Calculus 1 Chapter 2 notes
Calculus 1 Chapter 2 notes MTH 161
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This 12 page Class Notes was uploaded by Jack Magann on Sunday February 15, 2015. The Class Notes belongs to MTH 161 at University of Miami taught by Dr. Beverly in Fall2014. Since its upload, it has received 119 views. For similar materials see 161 Calculus 1 in Mathematics (M) at University of Miami.
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Date Created: 02/15/15
Chapter 21 Derivatives and rate of change Tangent line A line which intersects a function at one single point In this graph line T is tangent to curve C at point P Though L looks like the tangent The tangent does not exactly have to pass through only one point on the graph Finding the equation of the tangent line Need the slope and at least one point since m M 962 961 When just given just the equation one cannot do this though Have to find slope of the original function at the point of intersection then ex find the equation of the tangent line to yx2 at point 11 For y x2 the slope is constantly changing so a certain method is needed to find the slope at a single point To start we will pick a second point in the graph to draw a similar line that s close to the tangent but not exact As seen below as we move the second point closer to the original point the line in between them is closer to the tangent line J nppunmlu x tlunl In thI A z A I V I I 1 quot I y fquot z 39 gt I 1 1 quotxx quotK 339 f I J Q Q I 1 Q d pl urtcl1c P lion lllc lcl39l This is then a limit problem where m y2y1 becomes mpq limxa yz yl 962 961 962 x1 2 For this problem m limx1 11 our second point is x x2 So m limx1 W 2 We can now find the slope using y y1 mx x1 y 12x 1 y2x 1 The tangent line to curve fx at point Pa f a is the line through P with slope fxfa ml W W If there are no points given ah is used to replace x the two points are the separated by distance h let ah approach a so the second point now approaches the first Can now write the slope of the tangent formula as mlimfah falimfah fa h gt0 ah a h gt0 h ex find f a ofy 3x2 4x 1 Here there are no points given so the second version of the formula is used Also it isn t asking for the equation of a line at a specific point but rather for an general equation that can be used to find the slope of the tangent at any given point So 1 3ah2 4ah1 3a2 4a1 1 3a2 3a26ah3h2 4a4a4h1 1 1mh gt0 h mlgt0 h 3h26ah4h llmh0 T llmh0 3h 661 4 661 4 f39a 1 Find the equation for the tangent ofy x3 2x 1 at point 322 Following the definition of a derivative 3 3 lima3 lima3 Through the process of synthetic division 2 lima3 W lima3 a2 3a 7 997 25 a 3 Chapter 22 There are two different ways to use the definition of a derivative each with slight variations f a limh0 W is used to find the slope at a fixed point f x limh0 W is used to find slopes at varying points along fx ex fxx3 x find f x 3 3 f39OC limit gt0 Hm xzh x X limh gtO 9633x3h3xh2 h3 x h x3x h 3x2h3xh2h3 h h3x23xhh2 1 h 3x2 1 h hmhao Since a point was not given an equation is the answer There are certain situations where a function is not differentiable at a point TA 39r A y nu cornr um discontinuil IL xt39rllcul tangent These three cases do not allow the function to be differentiable There are many different notations for derivative d d d f x y39 g g are Dfx Dxfx All these are variations on how to notate a derivative There are also notations for second third fourth etc derivatives d f x f x wh1ch 1s the second der1vat1ve Th1s same notatlon 1s repeated for dx higher derivatives Chapter 23 There are certain generalized derivatives based on the definition that make it easier to take the derivative Derivative of a constant since f x c is a horizontal line f x 0 d c 0 dx Derivative of Power Function d d d d x1 x22x x33x2 x44x3 dx dx dx dx d The pattern seen here 15 a x39 nxquot 1 This also works for negative powers and square roots ex fo a xi W g x39 Constant multiple rule If c is a constant and fx is a differentiable function d d Ewen came This can be combined with the power function rule d d dx cx c x dx x cnx Sum ruIe di erence rule d d d a fx i 900 E x i E906 Derivative of Sine cosine These can be thought of a pattern almost since icosbc sinx j x sinx cos x j x cosx sinx sinOc cos x Due to this pattern it is easy to predict very high derivatives of sine or cosine EX 27 d27x cosx sin x Chapter 23 continued Derivatives are most commonly applied to rates of change The original function can be thought of as the position function The first derivative is then velocity Velocity has a direction while speed does not Where velocity is positive position is increasing Where velocity is negative position is decreasing Second derivative is acceleration Where acceleration is positive position is increasing Where acceleration is negative position is decreasing Chapter 24 Product Rule dm 1 d gt d dx x g x f x dxg x gxdxfx use definition of derivative for proof Quotient Rule 1 m gxfx fltxf xgcx dx gx 9062 use definition of derivative for proof Trig functions all proven using the definition of derivatives d t d sinx c032xsin2x 1 2 an x 2 sec x dx dx cosx coszx coszx j x cscx cscx cot x sedx SECOC tan 9 i 2 dx cotx csc x Chapter 25 F x sz 1 This can t be easily differentiated using the rules from previous chapters However Fx is a composite function where fu xEandu gx x2 1 so F x f g 96 We can now find the derivative much easier