Math 31A Discussion Notes Week 4 October 20 and October 22, 2015

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1 Mat 3A Discussion Notes Week 4 October 20 and October 22, 205 To prepare for te first midterm, we ll spend tis week working eamples resembling te various problems you ve seen so far tis term. In tese notes we reference various results witout eplaining tem; you can find te eplanations in te tetbook, Dr. Heilman s notes, or our old notes. Te questions and solutions are listed separately so tat you can work eac problem, and ten click on (Solution) to see te solution. Finally, understand tat tis is just a sampling of problems tat you sould know ow to do, and sould not be considered an indication of te problems you ll see on te eam. Limits Evaluate, if it eists. (Solution) Evaluate, if it eists. (Solution) Find two functions f, g suc tat a f() does not eist, a g() does not eist, but suc tat (f() + g()) a does eist. (Solution) sin 2 (θ) 4. Evaluate, if it eists. (Solution) θ 0 cos(θ) 5. Evaluate 0, if it eists. (Solution) 6. Evaluate, if it eists. (Solution) ( 7. Evaluate t 0 t ), if tis it eists. (Solution) t 2 + t 8. Evaluate, if tis it eists. (Solution) Continuity 9. Find values k, m R so tat f() := m( + ) + k < <

2 is continuous, if possible. (Solution) 0. Find two functions f, g suc tat neiter f nor g is continuous at a, but suc tat f + g is continuous. (Solution) Te Squeeze Teorem and Limits at Infinity. Find θ 0 θ 2 cos(/θ), if tis it eists. (Solution) sin(2) sin(3) 2. Evaluate, if tis it eists. (Solution) 0 sin(5) sin(7) sin(2 4) 3. Evaluate, if tis it eists. (Solution) cos( 2 7) 4. Evaluate, if tis it eists. (Solution) Te (Limit Definition of te) Derivative 5. Use te it definition of te derivative to find f (a), were f() = and a 0. (Solution) 6. Use te it definition of te derivative to find g (t), were g(t) = 0 + 5t 4.9t 2. (Solution) Computing Derivatives 7. Find dy d, were y = 2 +. (Solution) 8. Find dy 33, were y = d (Solution) 4 9. Find g (), were g() = (Solution) 20. Find ds dt and ds da, were s = t a 2 + t 2. (Solution) 2. Find dy d, were y = a a +. (Solution) 22. Find dy d, were y = c + c. (Solution) 2

3 Solutions Evaluate, if it eists. 2 2 (Solution) Doing some factoring, we ave ( 2)( + 3) Evaluate, if it eists (Solution) We ave 0 2 ( + 3) = ( ) ( ) ( ) 0 ( ) = Find two functions f, g suc tat a f() does not eist, a g() does not eist, but suc tat (f() + g()) a does eist. (Solution) Tere are many different solutions to tis problem, but one pair of functions in particular is given by f() := a and g() := a. You can ceck tat neiter a f() nor a g() eists, but f() + g() = 0 for a, so a (f() + g()) = 0. sin 2 (θ) 4. Evaluate, if it eists. θ 0 cos(θ) (Solution) We can use te identity sin 2 (θ) + cos 2 (θ) = to find 5. Evaluate θ 0 0 sin 2 (θ) cos(θ) cos 2 (θ) θ 0 cos(θ) ( cos(θ))( + cos(θ)) θ 0 cos(θ) ( + cos(θ)) = 2. θ 0, if it eists. 3

4 (Solution) Notice tat and tat {, < 0 =, > 0, is undefined wen = 0. So we see tat 0 = and 0 Since te left and rigt its are not equal, 0 6. Evaluate, if it eists. 0 + =. 0 + does not eist. (Solution) Since bot te numerator and denominator approac 0 as approaces, we must find a way to rewrite te fraction. We do tis by rationalizing te denominator: = + = ( )( + ). + Ten ( )( + ) ( 7. Evaluate t 0 t ), if tis it eists. t 2 + t ( + ) = 2. (Solution) Notice tat neiter quotient as a it as t 0, so we find a way to combine te terms: ( t 0 t ) ( ) ( ) t 2 + t t t 2 t 2 + t t 0 t(t 2 + t) t 0 t 2 (t + ) ( ) =. t 0 t + 8. Evaluate, if tis it eists (Solution) Again, we need to rationalize te denominator = ( ) 0 ( + 3) 4

