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Math 103, Week 7 notes

by: Cambria Revsine

Math 103, Week 7 notes MATH 103 001

Cambria Revsine

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These notes cover material from chapters 3.9-3.11 from Thomas' Calculus
Intermediate Algebra Part III
William Simmons
Class Notes
Math, Calculus
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This 6 page Class Notes was uploaded by Cambria Revsine on Wednesday March 9, 2016. The Class Notes belongs to MATH 103 001 at University of Pennsylvania taught by William Simmons in Spring 2016. Since its upload, it has received 15 views. For similar materials see Intermediate Algebra Part III in Mathematics (M) at University of Pennsylvania.

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Date Created: 03/09/16
Math 103—Week 7 Notes—3.9­3.11 3.9: Inverse Trig Functions: Graphs of Inverse Trig Functions: *Inverse and regular trig functions have opposite inputs/outputs **Graphs of inverse trig functions are (one cycle of) trig function graphs rotated clockwise 90 Derivative Rules: If x falls within each inverse trig function’s domain… 1 du √(¿−u 2) dx d −1 1 dx(sin u = ¿ 1 2 du √(¿−u ) dx d −1 (cos u )= dx ¿ d −1 1 du (tan u )= 2 dx 1+u dx d −1 −1 du dx(csc u )= u √(u −1) dx | | d −1 1 du dx sec u = 2 dx | |√(u −1) d −1 −1 du dx cot u = 1+u 2 dx 3.10: Related Rates: Related rates problems: Finding a rate of change from other known rates of change, by differentiating multiple variables that are functions of time. Steps: 1.   Draw a picture and name the variables and constants (t= time, all variables are  differentiable functions of t) 2.   Write down additional numerical information. 3.   Write down what you are asked to find  usually a rate (derivative) 4.   Write an equation that relates the variables. (Might have to combine equations to  eliminate variables) 5.   Differentiate with respect to t. 6.   Plug in the known rates, and then evaluate for unknown rate. Ex: A hot air balloon rising straight up from a level field is tracked by a range finder 150 m from the liftoff point. At the moment the range finder’s elevation angle, the angle is  4 increasing at the rate of 0.14 rad/min. How fast is the balloon rising at that moment?                        1.                                         Balloon   dθ =¿ dt dy          .14 rad/min y   =?                dt   Range         θ       finder                        150 m Variables: θ  (angle in radians the range finder makes with the ground) and y (height in  meters of the balloon off the ground) dθ π 2. More information:  =0.14 rad/min when  θ= dt 4 dy π 3. We want to find: dt when  θ= 4 y 4. Equation:  150 =tanθ    y=150tanθ dy 2 dθ 5. Differentiate: dt=150(sec θ) dt 6. Plug in known rates and evaluate:dy=150 sec 2π 0.14 dt ( ) 4 2 √2¿ 0.14 =42 dy dt=150¿  At the moment in question, the balloon is rising at the rate of 42 m/min. Ex: A police cruiser, approaching a right­angled intersection from the north, is chasing a  speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6  mi north of the intersection and the car is 0.8 mi to the east, the distance between them is  increasing at 20 mph. If the cruiser is moving at 60 mph at the instant of measurement, what is  the speed of the car?              Police               cruiser ds =20        Situation when x=0.8, y=0.6 dt dy   =−60 dt dx                    dt                Car   Variables: x (position of car at time t), y (position of cruiser at time t) and s (distance between  car and cruiser at time t) dx dy ds We want to find   when:  x=0.8mi ,  y=0.6mi ,   =−60mph ,  =20mph dt dt dt dy ** dt  is negative because y is decreasing 2 2 2 Equation:  s =x +y ds dx dy 2s dt=2x dt+2y dt ds 1 dx dy = (x +y ) dt s dt dt ds 1 dx dy dt= √(x +y ) (x dt+y dt) 1 dx 20= 2 2(0.8 dt+0.6(−60)) √ 0.8 +0.6 dx 20=0.8 dt−36 56=0.8 dx dt dx =70 dt  At the moment in question, the car’s speed is 70 mph. 3.11: Linearization: Approximating complicated functions with simpler functions (linearizations) at a point   Results in same equation as tangent line, but found in different way Linearization of f at a:  L(x)= f a + f '(a)(x−a) 1 Ex: Find the linearization of (¿+x)  at  x=0 f x = √ ¿ −1 1+x¿ 2 f ( ) ¿1 2 −1 2 1 1+0¿ = 2 f'0 = ¿ 2 1 (¿+0)=1 f ( ) √ ¿ Linearization:  L (x)= f a + f '(a)(x−a) L x =1+ 1(x−0 =)+ x 2 2  Plugging in numbers close to 0 show that the linearization is more accurate the closer to 0 the  input is k 1+x¿ ≈kx    (x near 0; any number k) ¿ Differentials: With the differentiable function  y=f (x) , the differentialdx  is an independent variable.  The differential  dy  is  dy= f ’(x)dx  (Sometimes also called  df = f ’(x)dx ) dy  different set­up for  = f '(x) dx Ex: Find  dy  if  y=x +37x 5x (¿¿4+37)dx d y=¿  E :  d(tan2x) 2 2 sec (2 x)d(2x )=2sec 2xdx  Same as derivative of the function times dx Estimating with Differentials: When  dx=∆ x ,  f (a+dx )= f(a +∆ y   The differential estimation gives:  f(a+dx )≈ f (a)+d y   a  is a (usually) whole number on the x­axis near point a+dx  that is used to estimate f (a+dx)  because  f (a)  is easier to compute.  a  plus the change in  x  equals the new  x. 1 Ex: Use differentials to estimate  3 7.97 1 dy= 3x2/3dx We make  a=8  because it is the closest whole number to 7.97 f a+dx )≈ f (a)+d y :  To make a+dx=7.97 ,  dx  is ­0.03. f (7.97)≈ f (8)+ f'(x)dx       because  dy= f ’(x)dx ¿8 +/3 1 (−0.03) 3 3(8) 1 ¿2+ (0.03 =).9975 12 1 *The true value of  3  is 1.997497… 7.97 Error in Differential Approximation: True change in  f :  ∆ f = f(a+∆ x )− f (a) Estimated change in  f :  df = f'(a ∆ x * ∆ f  and  df  are not the same Approximation error ¿∆ f −df   ¿ f(a+∆ x − f (a − f 'a )∆x ¿( f(a+∆ x )− f a )− f '(a))∙∆x ∆x ¿ϵ∙∆x   f(a+∆ x )− f a )  (  − f'(a) ) is called  ϵ ∆ x Change in  y= f (x)  near  x=a: ' ∆ y= f (a)∆ x+ϵ∆x true change = estimated change + error  * ϵ→0  as  ∆ x→0


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