using the chain rule Chain Rule if f and g are differentiable F x f gx x 93906 ex find F x if Fx 2 sz 1 fuE and gxx21 2x 1 Fx 2m m f39u 5 and g x 2x Chain rule and power rule d I a 900 ngxquot 1 x g 6X differentiate y x3 D100 yr 2 1993x2 300362063 199 Extended Chain Rule fx sin costanx f u sin u u 2 gt cos t t hx tan x f x cos costanx sintanxsec2x Chapter 26 Implicit differentiation An implicit function is one not explicitly defied like all the equations in the previous chapters Explicit function is one such as yx ex The implicit function x2 y2 25 circle not a function and not explicitly defined for y need to start by solving for y 31225 x2 y ixZS x2 Can see that there are two separate equations that make up the original equation Find the equation of the tangent to the circle x2 y2 25 at 34 since the point is part of the top of the circle you must use positive equation of the circle 1 x y39 2x 2x25 x2 xZS x2 So m and can be plugged into the point slope form 3 9 3 25 y 4 TXZ y Tx However not every equation can be set to solve for y or split into two equations must use implicit differentiation Still using x2 y2 25 take derivative of both sides like usual the derivative of x can be taken normally but y represents its own function 2x 2yy 0 Now we can solve for y I 2x x y 2 2 2y y This method is need when you can t solve for y Explanation i means to take the derivative in terms of x of the function that follows 312 doesn t match the definition though El y 2y 2 d zdy d We see y y s1nce dy crosses out dx dy dx d dy CW 1 53 2 a 23 2300 ex 1 Find the derivative of x3 y3 2 69 can not isolate y left side uses the power rule and right uses product rule 3x2 3y2y 6xy 6y x2 3101 ny 2y Now solve for y 962 2y 2901 320quot x2 2y I 3 2x y2 2 Differentiate sinx y yzcos x cosx y 1 y y2 sinx 2yy cosx cosx y y cosx y y2 sinx 2yy cosx cosx y y2 sinx 2yy cosx y cos x y cosx y y2 sinx y 2y cosx cos x y Chapter 27 Related rates As one thing changes usually another is also changing Balloon sphere being blown up volume increasing Surface area increasing Diameter radius increasing thickness of balloon decreasing To find one of these rates of change when we know another we use derivatives to relate them ex Air is being pumped into a balloon The volume of the balloon is increasing at a 3 rate of 100 How fast is the radius increasing when the diameter is 50cm 1identify the known and unknown 3 Known Volume increasing at rate of 100 S Unknown Rate radius is increasing when diameter is 50cm 2Identify formula which relates variables Volume of sphere V 2 gm The knowns should not be plugged in yet 3Use implicit differentiation to take derivative to relate rates of change dV 2 dt 4TH dt 4plug in knowns and solve for unknown 100 M252 9 dt 3 100 i and if we add the units it is i cms Memorize these common formulas used in many related rates problems Pythagorean Theorem a2 b2 c2 Perimeter and area of Triangle A bh P a b c Perimeter and area of rectangles A bh P 2l w Perimeter and area of circles A nrz P 2mquot Given Volume and Surface Area of cones cylinders spheres still should look them up ex 2 10ft ladder is set against a vertical wall The bottom of the ladder slides away at 1 foots What rate is the top of the ladder dropping when the bottom is 6 feet away Usesazb2c2 since 36b2100 b648 Take the derivative Zad a 219 ZCdC dt dt dt dc and s1nce c 1s a constant wh1ch never changes 2c a 0 Plug in and solve for unknown 261 28 0 db 3 fts dt 16 negative since the side of the triangle is shrinking question can imply the direction of the change and answer then might not need negative sign If the previous question asked At what rate was the ladder s distance from the ground decreasing It is clear from the question that the rate is going to be negative and is not really needed in the answer EX3 Two cars are traveling towards the same intersection Car A is going west at 60miles per hour Car B is going north at 50 miles per hour What rate is the distance between the cars changing when car A is 3 miles from the interaction and car B is 4 miles from the intersection The diagram would look like X3 60 y4 d y 50 dt dt Since y2 x2 22 z5 l h 1 The rates are negative since the triangle s area is decreasing 2xE 2 d y ZzE b l in in the knowns and sol in forE dt ydt dt yp ugg g V g dt E 78 mih dt EX4 PartA There is baseball 90 by 90 baseball diamond and a player is running from home plat to first base at a rate of 29 fts When the runner is half way to first at what rate is their distance from second base decreasing Picture is Al If we draw out the triangle we see that d x E2 x45 dt 29 y 90 dt o z 45xE dt Here we use the Pythagorean thermo to solve for g which is rounded to 13 fts Here though the negative isn t need since the question implies it Part B Then the runner is half way to first what is the rate the distance between him and third base increasing If we draw the triangle for the problem we see that it is exactly like part a but 29 so the answer is 13 fts EXS A ladder slides down a vertical wall at 375ms The bottom of the ladder is 5m away sliding away at 9ms How long is the ladder We know that the Pythagorean theorem is used dx dy dz Known X 5 dt 9 y dt 375 z dt 0 dx dy 2 E In the equatlon 2x dt 2y dt Zz dt szt 0 If we then solve for y y12 Plugging that in to the original formula z13 All these notes are from a Calculus 1 class run by Dr on Beverly at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print