2 MAT1330X CALCULUS FOR THE LIFE SCIENCES I 9. Derivatives: The Definition and Basic Rules Lec 8 mini review. continuous at a number x = a lim f (x) = f (a) x!a a new limit law: if f (x) is continuous ⇣ at x = g(a), ⌘ then lim f (g(x)) = f lim g(x) x!a ( May. 27 ) discontinuous at a number x = a I f (a) undefined (fails 1) I x!a lim f (x) DNE (fails 2) I x!a lim f (x) 6= f (a) (fails 3) x!a p Exercise 9.1. From the definition, calculate the derivative of f (x) = 8×2 + 1 at x = 0 and at x = 1. When calculating these derivatives at different points for the same function, do you notice any patterns or similarities? D EFINITION OF THE D ERIVATIVE AS A F UNCTION The derivative of f (x), denoted f 0 (x), is the function ‘ f (× ) = fkth ) him h→0 f- – FK ) • The domain of f 0 (x) is the set of all x in the domain of f , for which the above limit exists. • The domain of f 0 can be smaller than the domain of f . • f (x) is differentiable at all x such that the limit f 0 (x) exists. • As a function of x, we can regard the value of f 0 (x) geometrically as the slope of the tangent line of f at x Example 9.2. Using the definition of the derivative, calculate the derivative of f (x) = f ‘ k) = = ⇤ 2 1 3x . lim f(xth)-fK)h→0 h leg Fzfxth) ⇒ – = These notes are solely for the personal use of students registered in MAT1330; their creator does not give permission for this material to be uploaded or shared online. 1 ⇒ Xta ) =knso⇐Eai – h→o(F3xshH-3xT )# deja 24-3×1 him = h) 24-3×-3 – him 24.3×1-24.3×-340 -3×-344-3×7 ( h ) = I = him = him 2 6×-2+6×+6 – h→o=3xshXt3x)(h h 6k h→oF3F3hXt3x)(h) 6 him = h→oF3xshXt3x) than = 6- = ( 1-3×4 Example 9.3. Using your derivative from Example 9.2, find the slope and equation of the 2 tangent line to the graph of f (x) = at x = 1. 1 3x From Example 9.2 %f ° : the The The slope value of the of fat We , ‘t 1) – found Tapp tangent x= -1 is point Slope equation – using m=E and that to line f- C- 1) of the = the off derivative Ep =¥µ fz = graph tangent ( xo ,yo)=f1 E) , the of , the line ‘s Exp of = fat x= is eqa Y is – y Equivalently , 2 = I – is fyt )=3g ¥=£ = line ‘ f (×) is Yo – =m(× k=3g(× y= – – xo ) Ct ) ) Ext } . O THER N OTATION FOR T HE D ERIVATIVE For y = f (x), here are several equivalent notations for the derivative f 0 (x). ‘ f (x ) ‘ Y = = =ad×f(xD doff Dxfk ) = H OW F UNCTIONS C AN FAIL T O B E D IFFERENTIABLE 🙁 Differentiable Discontinuities. Recall % IF f- Functions ⇒ Atta discontinuous is are -2 × a) ( not be at continuous # differentiable a) at ×=a asymptotes orjumps )=×,E}×- forfk • < ⇒ holes , vertical at £ E±f(×)=x2-2X_ # then f- will , differentiable not at ‘ f (2) , DNE > is at ) differentiable Corners. Functions are not ( corners or ” cusps because ” the slope from dteheitosinaeeroDt@fiEhoEEnaIFoD0othederivatiVeCwhichnisalimitdescribingslopesjdoesnotexist.t # FK ) = for 1×1 < ✓ Vertical Tangents. Recall : the slope of FK )=/X| ‘ , f ( o ) DNE . > a vertical line is undefined EgMnYeaYtahxnEsasenft-s@enftaiDnetforffy-Fx.f E±fktF× < ¥rticaltangentatxo 3 ‘ (o) DNE . . B ASIC R ULES OF D IFFERENTIATION : P OWERS Constants Letter . da,G]=lf .mg#=h&nsoF=lhtIo0=0 ” l horizontal f×k9=gt×[ 11=0 da,E]=o -09 lineytchasslopem Powers n=0 . ( constant ddyfxoto rule ) n=1 .gg#hfI=fimohF=hhIot=t dg×[×]=lg da,[x]=1 n=2 dQ×[×2]=qhnjoQthf÷X2=fignox2t2xh#2×2 h =lfnfo2×hg_+h2=lfmgo2xth df×Q4=2× =2× n=3 at,fx3]=lfngo(×+h)q=x3=lg .gg#h)KthX=h)x3h=qlifno(xth)(x2t2qxh_h2)X3=lqignox3t3x2ht3x=h2h3X3=lim3x2ht3qxh2+h3_=qlingo3x2t3xhth2 h→o = I DX 3×43×61+02=3×2 The Power Rule (n 2 ) Foranypowerne _R , ft×[xn]=nxm 4 ‘ [×3]=3×2 B ASIC R ULES OF D IFFERENTIATION : C ONSTANT