Math 250 Skills Assessment Test
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1 Math 5 Skills Assessment Test Page Math 5 Skills Assessment Test The purpose of this test is purely diagnostic (before beginning your review, it will be helpful to assess both strengths and weaknesses). The test problems cover Calculus I concepts that are essential for second semester calculus. s are provided, and each answer has references to relevant review topics either in Math 5 or earlier courses. If anything is unclear, the review material should help. You may click on the blue words if you wish to jump to an answer or the review topics. If you would like to print the Skills Assessment so you can work it out on paper, please click Print.. a secant line f(x) = x a + h Find the slope of the secant line for f(x) = between the two points x whose x-coordinates are a and (a + h).. Consider a circle of radius centered at the origin. Find the area of a sector of the circle determined by an angle θ = π 6 radians. 3. Evaluate the following limits. a) lim x x 4 x +6x b) lim ln(x ) x +
2 Math 5 Skills Assessment Test Page c) lim x x + x +3 d) lim x sin 4x x 4. Find y for the following functions. a) y =e x + π tan x 7cosx b) y =3 x + 7 x sin x + 3 cot x c) y = 4 ln x sec x cscx 3 d) y =(x +x )(e x + tan x) e) y =(4x /3 3 x)(x 3 +3x ln x) f) y = x x + tan x +cotx g) y = xex +cosx 4 x 4 h) y =(x +3) / sin(x)+e tan x i) y =sin x +sinx j) y =csc x + k) x y + xy =4 l) e xy =ln(x y) m) y = x cos x 5. Find the equation of the tangent line to the curve y =+lnx where y =+lnx crosses the x-axis.
3 Math 5 Skills Assessment Test Page 3 6. Evaluate: a) x dx b) c) ln π/4 e x dx cos xdx d) e) π/3 π/6 tan xdx (e x +cosx)dx f) π/ (csc x cot x + 3 x)dx g) h) (x 3 +) dx x(x +) / dx i) dx x + j) k) π/4 sec xe tan x dx e πx dx l) x x +dx
4 Math 5 Skills Assessment Test Page 4 m) x +3 x +6x dx 7. Find the area bounded by the curves y = tan x, x = π 6, x = π 4,and y =. 8. Find the area bounded by the curves y = x and y =4x 3. s Top of File Math 5 Review Topics
5 Math 5 Skills Assessment s Page 5 ANSWERS to SKILLS ASSESSMENT. To find the slope, we need two points. If x = a then y = f(a), and so one point is (a, f(a)). The other point is (a + h, f(a + h)). Then slope = m = y y f(a + h) f(a) f(a + h) f(a) = =. x x a + h a h (This quotient has a special form and is called the difference quotient.) In this problem f(x) = x,orf( ) = ( ). Thus, f(a + h) f(a) slope = = h h = ah(a + h) = a(a + h). a + h a h = a (a + h) a(a + h) h Review Topic. The area A of a sector of a circle of radius r subtended by a central angle θ is given by the formula A = r θ.thus, A = ( ) π 6 = π 3. Review Topic 6 x 4 3 a) Find lim x x +6x. This has form must manipulate. Let s try factoring. lim x x 4 (x +3x ) = lim x (x )(x +) (x )(x +5) = lim x at x =,andsowe x + (x +5) = 4 4 = 7 Review Topic B
6 Math 5 Skills Assessment s Page 6 3. b) Find lim ln(x ). The graph of x + ln(x ) has the following appearance, which is a shift of ln x one unit ln(x ) to the right. Thus, lim ln(x ) =. x = x + Review Topic A 3. c) x + Find lim. Let s factor out the highest power of x that x x +3 we can in the numerator and also the denominator. ( x + ) ( x + ) x + lim x x +3 = lim x x ) x (+ = lim x 3x x x + 3 x = lim x ( + ) x + 3x. As x, x and 3. Thus, x lim x x + x +3 = =. Review Topic C
7 Math 5 Skills Assessment s Page 7 sin 4x 4 sin 4x sin 4x sin x 3. d) lim = lim = lim 4. We know lim x x x 4 x x 4x x x = sin(θ) (See Example B.7). This really says lim =, where θ is θ (θ) the same for the numerator and the denominator. Thus, sin 4x lim x 4x =. This means sin 4x lim x x sin 4x = lim 4 x 4x =4 =4. Review Topic B 4. a) y =e x + π sec x +7sinx Review Topic B 4. b) Rewrite y as y =3x / +7x sin x + 3 cot x ( ) y =3 x / 7x cos x 3csc x Review Topic B 4. c) y = 4 3 sec x tan x + csc x cot x x Review Topic B