5 9. Find values k, m R so tat f() := m( + ) + k < < is continuous, if possible. (Solution) Notice tat all tree pieces of f are continuous functions, so te only points were we ave to worry about continuity are at = and = 2. For eac of tese points, we ll need to ensure tat te left- and rigt-its agree. First we ave wile So k = 4. Ten at = 2, and f() + + 7) = = 4, (23 f() + +(m( + ) + k) = k. 2 f() (m( + ) + k) = 3m So 3m + 4 = 9, meaning tat m = 5/3. f() + 5) = (2 0. Find two functions f, g suc tat neiter f nor g is continuous at a, but suc tat f + g is continuous. (Solution) Te two functions we used for te it version of tis question won t work, since teir sum isn t defined at = a, but te following pair of functions will do te trick: { { 0, a, a f() := and g() :=, > a 0, > a. Find θ 0 θ 2 cos(/θ), if tis it eists. (Solution) We know tat cos(/θ) oscillates as θ 0, but tis won t be a problem ere. Since cos(/θ) for all θ 0, we ave θ 2 θ 2 cos(/θ) θ 2. But θ 0 ( θ 2 ) = 0 and θ 0 θ 2 = 0, so te squeeze teorem allows us to conclude tat θ 0 θ2 cos(/θ) = 0. 5

6 sin(2) sin(3) 2. Evaluate, if tis it eists. 0 sin(5) sin(7) (Solution) We want to take advantage of te fact tat sin(u) u 0 u =, so we rewrite our it as ( ) ( ) ( ) ( ) sin(2) sin(3) 0 sin(5) sin(7) 6 sin(2) sin(3) sin(5) sin(7) = sin(2 4) 3. Evaluate, if tis it eists (Solution) As wit te previous problem, we ll rewrite tis it to look like someting we know. First, notice tat sin(2 4) sin(2 4) Since te argument of sine and te denominator in our first fraction agree and approac 0 linearly, we may use te same rule we used above to find tat sin(2 4) = cos( 2 7) 4. Evaluate, if tis it eists = 2. (Solution) Since cos( 2 7), we see tat te numerator of tis quotient remains bounded wile, but te denominator grows witout bound. So te quotient approaces 0 as. More rigorously, cos( 2 7). Since bot bounding functions approac 0 as, our it is Use te it definition of te derivative to find f (a), were f() = and a 0. 6

7 (Solution) We ave f (a) Here we ll rationalize te numerator to find f (a) = 2 a. a + a ( a + + a) a + a. a + + a a + + a a + + a (a + ) a ( a + + a) 6. Use te it definition of te derivative to find g (t), were g(t) = 0 + 5t 4.9t 2. (Solution) We ave 7. Find dy d, were y = g (t) g(t + ) g(t) (0 + 5(t + ) 4.9(t + ) 2 ) (0 + 5t 4.9t 2 ) 0 + 5t t 2 9.8t t + 4.9t t = 5 9.8t (Solution) According to te quotient rule, (5 9.8t 4.9) dy d = (2 + )() (2) ( 2 + ) 2 = 2 ( 2 + ) Find dy 33, were y = d (Solution) First, we rewrite y as y = 3 3 2/5 7 4/ /7 = 3 3/5 7 / /7. We may now apply te power rule to find tat dy d = / / /7. 7

8 9. Find g (), were g() = (Solution) Notice tat g() = f(()), were f() = 3 and () = 4 9. Ten te cain rule gives g () = f (())() = 3 (4 9) 2 3 ( 9) = 3. (4 9) Find ds dt and ds da, were s = t a 2 + t 2. (Solution) First, we ll use te product and cain rules to see tat ds dt = 2t a 2 + t 2 + t 2 a 2 + t = a2 + t 2 + t 2 = a2 + 2t 2 2 a2 + t 2 a2 + t. 2 On te oter and, ds da = t(2a) 2 a 2 + t 2 = at a2 + t Find dy d, were y = a a +. (Solution) By te quotient rule, dy d 22. Find dy c d, were y = + c. (Solution) Using te cain rule we find tat = (a + )( ) (a )() (a + ) 2 = 2a (a + ) 2. dy d = ( ) /2 c ( + c)( c) ( c)(c) 2 + c ( + c) 2 ( ) c ( c) /2 ( + c) + ( c) /2 =. 2 ( + c) 3/2 8