M ULTIPLES AND S UMS Letke .IR Constant Multiples adyfkfk)] = dd×[kfC×D=kf’ H . bgfkfktghtkfI-kffmjofkt.hn#=h.f'( ) × Sums and Differences )tgKD=f’ )tg’ ddxffcx ) (×) (x ddxffcx – gKD=f’ ( x ) – g ‘K ) Example 9.4. Differentiate each of the following functions. Bonus: Find their 2nd derivatives. p a. g(x) = 2×3 + x 5 gk ) ⇒ – ¥42 tat gyx )=6×2 2×2×42-5 g ‘Cx)=2jd×[x3]ad×[×”2]ddx[5] =2( 3×9-(3×42) – => b. f (x) = x ) – ‘zx’ %) = – p 3 “(x)=6(2x 94×7=12×+4×+3,2 12×+415312 0 gYx)=6X2ztrx ⇡+ g – x8 x4/3 -152-231×1713 (-4×-3.213413)×1413=-415×-713 f “Q)=-4Ezf})×- 93 fkkkt.nl#3X8=hx’4ts+xt_xao/3fYx)=-4zEXM3 (20-3×143) ⇒ fyx )=If¥x→B)tfx’ 2) ‘ – – =28gIX” -152-231×1713 5 43+2×3 -351×1413 T HE P RODUCT R ULE fxtffCxjgCxD-hkfnofkthtgCxthIgCyh-qlignofQthKgCxth1-gHtgQD-ffygfyClevertricK1_subtraotandaddgHh-hhmfofKth1gKthtfCxth1gfytFKth1gH-HAgCxj-h-lfnfofCxthY9kthf_9EytgcxsfkthfffY-fhnfofkthD.fhmgo9kthf_9cIJt@gmail.com h→o ftp.gyx ) = =f’ ( x )gk)tFQ)g’ + ) gQ)f’ ( × ) ( ×) Important Remark. ⇥ ⇤ d • We might wish the Product Rule was dx f g = f 0 g 0 , just like we might sometimes wish p p p that a + b = a + b, but both of these wishes are actually false in general. • The (actual) Product Rule is the correct rule that describes how rates of change are related in a product. The Product Rule =F’K)gk)+fK)g’ k ) dd×[fktg(xD 6 Example 9.5. Find the derivative of f (x) = (x7 + 8×3 Then find f 0 (1). f ‘ F (×) ‘ ( 7×6+24×2 = (1) – I 6)(15 X ‘ – 4×3 = (749+2442)-1 6) ( 154T 44 = (7+24-16) ( 15-4+1 ( I 5) ( 12 180 = ‘ – = = tx 190 – ln l5ln4 – 4) + In f 5) f 4) 3) + – t bn 4) 1 – 16x + 2)(15x + ln 4) 1 4x 3 +x ln 4). ( x +8×3-16×+2) (-15×2+12154+1) ‘ + (17+843)-164 ) + 2) (-1549+12 (541+1) (1+8-16+2) (-15+12+1) 2) +10 15h 4 – S TUDY G UIDE Important terms and concepts: ⇧ Definition of the Derivative at a: slope of tangent line f (a+h) f (a) h h!0 f 0 (a) = lim instantaneous rate of change derivative at a point ⇧ Definition of the Derivative as a Function: f 0 (x) = lim h!0 f (x+h) f (x) h ⇧ Relationship between the graph of f and the graph of f 0 ⇧ Other notation for derivative of y = f (x): f 0 (x) ⇧ Differentiability =) Continuity y0 dy dx df dx d f (x) dx Dx f (x) ⇧ Failing to be differentiable (discontinuity, corner, vertical tangent line) Constant Multiple Rule: for any k 2 ⇤ d⇥ , kf (x) = kf 0 (x) dx Sum/Difference Rule: ⇤ d⇥ f (x) ± g(x) = f 0 (x) ± g 0 (x) dx Constant Rule: Power Rule: for any c 2 for any n 2 d⇥ ⇤ , c =0 dx d ⇥ n⇤ , x = nxn dx Exercises: Adler & Lovrić textbook, 2nd ed. Prof. Lutscher’s Notes (posted on Virtual Campus) 7 Product Rule: 1 d⇥ ⇤ f g = f 0g + f g0 dx §4.5 # 8–28, 39–53 §5.1 #1–26 §8.1 pg. 31 # 1–2 MAT1330X CALCULUS FOR THE LIFE SCIENCES I 10. More Derivative Rules: Quotient Rule, Chain Rule, ( June. 1) Rules for Exponential, Logarithmic, and Trigonometric Functions Lec 9 mini review. Definition of the Derivative at a number x = a: (a) f 0 (a) = lim f (a+h)−f h h→0 Constant Multiple Rule: for any k ∈ R, d kf (x) = kf 0 (x) dx Constant Rule: Definition of the Derivative as a Function: (x) f 0 (x) = lim f (x+h)−f h h→0 Sum/Difference Rule: d f (x) ± g(x) = f 0 (x) ± g 0 (x) dx Power Rule: for any c ∈ R d c =0 , dx When f 0 (a) DNE: discontinuous at a corner at a vertical tangent line at a Product Rule: for any n ∈ R d n x = nxn−1 , dx d f g = f 0g + f g0 dx Q UOTIENT R ULE Exercise 10.1. Derive the Quotient Rule from the Product Rule as follows: 1. Let Q(x) = N (x) . D(x) 2. Isolate N (x) and find N 0 (x) using Product Rule. 3. From the expression found in step 2, isolate Q0 (x) from your equation in order to find an expression for the derivative of the quotient Q(x). The Quotient Rule ∗ These notes are solely for the personal use of students registered in MAT1330; their creator does not give permission for this material to be uploaded or shared online. 