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# MTH 132 Week 5 Lecture Notes Calculus 1 MTH 132

MSU

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This 3 page Class Notes was uploaded by Ren K. on Sunday October 2, 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 2 views. For similar materials see Calculus 1 in Mathematics at Michigan State University.

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Date Created: 10/02/16

MTH 132 - Lecture 12 - Position Functions Summary ● s(t) - position function ● v(t) - s’(t) ● a(t) - v’(t) - s’’(t) ● If velocity and acceleration are the same - speed up. ● If velocity and acceleration are different - slow down. Example ● A ball is thrown upward from the top of a 350 foot building, with the ball's initial velocity 10 foot/second. ● How high can the ball get? ○ s(t) = 350 + 10t -16t^2 ○ T1 = highest point = s’(t) = 10 - 32t = 0 ○ T11 = 10/32 = 5/16. ○ 350+10(5/16)-16(5/16)^2 ○ 350 +25/16 = highest point. ● When does the ball hit the ground? ○ Let s(t) = 0. ○ 350 + 10t - 16t^2 = 175 +5t - 8t^2 √ ○ T = (-5 + or - 25 −*4 17* − 8 )/2*-8 = 5 seconds ● Velocity when the ball hit the ground? ○ 10-32*5 ○ 10- 160 ○ -150 ft/sec MTH 132 - Lecture 11 - Rate of Change Velocity ● s(t) = position of a particle ● s’(t) = limit h as it approaches 0 [s(t+h)-s(t)]/h ○ s’(t) = rate of change of position ● Instantaneous velocity = v(t) Direction ● v’(t) > 0 ○ Movement in a positive direction ● v’(t) < 0 ○ Movement in a negative direction Acceleration ● v’(t) = [v(t+h)-v(t)]/h ○ v’(t) = a(t) = Acceleration Speed ● a(t) > 0 and v(t) > 0 ○ Speed up ● a(t) < 0 and v(t) > 0 ○ Slow down ● If a(t) ≠ v(t) ○ Slow down Jerk ● a’(t) = jerk = third derivative of speed Derivative writing 5 ● Once you go past the fourth derivative, write out a (t) instead of a’’’’’(t). MTH 132 - Lecture 13 - Implicit Derivatives Implicit Differentiation ● F(x,y) = 0 ● Y = y(x) ○ Use chain rule to find y’ Example ● Find the derivative on both sides, think y = y(x), use the chain rule. ● x^3+y^3 = 9xy ● Find the tangent line at (2,4) ● (x^3)’ + (y^3)’ = 9xy ● 3x^2 + 3y^2 *y’ = 9[y+xy’] ● 3y^2*y’- 9xy’ = 9y - 3x^2 ● (3y^2-9x)* y’ = 9y - 3x^2 ● Y’ = (9y-3x^2) / (3y^2-9x) ● y’ = (3y-x^2) / (y^2-3x) ● Plug in points (2,4) ● y’(2) = (3 * 4 - 4 ) / (16 - 3 * 2) = 8/10 = ⅘ ● Tangent line = ○ Y - y0 = m(x-x0) ○ Y - 4 = ⅘(x-2)

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