Matematics 3.3: Solutions to Lab Assignment #5 Find te derivative of te given function using te definition of derivative. State te domain of te function and te domain of its derivative..: f(x) 6 x Solution: f (x) f(x+ ) f(x) 6 (x + ) 6 x 6 (x + ) 6 x (6 (x + )) (6 x) ( 6 (x + ) + 6 x) ( 6 (x + ) + 6 x) 6 (x + ) + 6 x ( 6 (x + ) + 6 x 6 (x + ) + 6 x ) 6 (x + 0) + 6 x 6 x Domain of f (, 6], Domain of f (, 6)
.: f(x) x + x Solution: f (x) f(x+ ) f(x) x++ x+ x+ x (x++)(x ) (x+)(x+ ) (x+ )(x ) (x + + )(x ) (x + )(x + ) (x + )(x ) (x + )(x ) + (x ) (x + )(x ) (x + ) (x + )(x ) (x + )(x ) (x + )(x ) (x + 0 )(x ) (x ) Domain of f (, ) (, ), Domain of f (, ) (, )
.:6 g(x) x Solution: g (x) g(x + ) g(x) (x+) x x (x + ) x (x + ) x x x x (x + ) x x (x + ) x x (x + ) x 0 x (x + 0) x x x x 3 Domain of f (, 0) (0, ), Domain of f (, 0) (0, ) 3
.:8 F(x) x Solution: F (x) F(x + ) F(x) (x+) x (x+) x (x+) (x ) ( (x+) + x ) x (x+ ) (x+ )(x ) ( (x+) + x ) (x+) + x (x+) + x (x + )(x )( (x+) + x ) (x ) x Domain of f [, ), Domain of f (, ).:3 Matc graps wit derivatives. Solution: (a) (ii), (b) (iv), (c) (i), (d) (iii)
.:58 (a) Use te definitions in Ex 57 to compute f () and f + () for te function 0 if x 0 f(x) 5 x if 0 <x< if x 5 x Solution: f f( + ) f() 5 ( + ) (5 ) () Solution: f +() f( + ) f() 5 (+) 5 + + ( ) + + + 6 5 3 y + 0-5 - -3 - - 0 3 5 6 7 8 9 0 - x - -3 - -5-6 (b) Sketc te grap of f. (c) Were is f discontinuous? Answer: 0, 5 (d) Were is f not differentiable? Answer: 0,, 5 5
.:60 Determine weter or not f (0) exists. f(x) Solution: Differentiate: { x sin x if x 0 0 if x 0 f f () f(0) sin 0 (x) sin 0.:6 y x x + Solution: y ( x + )( x ) ( x )( x + ) ( ( ( ) x + ) ( ( ) x ) x x x + ) ( x + ) ( x + ) ( x ) x( x + ).: y t + 5 3t x( x + ) Solution: y ( 3t)(t + 5) ( 3t) (t + 5) ( 3t) 8 t + t + 5 3 ( 3t) ( 3t) ( 3t)() ( 3)(t + 5) ( 3t) 6
.: Find te equations of te tangent lines to te curve y x tat are parallel x + to te line x y. Solution: Te line x y as slope. On te curve, we ave y (x + )(x ) (x + ) (x ) (x + )() ()(x ) (x + ) (x + ) (x + ) Te points (x, y) on te curve were te slope equals must terefore satisfy (x + ) or (x + ). Tese two points tus ave x coordinates 3 and. Teir respective y coordinates are and 0, so te two points of tangency are ( 3, ), and (, 0). te equations of te lines troug tese points wit slope are: y (x ( 3)) and y 0 (x ) wic simplify to y x + 3 and y x or x + y 7 and x + y y 0 9 8 7 6 5 3 0-0-9-8-7-6-5--3--03567890 - -3 - -5-6 -7-8 -9-0 x 7
.:8 Draw a diagram to sow tat tere are two tangent lines to te parabola y x tat pass troug te point (0, ). Find te coordinates of te points were tese tangent lines intersect te parabola. Solution: Te tangent lines to y x at te point (a, a ) as slope a and equation y a a(x a) or y ax a. In order for teir y intercept to be, we must ave a, so we ave a ora. Te desired points are terefore (, ) and (, ). y 0 9 8 7 6 5 3 0-5 - -3 - - 0-3 5 x - -3 - -5-6 8
.:50 Find te equations of bot lines troug te point (, 3) tat are tangent to te parabola y x + x. Solution: Te tangent line to y x + x at te point (a, a + a) as slope a +, so its equation is y (a + a) (a + )(x a). To pass troug te point (, 3) it must satisfy 3 (a + a) (a + )( a) or 3 a a a + 3a + ora a 5 0or (a 5)(a + ) 0. Tis is satisfied wen a ora 5, so te equations of te two tangent lines are: y (( ) + ( )) (( ) + )(x ( )) and y (5 + 5) ((5) + )(x 5) or y (x + ) and y 30 (x 5).:5 A manufacturer of cartridges for stereo systems as designed a stylus wit parabolic cross-section as sown in te figure in te text. Te equation of te parabola is y 6x, were x and y are measured in miletres. If te stylus sits in a record groove wose sides make an angle of θ wit te orizontal direction, were tan θ.75, find te points of contact P and Q of te stylus wit te groove. Solution: P and Q will ave coordinates (a, 6a ) and ( a, 6a ), were a>0. Te slope of te te tangent line at P is 3a, and tis must equal.75 or 7. Tus 3a 7 ( ) ( ) so a 7 8, and tus P 7 8, 9 and Q 7 0 8, 9 0.:5 Find te equation of te normal line to te curve y at te point (, ). x Solution: Te slope of te tangent line at x, x ) is y, wic equals - (x ) wen x. Te normal line terefore as slope, so its equation is y x 9