8 Math 5 Skills Assessment s Page 8 4. d) y = [ ] d dx (x +x ) (e x + tan x) +(x +x ) d dx (ex + tan x) =(x + )(e x + tan x)+(x +x )(e x +sec x) Review Topic C 4. e) y = ( 4 ( ) x /3 3 ) 3 x /3 (x 3 +3x ln x) +(4x /3 x /3 )(3x +3 x ) Review Topic C 4. f) y = [ ] d dx (x x +) (tan x +cotx) (x x +) d (tan x +cotx) dx = (tan x +cotx) = (x )(tan x +cotx) (x x + )(sec x csc x) (tan x +cotx) Review Topic D 4. g) y = (e x + xe x sin x)(x /4 4) (xe x +cosx)( 4 x 3/4 ) (x /4 4) Review Topic D
9 Math 5 Skills Assessment s Page 9 4. h) y = (x +3) / (x) (cos x) +e tan x sec x Review Topic E 4. i) Rewrite y =(sinx) +sin(x ). y =sinx cos x +cos(x ) x. Review Topic E 4. j) y = [ ( ) csc(x +) / cot(x +) /] (x +) / x Review Topic E 4. k) Use implicit differentiation. xy + x y + y + x(yy )= y (x +xy) = xy y y = xy y x +xy Review Topic F
10 Math 5 Skills Assessment s Page 4. l) Use implicit differentiation. e xy [y + xy ]= x y (x y ) e xy xy + x y y = y = x x y yexy x y + xexy x x y yexy Review Topic F 4. m) Take the natural log of both sides to arrive at ln y =lnx cos x =cosx ln x. Now use implicit differentiation Then y y = sin x ln x + (cos x) x. ( y = y sin x ln x + cos x x ) = x cos x [ sin x ln x + cos x x ] Review Topic G
11 Math 5 Skills Assessment s Page 5. To find the equation of the tangent line we need the point and slope. The curve crosses the x-axis at f ==+lnx ln x = x = e. The point is (e, ). To find the slope we need to take the derivative and evaluate the derivative at the point. Thus f = x f (e )= e = e. Then y =e(x e ) y = ex. Review Topic H 6. a) x C Review Topic 3A 6. b) e x ln = e ln e = = Review Topic 3A 6. c) sin x π/4 =sin π 4 sin = Review Topic 3A 6. d) π/3 π/6 tan xdx= π/3 π/6 (sec x )dx = (tan x x) ) = tan π 3 π 3 ( tan π 6 π 6 π/3 π/6 = 3 3 π 6 Review Topic 3B
12 Math 5 Skills Assessment s Page 6. e) e x +sinx + C Review Topic 3A 6. f) Rewrite integral as = csc x + x4/3 4 3 = + 3 ( π 4 π/ π/ (csc x cot x + x /3 )dx = csc π ) 4/3 +csc 3 4 = ( π ( π ) ( 4/3 csc + 3 ) 4 ) 4/3 +csc Review Topic 3B 6. g) ( ) x (x 6 +x 3 7 +)dx = 7 + x4 4 + x = ()7 7 + [ ( ) ] 7 ( )4 +( ) = 95 4 = Review Topic 3B
13 Math 5 Skills Assessment s Page 3 6. h) Let u = x +,du =xdx. (x } {{ + } ) / }{{} xdx= u du u/ du = u 3/ 3 + C = 3 (x +) 3/ + C Review Topic 3C 6. i) u = x +,du= dx. du {}}{ dx (x }{{ + } ) = u du u =ln u + C =ln(x +)+C Review Topic 3C 6. j) u = tan x, du = sec xdx. We now need to change the limits of integration; when x =, u = tan = ; when x = π 4, u = tan π 4 =. u= π/4 u= u {}}{ e tan x sec } {{ xdx} = du ( e u du) = e u = e ( e )= e Review Topic 3C
14 Math 5 Skills Assessment s Page 4 6. k) u = πx, du = πdx, u {}}{ e πx }{{} dx = π du π e u du = π eu + C = eπx π + C Review Topic 3C 6. l) u = x +, du = dx x = u = += x = u =+= u= }{{} x u= u =x = (x }{{ + } ) / }{{} dx u du = (u )u / du (u 3/ u / )du = 5 u5/ 3 u3/ = 5 3 ( 5 3 ) = 4 5 Review Topic 3C
15 Math 5 Skills Assessment s Page 5 6. m) u = x +6x, du =(x +6)dx =(x +3)dx du {}}{ (x +3)dx x +6x }{{} u = du u = du u = ln u + C = ln x +6x + C 7. Area = π/4 π/6 π/4 (tan x ( ))dx =ln sec x + x π/6 ( =ln sec π ) + π ( ( 4 4 ln sec π ) + π ) 6 6 =ln ln π 3 ( ) 6 =ln π Review Topic 3C y =tanx x = π 6 π 4 = x y = Review Topic 3D
16 Math 5 Skills Assessment s Page 6 y =4x 3 8. y = x We must first determine where the curves cross. Set x = 4x 3, then 4x 3 x = x(x )(x+) = so x =,,. We can use symmetry and find Area = [ ( ) / ( ) x (x 4x 3 / )dx = x4 ( ) 4 ( 4 )] = [ 8 ] = 6 8 Review Topic 3D Math 5 Web Page Top of File Math 5 Review Topics
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