Total Score Maximum

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1 Lst Nme: Mth 8: Honours Clculus II Dr. J. Bowmn 9: : April 5, 7 Finl Em First Nme: Student ID: Question Totl Score Mimum No clcultors or formul sheets. Check tht you hve 6 pges.. Find () ( + ) d Prtil frction decomposition my be used here, but it is esier to use the substitution u = +. u u du = u du u du = log u + + C = log + + u + + C. (b) cosh udu e cosh u + + e u u du = du = [ e u 4 4 Equivlently, one cn use the fct tht cosh u = coshu +. (c) e ] e u + u + = 4 sinh(u) + u + C. Let u = e. Then du = u d nd e u + e d = du + u u = + e d 4 u u(u + ) du. We cn re-epress u u( + u) = A u + B A(u + ) + Bu = u + u(u + )

2 by equting like powers of u in the polynomils u nd A(u + ) + Bu. Tht is, = A, = A + B. The solution to this simultneous system of equtions is A =, B =. Thus ( u u(u + ) du = u ) du = log u log u + + C = log(e + ) + C. u + Alterntively, one my rewrite the integrl s ) ( e + e d = + u du = log + u + C = log(e + ) + C. Alterntively, one my rewrite the integrl s tnh sinh ( d = cosh d = log cosh ) + C.. Consider right circulr cone of rdius r nd height h. () Use the method of cross sections to compute the volume of the cone. On slicing the is of the cone we find, using similr tringles, tht the rdius of slice t distnce z from the pe is rz/h, yielding n re A(z) = π = πr z /h. The volume is then given by h A(z)dz = πr h h z = πr h [ z ] h = πr h. (b) Use the method of shells to compute the volume of the cone. We now slice the is to obtin shells of rdius, height h z = h( /r), nd thickness d. The volume is then r ( π h ) r [ ] d = πh r d = πh r r = r πr h.. Consider the curve γ described in polr coordintes s r(θ) = cos θ. () Sketch γ in the y plne. y (,)

3 (b) Find the totl re enclosed by γ. π π π ( ) + cosθ dθ A = = 8 π r (θ)dθ = ( + cosθ + + cos4θ cos 4 θ dθ = ) dθ = ( 8 )( ) π = π Determine which of the following improper integrls converge nd which diverge. () 5/ d This integrl diverges since (for emple) the limit does not eist. (b) lim t + t [ 5/ d = lim ] t + / t + e d Since e = e ( log ) is continuous nd bounded on (, e ], the integrl e + e d eists. For e, we know tht log so tht log nd hence e ( log ) e. Since e d (nd hence e e d) converges, we conclude by the Comprison Test tht e e d lso converges. Hence + e d converges s well. 5. A piece of wire is bent into qurter unit circle centered bout the origin with endpoints t (,) nd (,). () Compute the coordintes (,ȳ) of the centroid of the wire using integrtion. A qurter circle cn be prmetrized s y() = for [, ], for which y () = / nd ds = / = /y. In terms of the length L = π/ of the wire, we find = L ds = L d = L [ ] = L = π nd, s epected from the symmetry principle, ȳ = L y ds = L d = L = π. (b) Use Pppus first theorem to check your nswers in prt (). We check the centroid by computing the surfce re π L = π generted by revolving bout the y is, which is indeed the surfce re of hemisphere. One cn lso check the y centroid by revolving round the is.

4 (c) Find the hydrosttic force on unit hemisphericl bowl filled with wter. Epress your nswer in terms of ρ, the density of wter, nd g, the ccelertion due to grvity. It is convenient to orient the verticl is, sy z, downwrd, with the origin t the surfce of the wter. Ech circulr slb of wter t depth z hs rdius of = z. The force on n re element ds t point (, y, z) is vector in the direction (, y, z). By the symmetry, y y, the nd y components of the totl hydrosttic force vnish. The z component is given by projecting the force contributions in the z direction nd integrting: F = π ρgzz ds = πρg z z + z dz = πρg z dz = z πρg. Alterntively, the totl hydrosttic force is just the weight πρg of wter contined in bowl of volume π. (d) A qudrtic Bezier pproimtion to qurter unit circle is described by the curve (t) = t nd y(t) = t t, with t [,]. Use Question (b) to show tht the rc length of this curve is +log(+ )/. Hint: Recll tht sinh = log(+ ). 4 L = = [ (t)] + [y (t)] dt = = 4 = sinh () = ( t t + dt = τ + 4 dτ = 4 [ t] + [ t] dt = t t + dt cosh u du = [ ( t τ + 4 dτ = ) + 4 dt = ] sinh () sinh () 4 sinh(u) + u sinh ) () + + = + sinh () = + log( + ), where we hve used the substitution t = τ followed by τ = sinhu. τ + 4 dτ sinh u + coshu du = [ sinhu + sinh u + u ] sinh () Note: The resulting rc length,.6, is bout % greter thn the rc length of true qurter unit circle. However, if one uses cubic Bezier pproimtion, the rc length is then ccurte to.%. 6. Use the Limit Rtio Test to find the rdius of convergence of the power series (n) n. n! n= Since ((n + )) n+ ( ) n (n + )! n + lim n (n) n = lim = e, n n n! we see from the Limit Rtio Test tht the rdius of convergence is /e. 4

5 7. Let f be function with continuous (n + )st derivtive on [,b] for some n N. Consider the function g(t) = (b t) n. () Find g (t). n(b t) n. (b) Find the kth derivtive g (k) (t) of g for ll k N. Wht hppens for k > n? ( ) k n! (n k)! (b t)n k for k < n, ( ) n n! for k = n, for k > n. (c) Show tht g (n) (b) = ( ) n n! nd tht g (k) (b) = for ll k n. This follows directly on substituting t = b in prt (b). (d) By multiplying f by g (n+) nd integrting, show tht b This follows from the fct tht g (n+) =. fg (n+) =. (e) Use induction on m to show for m =,,...,n,n + tht m fg (n+) = ( ) k f (k) g (n k) + ( ) m k= The cse m = is just the integrtion by prts formul: fg (n+) = f () g (n) f () g (n). f (m) g (n+ m). Moreover, if the sttement holds for the cse m then n dditionl integrtion by prts in the finl term leds to m fg (n+) = ( ) k f (k) g (n k) + ( ) m f (m) g (n m) ( ) m f (m+) g (n m) = k= m ( ) k f (k) g (n k) + ( ) m+ f (m+) g (n m), k= which is precisely the desired result with m replced by m +. We cn continue this induction process only until m = n +, when the g in the integrnd of the desired result is differentited times. 5

6 (f) Consider now the impliction of prts (d) nd (e) for m = n+, when the indefinite integrls re evluted between nd b: = [ n k=( ) k f (k) g (n k)]b b + ( ) n+ f (n+) g (). Use this result to obtin version of Tylor s theorem where the reminder is epressed in integrl form: n (b ) k f(b) = f (k) () + k! n! k= We find from prts (b) nd (c) tht b (b t) n f (n+) (t)dt. = f () (b)( ) n n! n ( ) k f (k) ()( ) k= n k n! k! (b )k ( ) n b f (n+) g (). On dividing by ( ) n n! we obtin the desired result. (g) Our derivtion etends to the cse n = simply by setting g(t) = on [,b]. Wht theorem does the result in prt (f) reduce to when n =? This is just the sttement of the Fundmentl Theorem of Clculus: f(b) = f() + b f (t)dt. 6