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# MTH 132 Week 6 Lectures MTH 132

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This 10 page Class Notes was uploaded by Ren K. on Saturday October 8, 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 5 views. For similar materials see Calculus 1 in Mathematics at Michigan State University.

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Date Created: 10/08/16

MTH 132 - Lecture 16 - Test Review 1.4-2.8 1. Limit 2. Continuity 3. Derivative 1: Limits ● Limit x approaches a f(x) = L ● f(x) gets closer to and closer to L (but not equal to) a ● Limit x→a+/- f(x) = L ● Limit x→a f(x) f(x) = + INF -INF Exact Limit Definition ● For any ε>0, you can find a delta>0 such that |f(x) - L| <|x − a|<▯ ● Important Limit ● Limit approaches 0 sin(x)/x = 1 Squeeze Theorem ● If H(x) ≤ f(x) ≤ g(x) ● And h(x) and g(x) both approach a = L ● f(x) also = L Limit Rules ● 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. 2: Continuity ● Def f(x) is continuous at a, if and only if limit x approaches a f(x) = f(a) ○ f(x) must be defined at a ○ F must have a limit ○ f(x) = f(a) ○ f(x) is continuous on (a,b) if and only if f is continuous on every point (a,b) Intermediate Value Theorem - Which Intervals are continuous? ● If f(x) is continuous on the closed interval on [a,b] and c is between a and b; then there exists a point (d) on the interval [a,b] such that f(d) = c. ○ Application of IVT: to prove the existence of a solution. ○ f(x) = g(x) on [a,b] ○ Step 1: h(x) = f(x) - g(x) ■ Check if h(x) is continuous on [a,b] ○ Step 2: H(a) and H(b) must have different signs. ○ This means that there is a 0 between h(a) and h(b) ● The function must be continuous = reason why the IVT works. 3: Derivative Derivative by definition: Do not use shortcuts! ● Definition: ○ f’(x) limit as h approaches 0 ○ [ f(x+h) - f(x) ]/h ● f’(x) = slope of y = f(x) at (x,f(x)) ● If f’(x) exists, then f is continuous at x Velocity / Acceleration ● If f is a position function f’(x) = velocity. ● If f is a position function f(x) = velocity, then f’(x) = acceleration. 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 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) 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) MTH 132 - 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) 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. 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: ● Air is pumped into a spherical balloon so that its volume is increased at a rate of 100 cm cubed. How fast is the radius of the balloon increasing when the diameter is 50 cm? ● Need precursory knowledge: ○ Volume formula is 4 * pi * r^3 ● V = 4(pi)r^3 ● 4pi/3 * 3r^2 * r’(t) ● r’(t) = v’(t) / (4 pi r^2) ● Diameter = 50 ○ Radius is half the diameter - 50/2 = 25 ● Plug in radius to derived equation. ● r’(t) = 100 / [ 4pi * (25)^2 ] ● = 1 / 25pi or 0.0127 cm/s. MTH 132 - Lecture 15 - Related Rates continued Example 1: ● Find the rate of change when r(t) when h = 15 ● r’(t) = -r1’(t) ● r1^2(t) = 4^2 +h^2 ● r1(t) = h(t)/ r1(t) * h’(t) ● When h = 15 2 2 ● R = √4 + 15 = 241 √ ● r1’(t) = 15/ 241 * 2 < 2 Example 2: Spill Radius ● A stream of water is spreading a circular puddle on the floor. If the puddle is 1 meter across, the stream increases the area at the rate of 2 meters squared per minute. How fast is the puddle widening? ● Widening = diameter rate. ● A(t) = area at time (t) ● r(t) = radius at time (t) ● We want r’(t) when r = 0.5 ● A(t) = pi*r^2 ● A’(t) = pi * 2r*r’(t) ● r’(t) = a’(t)/ (pi * 2r) ● At r = 0.5 ● r’(t) = 2/ pi ● The puddle is widening at 4pi meters per minute Example: ● A headlight is rotating at 5 degrees per second. ● What is the speed of the spotlight image moving along the wall when theta = 30 degrees? ○ Theta is a variable. ● s(t) / 20 = tan(theta(t)) ● s’(t) =sec^2(theta(t)) ● s’(t) = 20sec^2(theta(t)) * theta’(t) ● theta’(t) = 5 degrees per second ● 5pi/180 ● 360 = 2pi ● 1 degree = 2pi/360 rad ● = pi/180 ● 1/20cos(30) * 5pi/180 ● = 20pi/27 m/s

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