Math 115 (W1) Solutions to Assignment #4
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1 Math 5 (W Solutions to Assignment #. ( marks Fin the erivative of the following. Provie reasonable simplification. a f( 3 + e sec ( ; ( ( b f( log + tan ; ( c f( tanh ; + f( ln(sinh. a f( ( ln( 3 ln( 3 ln(3 ( e sec ( ( 3 + e sec ( by linearity rule (sec ( ( + e sec ( 3 ln( 3 ln(3 + e sec ( by chain rule an erivatives for a an e by chain rule an erivatives for a an sec ( by erivative power rule. b f( [ ( ( ] ln ln + tan by log b c log a c log a b ln c ln b [ ( ] ln ln + ( tan by linearity rule ln ( + ( by chain rule an + erivatives for ln an tan 6 ln ( + by the quotient rule an erivatives ( + 6 ( ln( + simplifying. 6 +
2 c f( ( + ( + by chain rule an erivatives for tanh ( + ( + ( ( ( + by quotient rule ( + ( ( + ( + ( ( ( simplifying an using a b (a + b(a b ( + + ( ((. simplifying f( ln(sinh ln( ln(sinh (ln by (ln(sinh sinh by (sinh ln(sinh (ln sinh cosh ln(sinh (ln coth. chain rule an erivative for a chain rule an erivative for ln by erivatives for sinh. ( marks Scientists iscover a fossil that contains 7% as much raioactive carbon, C, as oes a living mammal. If the half-life of C is about 573 years, how ol is the fossil? We use the SIMPLE proceure. Step : Set T / 573, H(t.7, t, r ln T /, H(t, b. Step : Integrate k(t r(t t ln 573 t. Step 3: Missing t. Step : Plug In Step 5: Locate t 573 ln ln(.7. H(t H e k(t since b.7 e ln 573 t
3 Step 6: Evaluate t 89.3 years. The fossil is approimately 89 years ol. 3. ( marks a Obtain a reuction formula for I n n sin ; b Use it to evaluate π sin. We use the DIRT proceure. a (Define I n n sin. (Initial I sin cos + C. (Reuction (Manipulate I n }{{} n f( sin( }{{} g ( (Eliminate I n vu with v n, u sin (Solve I n uv uv where u cos, v n n (Substitute I n n cos + n n cos But so (Manipulate }{{} n f( cos( }{{} g ( (Eliminate vu with v n, u cos (Solve uv uv where u sin, v (n n (Substitute n sin (n n sin I n n cos + n n sin n(n n sin } {{ } I n 3
4 b (Track I cos + 3 sin [ cos + sin ( cos ] + C Now, by FTC part II, we have π sin [ cos + 3 sin + cos sin + cos ] π (π + π ( + π π 8.. ( marks Evaluate the following integrals: a ; b ; c + 3 ; +. a We use the MESS proceure of substitution. (Manipulate I ( (Eliminate t with t t completing the square (Solve arcsin(t + C (Substitute arcsin( + C by antierivatives since arcsin(t t t b We use the MESS proceure of substitution. (Manipulate I completing the square ( + (Eliminate t with t + t (Solve ( t by antierivatives tanh ( since + C tanh t (Substitute tanh ( + + C t t
5 c We use the MESS proceure of substitution twice, meaning iterate MESS. (Manipulate I (Eliminate I ( + completing the square ( + t + t with t t + t t + t + t + t by linearity t (Manipluate I t + t t + tt for the first integral (Eliminate I u u with u t + (Solve I ln u + C by known antierivative (Substitute I ln(t + + C (Solve I ln(t + + ( by the inner MESS an t the antierivative arctan + C ( since t arctan (Substitute I ln( ( arctan + C. t + t I ( + factoring the enominator cosh ( sinh ( + C by linearity rule an antierivatives. Alternatively, ( I ln + + ln( C ( marks Evaluate the following integrals: 5
6 a I π/ sin ; b ln( ; c sin( ln (Hints: Consier sin( ln as v when applying integration by parts on. vu; a Using the MESS proceure for integration by parts, we have (Manipulate & Eliminate I }{{} sin }{{ } v u (Solve vu uv uv where u cos, v (Substitute I [ cos ] π/ + π/ cos [ cos + sin ] π/ by antierivatives [( + ( + ]. b Using the MESS proceure for integration by parts, we have (Manipulate & Eliminate I }{{} ln }{{} v u (Solve vu [ ] uv uv (Substitute I [ ln ] where u, v ( ln by antierivatives [ ( ( ln ln ] ln( 3. 6
7 c Using the MESS proceure for integration by parts, we have (Manipulate & Eliminate I sin( ln }{{}}{{} v u (Solve I uv uv where u, v cos( ln (Substitute I sin( ln cos( ln sin( ln cos( ln } {{ } J We use the MESS proceure again to evaluate J. (Manipulate & Eliminate J cos( ln }{{}}{{} v u (Solve J uv uv where u, v sin( ln (Substitute J cos( ln + sin( ln cos( ln + sin( ln } {{ } I Therefore, I sin( ln cos( ln I I [sin( ln cos( ln ] + C. 5 Using the MESS proceure for integration by parts, we have (Manipulate & Eliminate I }{{} }{{} v u (Solve I uv uv where u, v ln (Substitute I ln ln ln ln }{{} J 7
8 We use the MESS proceure again to evaluate J. (Manipulate & Eliminate J }{{} }{{} v u (Solve J uv uv where u, v ln (Substitute J ln ln ln ln ln by antierivatives for a ln ln Therefore, I ln ( ln ln + C. (ln 6. ( marks Fin the arc length of the curve y ln(coth(/, 3. For sketching the curve, use the points in the following table. We use the SAFE proceure. y Step : Sketch Using the given points, we have the following sketch. y y ln(coth(/ 3 Step : Assign We use the following parameterization for the given curve. } (t t t 3 y(t log(coth(t/ The limits of integration are a, b 3. 8
9 Step 3: Form Since (t an y (t coth(t/ ( csch (t/ ( sinh( t cosh( t cscht, by the chain rule an antierivatives for ln, coth we have Step : Evaluate L[, 3] L[, 3] b a ( (t + (y (t t coth t t + csch t t since coth t + csch t coth tt since coth t > t [, 3] Using the MESS proceure for substitution, we have (Manipulate L (Eliminate L 3 g(3 g( cosh t } sinh {{} t }{{} t g (t g(t u u (Solve L [ln u ] sinh 3 sinh (Substitute L ln(sinh 3 ln(sinh ln ( sinh 3 sinh. since coth t cosh t sinh t with u g(t sinh t by antierivatives 9
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