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by: Ren K.

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# MTH 132 Exam 1 Study Guide MTH 132

Marketplace > Michigan State University > Mathematics > MTH 132 > MTH 132 Exam 1 Study Guide
Ren K.
MSU
GPA 4.0

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These are a summary of key concept from all past lectures 1.4 - 2.8
COURSE
Calculus 1
PROF.
Z. Zhou
TYPE
Study Guide
PAGES
8
WORDS
CONCEPTS
Calculus, Calc, MTH, 132, Engineering, Math, Z.Zhou, Zhou
KARMA
50 ?

## Popular in Mathematics

This 8 page Study Guide was uploaded by Ren K. on Thursday October 6, 2016. The Study Guide 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: 10/06/16
MTH 132 - Exam 1 Review Guide Lecture 1 - Tangents Tangents ​ ​ y ​ 1​y​ =​ m(​1​–​ x ) Secants ● f(x൦+h)-f(x൦) / x൦+h-x൦ or simply f(x൦+h)-f(x൦) / h ● The limit of a slope of a secant line is when h approaches 0. Velocity ● Y = f(t) = position of a particle. ● velocity(t൦) as the limit h approaches 0 = f(t൦+h) - f(t൦)/h = change in position/ change in time. Lecture 2 + 3- Limits ● We are interested in the behaviour of f(x൦) near a. ● Finding the limit has nothing to do with the value f(a). Both Sided limits ● 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. ● 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. ● Limit x→a f(x) = L ● f(x) approaches L when x approaches a, but x = / a. ● f(x) doesn’t necessarily have to be defined. ● Limit x → a+ ○ f(x) = L ● Limit x → a- ○ f(x) = L ● If x→a+ and a- are not equivalent then x→a limit DNE (Does not exist) Limit Calculations ● If Limit x→a f(x)= L ● Limit x→a g(x) = M ● Sum / Difference ○ Limit x→a f(x) ± g(x) = L ± M ● Constant Multiple ○ Limit x→a [c*f(x)] = C * L ● Product ○ Limit x→a f(x)g(x) = L*M ● Quotient ○ Limit x→a f(x)/g(x) = L/M as long as M does not equal 0. ● Power n n ○ Limit x → a f(x) = ( Limit x → a f(x)) ● Root n n ○ Limit x → a√ f(x) √Limit x → a f(x) ○ When L doesn’t equal 0, you can have 3 different solutions ■ If positive - positive infinity ■ If negative - negative infinity ■ Or a DNE. Squeeze theorem ● If g(x)≤ f(x) ≤ h(x) and Limit x → a g(x) = Limit x → a h(x) = L ● Then f(x) = L Lecture 4 - Epsilon Deltas ● For any ε > 0, we can find ▯>0 such that |f(x) − L|< ε for all 0 <|x−a|<▯ ● Example: ○ Limit x → 2 (3x+2) = 8 ○ |f(x) - 8| ＜ ε ○ ⇔|3x + 2 - 8| <ε ○ ⇔|3x - 6| < ε ○ ⇔3|x-2|<ε ○ ⇔|x-2|<ε/3 ○ ⇔-ε/3 < |x-2 < ε/3 Lecture 6 - Continuity ● Definition: f(x) is continuous at x = a, if and only if lim x→a f(x) = f(a) 1. Limit exists. 2. f(a) = defined. 3. f(x) = f(a) Between [a,b] ● Definition:​ if f(x) is defined in (a,b) we say that it is continuous if f(x) is continuous at every point of (a,b). ● If continuous at [a,b] a​___________​_b 1. If continuous at m(a,b) 2. Limit x→a f(x) = f(a) 3. Limit x→b f(x) = f(b) Lecture 6, 7, 8 - Derivatives Derivatives ● The hard way - limit of x as it approaches a [f(x+h)-f(x)]/h ● The easy way- ○ f(x) = constant f’(x) = 0 ○ f(x) = mx f’(x) = m ○ f(x) = x^2 f’(x) = 2x ■ Because 2*x^(2-1) = 2x ■ Formula power*x^(power - 1) = derivative version ○ f(x)= 1/x f’(x) = 1/(x^2) ■ Because 1/x = x^-1 ■ = x^(-1-1) ■ = x^-2 = 1/x^2 ○ f(x) =√x f’(x) = 1√(2 x ) ■ Because x√= x^½ ■ x^(½) - 1 = x^-½ ■ x^-½ = 1/2√x Theorem ● If f(x) is differentiable at a, meaning that f’(a) is possible then