.:70 Were is te function (x) x + x + differentiable? Give a formula for and sketc te graps of and. x if x Solution: (x) 3 if <x< x + if x> if x< so (x) 0 if <x< if x> y 0 9 8 7 6 5 3 0-5 - -3 - - 0-3 5 x - -3 - -5-6 0
.3:6 If a tank olds 5000 gallons of water, wic drains from te bottom of te tank in 0 minutes, ten Torricelli s Law gives te volume V of water remaining in te tank after t minutes as ( V 5000 t ) 0 t 0 0 Find te rate at wic water is draining from te tank after (a) 5minutes, (b) 0 minutes, (c) 0 minutes. ( )( ) ( ) ( ) Solution: V (t) 5000() t 0000 t 50 t 0 0 0 0 0 ( ) ( ) Tus V(5) 50 5 7 50 875 0 8 8.75, ) ( V(0) 50 0 0 ( ) V(0) 50 0 0 50 50 ( ) 3 375 ) ( 5 87.5
.3:0 Te data in te following table concern te lactonization of ydroxyvaleric acid at 5 C. Tey give te concentration C(t) of tis acid in moles per litre after t minutes. t 0 6 8 C(t) 0.0800 0.0570 0.008 0.095 0.00 (a) Find te average rate of reaction for te following time intervals: (i) t 6 (ii) t (iii) 0 t Solution: (a)(i) (0.095 0.0570)moles litre (6 ) minutes 0.075moles litre minutes moles 0.006875 litre minute (ii) (0.008 0.0570)moles litre ( ) minutes 0.06moles litre minutes moles 0.008 litre minute (iii) (0.0570 0.0800)moles litre ( 0) minutes 0.030moles litre minutes moles 0.05 litre minute (b) Plot te points from te table and draw a smoot curve troug tem as an approximation to te grap of te concentration function. Ten draw te tangent at t and use it to estimate te instantaneous rate of reaction wen t. Solution: about 0.0 moles litre minute y 0. 0.09 0.08 0.07 0.06 0.05 0.0 0.03 0.0 0.0 0.0 0 6 8 x
sin 5.: Find te it tan 3 Solution: cos 3 sin 5 tan 3 sin 5 5 sin 3 5 sin 5 sin 3 3 3 cos 3(0)5 3 cos 3 sin 5 cos 3 sin 3 sin 5 5 sin 3 3 5 3 ( ) 5 3 sin 5 cos 3 sin 3.:3 Find te equation of te tangent line to te curve y sec x cos x at te point(π/3, ). Solution: y sec x tan x + sin x, so wen x π/3, y sec π/3 tan π/3 + sin π/3 3 + 3/ 3/ 3 Te equation of te tangent line is terefore y 3 3(x π/3).:5 A semicircle wit diameter PQ sits on an isosceles triangle PQR to form a region saped like an ice cream cone, as sown in te text. If A(θ) is te area of te semicircle A(θ) and B(θ) is te area of te triangle, find θ 0 B(θ) Solution: Let r be te radius of te semicircle, let x be te lengt of te two equal sides of te isosceles triangle and let be te eigt of te triangle wit respect to te base PQ, so tat sin θ r x and cos θ x,or r x sin θ and x cos θ. Ten A(θ) πr π (x sin θ ) π x sin θ and B(θ) r (x sin θ )(x cos θ ) x sin θ cos θ, A(θ) so θ 0 B(θ) θ 0 π x sin θ x sin θ cos θ π sin θ θ 0 cos θ 0.5: Find te derivative of te function y sin(sin(sin x)) Solution: y cos(sin(sin x))(sin(sin x)) cos(sin(sin x))cos(sin x)(sin x) cos(sin(sin x))cos(sin x)cos x 3
.5:8 Find te derivative of te function y cos(sin x) Solution: y ( cos(sin x) cos(sin x) ),so y ( ) ( ) cos(sin x) cos(sin x) ( ) cos(sin x) ( sin(sin x)(sin x) ) ( ) cos(sin x) sin(sin x) sin x(sin x) ( ) cos(sin x) sin(sin x)sin x cos x sin(sin x)sin x cos x cos(sin x).5:60 Suppose tat w u v and u(0), v(0), u (0) 3, u (), v (0) 5, and v () 6. Find w (0). Solution: Since w(x) u(v(x)) and w (x) u (v(x))v (x), we ave w (0) u (v(0))v (0) u ()(5) (5) 0.5:6 Te fequency of vibrations of a vibrating violin string is given by f T L ρ were L is te lengt of te string, T is its tension, and ρ is its linear density. Find te rate of cange of te frequency wit respect to (a) te lengt(wen T and ρ are constant), Solution: df dl T L ρ df (b) te tension( wen L and ρ are constant), Solution: dt L Tρ df T (c) te linear density (wen L and T are constant). Solution: dl Lρ 3 Solution:.5:70 Suppose tat y f(x) is a curve tat always lies above te x axis and never as a orizontal tangent, were f is differentiable everywere. Fow wat value of y is te rate of cange of y 5 wit respect to x eigty times te rate of cange of y wit respect to x? Solution: (y65) 5y y 80y if y 6, or y
.5:80 If f and g are te functions wose graps are sown, let u(x) f(g(x)), v(x) g(f(x)), and w(x) g(g(x)). Find eac derivative, if it exists. (a) u () (b) v () (c) w () 6 y 5 3 0-0 3 5 6 7 x - Solution: (i) u () f (g())g () f (3)( 3) ( /)( 3) 3 (ii) v () g (f ())f () g ()() DNE (iii) w () g (g())g () g (3)( 3) (/3)( 3) 5