1 Example 10.2. Differentiate each of the following: a. q(x) = 4×5/2 + x3 + 2x − b. 6 x4 g(x) = h(x) √ 2 x2 f (x) where h(x) and f (x) are differentiable functions. If h(2) = 10, f (2) = 1, h0 (2) = 2, and f 0 (2) = −1, then what is g 0 (2) ? T RIG D ERIVATIVES investigation of sine’s derivative rough sketch of sine’s derivative 2 sin(h) cos(h) − 1 and lim h→0 h→0 h h Two important limits lurking within the derivative of sin(x): lim h sin(h) h cos(h) − 1 h Derivative of Sine 3 Derivative of Cosine Derivative of Tangent Derivative of Secant Derivative of Cosecant Derivative of Cotangent 4 R ULES FOR D ERIVATIVES OF E XPONENTIAL F UNCTIONS Let f (x) = bx for some base b > 0, b 6= 1. Then bx+h − bx d f (x) = lim = lim bx h→0 h→0 dx h bh − 1 h = lim b h→0 x b0+h − b0 h =b x b0+h − b0 lim h→0 h = bx f 0 (0) Derivative of the Natural Exponential Function Example 10.3. At which values of t (if any) does the graph of g(t) = et (t3 − 3t2 ) have a horizontal tangent line? 5 T HE C HAIN R ULE Some special examples of the Chain Rule: Power Chain Rule: Example 10.4. Differentiate: y = p x3 + 4ex − cos(x) Example 10.5. Differentiate: f (x) = cos3 (x) 6 Example 10.6. Differentiate: f (θ) = tan(5θ + 3) − θ 100 2 Exponential Chain Rules: Example 10.7. Differentiate: g(s) = −es 3 −s f (x) = e5x + 3x + 24 Example 10.8. Find the slope of the tangent line to the graph of y = esin x at x = 0. 7 Chain Rule for Inverse Functions: R ULES FOR D ERIVATIVES OF L OGARITHMIC F UNCTIONS S TUDY G UIDE quotient rule: d f (x) f 0 (x)g(x) − f (x)g 0 (x) = 2 dx g(x) g(x) derivative rules for trig functions: d [sin x] = cos x dx exp. rules d x [e ] = ex dx d [cos x] = − sin x dx d x [b ] = ln(b)bx dx d [tan x] = sec2 x dx (and others!) the chain rule: d d 1 [f (g(x))] = f 0 (g(x))g 0 (x) log. rules [ln(x)] = dx dx x Exercises Adler & Lovrić textbook, 2nd ed. Prof. Lutscher’s Notes (posted on Virtual Campus) 8 §4.5 # 8–28, 39–53 §5.1 pg. 272 #27–31, 43, 47 §5.3 pg. 294 #1–14, 17–45 §9.1 pg. 37 # 1–2 d 1 [logb (x)] = dx ln(b)(x) §5.2 pg. 280 #1–43 §5.4 pg. 305 #1–31 — C ALCULUS FOR THE L IFE S CIENCES I — MAT1330 C OURSE G UIDE The course is based on the book: Calculus for the Life Sciences: Modeling the Dynamics of Life by F.R. Adler and M. Lovric. These notes were developed at the University of Ottawa for MAT1330, originally by Frithjof Lutscher, with modifications by Monica Nevins, including themes from the Self-Regulated Learning Module developed by Alison Flynn. Preliminaries and expectations To be good at math (in fact at anything): practice, and know yourself! Find your resources. Do 5 problems every day. If you miss a day, do 10 the next. Check with your friends, your TA, the math help centre, and with me. I am here to facilitate your learning. If you think you understand everything in class, think again. Learning math is learning a language, and it’s powerful one, as you’ll see over time. You need to build vocabulary and learn the grammar. Practice! School is more about“how” and less about “why”; University is more about “why” and less about “how”. It is your responsibility to practice the “how” until you know. Draw a picture whenever you can! This document provides a short summary for each class with reference to the corresponding book chapter and practice problems. The problems in these notes are taken from previous MAT1330 assignments and exams. Use it w…

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