f(x) must be continuous at point a. Exceptions ● An equation can be continuous at all points, but cannot be differentiated. ● Example: ○ f(x) = |x| ■ X if x≥ 0 ■ -x if x≤ 0 ○ Because the slope alternates, you cannot derive it. ○ H tends to 0 must be the same from both sides. Basic Rules ● In general: ○ x^n = n*x^(n-1) ● Sum: ○ (f(x)+g(x))’ = f’(x) + g’(x) ● Difference: ○ (f(x)-g(x))’ = f’(x)-g’(x) ● Constant: ○ (cf(x))’ = cf’(x) ● Product: ○ (f(x)g(x))’ = f’(x)g(x) +f(x)g’(x) ● Three Products: ○ (f1*f2*f3) = f1’f2f3 + f1f2’f3 + f1f2f3’ ● Quotient: ○ (f(x)/g(x))’ = f’(x)g(x)-f(x)g’(x) / g(x)^2 Lecture 9 - Trig Derivatives Sine and Cosine addition ● sin(a+b) = sin(a)cos(b)+cos(a)sin(b) ● cos(a+b) = cos(a)cos(b)-sin(a)sin(b) Derivative of Sine ● sin(x)’ = cos(x) Cosine Derivative ● cos(x)’ = -sinx Tangent Derivative ● tan(x)’ = sec^2x Cotangent Derivative ● cot(x)’ = -csc^2x or -1/sin^2x Chain Rule ● f(g(x))’ = f’(g(x))*g’(x) Lecture 10 - Chain Rule Formula Chain Rule ● f’(u)g’(x) u = g(x) ● = f’(g(x)) * g’(x) Alternate Phrasing Formula ● d/dx [f(g(x)] = df/du * du/dx Power Rule ● A(x) = [f(x)]^n ● A(x) = u^n ○ U = f(x) ● a’(x)= u^n * u’ = n * u ^(n-1) f’(x) ● = nf(x)^(n-1)f’(x) ● (-x^-n)’ = -nx^(-n-1) Lecture 11 - Velocity 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) Acceleration ● v’(t) = [v(t+h)-v(t)]/h ○ v’(t) = a(t) = Acceleration Direction ● v’(t) > 0 ○ Movement in a positive direction ● v’(t) < 0 ○ Movement in a negative direction 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 Lecture 12 - Position Functions ● 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 Lecture 13 - Implicit Derivatives Implicit Differentiation ● F(x,y) = 0 ● Y = y(x) ○ Use chain rule to find y’ Example: ● x^3+y^3 = 9xy ● (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) Lecture 14 - Related Rates Related Rates ● Related rates - when two or more variables are related to each other in a problem; ○ F(x,y) = 0. ○ We know x’(t) and want to solve for y’(t) Steps for related rates problems: 1. Read the problem carefully. 2. Draw a picture if possible. 3. Name all variables depending on t. 4. Use an arrow to indicate the unknown rates of variables and unknown rates. 5. Find relationship between the variables a. Typically this is used in geometry, physics - many fields. b. F(x(t),y(t)) = 0 6. Take the derivative with respect to t on both sides of F(x(t),y(t)). a. Use the chain rule. 7. Find the relations between rates x’(t) and y’(t) and positions x(t) y(t). a. Solve for unknown rates. 8. Plug in the number for solution. a. Typically not stated outright - read carefully. Example: ● A ten foot ladder is resting on the side of a building wall. If the bottom of the ladder slides away from the wall at 1 foot per second, what will the rate be of the bottom of the ladder when it is six feet away from the wall? ● x + y = 10 = 100 ● Take the derivative with respect to time (t). ● 2x(t) * x’(t) + 2y(t) * y’(t) = 0 ● Use implicit derivative to solve for y’(t) ● y’(t) = [ x(t) x’(t) ] / y(t) ● = ( -x / y )*(x’(t)) ● Simplified version of the derivative - again, via implicit derivation. ● Plug in the number given - 6 feet. ● y’(t) = − 6/ 100 − 3* − 1 ● = -6/8 = -¾. ● The rate of the ladder when it is six feet away from the wall is -¾ feet per second.

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