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MTH 132 Calculus Week one notes

by: Ren K.

MTH 132 Calculus Week one notes MTH 132

Marketplace > Michigan State University > Mathematics > MTH 132 > MTH 132 Calculus Week one notes
Ren K.
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These notes include lecture one, from August 31st and lecture two on September 2nd. These are free as part of a promotion! These lectures are guaranteed to help you in your studying.
Calculus 1
Z. Zhou
Class Notes
Calculus, MTH, Math, Zhou, Z.Zhou




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This 6 page Class Notes was uploaded by Ren K. on Thursday August 25, 2016. The Class Notes belongs to MTH 132 at Michigan State University taught by Z. Zhou in Fall 2016. Since its upload, it has received 36 views. For similar materials see Calculus 1 in Mathematics at Michigan State University.


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Date Created: 08/25/16
MTH 132 ­ Lecture 1 ­ Tangents     Tangents  ● Tangent lines are lines that that touch a curve at one point.  ● Here’s how to find the tangent line at the point (x൦,f(x൦)).  ● It’s the same as finding the slope! The change in x / the change in y.  ​ ​● y​ ​ 1​​  m​  –​ 1​)  Secants  ● The same rule applies for secant lines.  ● Secant lines touch a curve at multiple points. (Two or more specifically.)  ○ The formula for secant lines is different.  ● It's the second y point ­ the first y point / the second x point ­ the first x point.  ● f(x൦+h)­f(x൦) / x൦+h­x൦ or simply f(x൦+h)­f(x൦) / h  ● The limit of a slope of a secant line is when h approaches 0.  Velocity  ● Basically, it’s the same idea for velocity.  ○ Y = f(t) = position of a particle.    ● Use it to find the average velocity from t൦ to t൦+h.  ● velocity(t൦) as the limit h approaches 0 = f(t൦+h) ­ f(t൦)/h = change in position/ change in  time.          Example:  2  ● Your precious laptop is dropped from a bridge has a position of y = f(t൦) = 16t .  ● It’s below the bridge after t seconds. Find the average velocity from [3,4] seconds.  ○ f(4)­f(3)/4­3  ○ 16*4 ­ 16 * 3 / 1  2 ○ 16[4 ­32 = 16x] = 112 ft/ sec  ● Find the velocity of your laptop at exactly 3 seconds.  ○ V = 3 as h →  0  ○ f(3+h)­ f(3) / h  ○ 16 * (3 + h) − 16 3 /h  * ○ 16 * (3 + h) − 3 /h  ○ 16 * 3  + h + 6h  −  3 /h  ○ 16(h + 6) = 16 + 6 = 96 ft/sec  ○ At exactly 3 seconds: 16(3+6) = 16*9 = 144 ft/ sec  ○ To miles: 96 * 3600/5280 = 65.5 miles.      MTH 132 ­ Lecture 2 ­ Limits    General definition  ● Let f(x൦) be a function near a. (a may not be in the domain)  ● We say that the limit of f(x൦) = L as x approaches a is denoted by limit f(x൦) = L  ● If f(x൦) is arbitrarily close to L by x sufficiently close to a (but not equal to) = the limit.       ○ We are interested in the behaviour of f(x൦) near a.  ○ Finding the limit has nothing to do with the value f(a).  Both Sides  ● For a limit to be defined ‘normally’ it has to have the same solution from both sides;  approaching from the positive side and approaching from the negative side.  ○ Otherwise it does not exist.   ○ Some can non exist because of oscillation.    ○ .    ● We can also define a one sided limit, where instead of approaching from both sides like  limit x approaches 1,  ○ we add a plus symbol to designate whether we’re approaching from the positive  side (the right)   ○ or a minus symbol to say we’re approaching from the negative side (the left).      ● If the limit exists from both sides, then the solution with the + and the solution with the ­  would be equivalent. Otherwise, if they’re different then the limit does not exist.  Finding limits!     


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