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Math 180 Calculus and Analytic Geometry

  • If you are considering taking a lower-level class than Math 180:
  • If you are considering taking a higher-level class than Math 180:
    • If you have taken an AP or IB Calculus exam, please take your exam results to the Assessment Center.
 
  • Sample of what you should know before taking Math 180:
    1. Find the unit vector in the direction of <-3,4>.  Then use a dot product to find the angle between <1,2> and <-3,1>.
    2. Let f of x equals x to the fourth minus x cubed minus 2 x squared minus 4 x minus 24
    Find all possible rational zeros.
      • Solution
        Use the rational zero theorem.  p, the divisors of the last term -24 are 1, 2, 4, 6, 8, 12, 24 and q, the divisors of the leading coefficient 1 is 1. Forming all possible quotients of  plus or minus p over q, possible rational roots are plus or minus 1 comma plus or minus 2 comma plus or minus 4 comma plus or minus 6 comma plus or minus 8 comma plus or minus 12 comma plus or minus 24
    3.  Use induction to show n cubed minus n  is divisible by 6.
      • Solution
        For n=1, 1 cubed minus 1 and 0 is divisible by 6.  Assume the statement is true for n=k; that is, k cubed minus k is divisible by 6.  Show the statement is true for n=k+1, that is, open parenthesis k plus 1 close parenthesis cubed minus open parenthesis k plus 1 close parenthesis is divisible by 6.
        But, 
        open parenthesis k plus 1 close parenthesis cubed minus open parenthesis k plus 1 close parenthesis 
        k cubed plus 3 k squared plus 3 k plus 1 minus k minus 1
        equals open parenthesis 3 k squared plus 3 k close parenthesis plus open parenthesis k cubed minus k close parenthesis
        k cubed minus k  is divisible by 6.  To show 3 k squared plus 3 k is divisible by 6, 3 k squared plus 3 k 3 k open parenthesis k plus 1 close parenthesis k times open parenthesis k plus 1 is divisible by 2 since either k is even or k+1 is even.  Thus 3 k times open parenthesis k plus 1 is divisible by 6 since it is divisible by both 2 and 3.  Therefore open parenthesis k plus 1 close parenthesis cubed minus open parenthesis k plus 1 close parenthesis is divisible by 6.
    4. Solve: sine of 2 x equals one half comma 0 less than or equal to x less than 2 pi
      • Solution
        Let capital a equals 2 x.  sine capital A equals one half right arrow capital a equals pi over 6 comma 5 pi over 6.
        Thus 2 x equals start fraction pi over 6 end fraction plus 2 pi k  right arrow x equals start fraction pi over 12 end fraction plus pi k and 2 x equals start fraction 5 pi over 6 end fraction plus 2 pi k  right arrow x equals start fraction 5 pi over 12 end fraction plus pi k.
        Picking k=0 and k=1, we get pi over 12 comma 13 pi over 12 comma 5 pi over 12 comma 17 pi over 12
    5. Identify the curve: 16 x squared minus 9 y squared plus 64 x minus 90 y equals 305
      • Solution
        16 x squared minus 9 y squared plus 64 x minus 90 y equals 305
        16 x squared plus 64 x minus 9 y squared minus 90 y equals 305
        16 open parenthesis x squared plus 4 x close parenthesis minus 9 open parenthesis y squared plus 10 y close parenthesis equals 305
        Next add, then subtract Open parenthesis start fraction b over 2 end fraction close parenthesis squared in the parentheses.
         16 open parenthesis x squared plus 4 x plus 4 minus 4 close parenthesis minus 9 open parenthesis y squared plus 10 y plus 25 minus 25 close parenthesis equals 305
        16 open parenthesis x squared plus 4 x plus 4 close parenthesis minus 64 minus 9 open parenthesis y squared plus 10 y plus 25 close parenthesis minus 225 equals 305
        16 open parenthesis x plus 2 close parenthesis squared minus 9 open parenthesis y plus 5 close parenthesis squared equals 144
        Now dividing each term by 16 to make 1 on the right side, we have
        start fraction open parenthesis x plus 2 close parenthesis squared over 9 end fraction minus start fraction open parenthesis y plus 5 close parenthesis squared over 16 end fraction equals 1
        This is a hyperbola with = 3 and = 4.  Thus, its center is (-2,-5) traversing a line parallel to the x-axis. Thus add/subtract a and c to the x-values.
  • Sample of what you will learn in Math 180:
    1. Determine whether f of x equals start set first row x cubed cosine start fraction 1 over x end fraction if x is not equal to 0 second row 0 if x is equal to 0 is differentiable at x = 0.  Justify your answer carefully.
      • Solution
        limit as h approaches 0 of start fraction f open parenthesis 0 plus h close parenthesis minus f of 0 over h end fraction
        equals limit as h approaches 0 of start fraction h cubed cosine start fraction 1 over h end fraction minus 0 over h end fraction
        equals limit as h approaches 0 of h squared cosine start fraction 1 over h end fraction
        Since negative h squared less than or equal to h squared cosine start fraction 1 over h end fraction less than or equal to h squared and , limit as h approaches 0 of h squared equals 0, limit as h approaches 0 of negative h squared equals 0, using the Squeeze Theorem, limit as h approaches 0 of h squared cosine start fraction 1 over h end fraction equals 0.  Thus f is differentiable at x = 0 and its derivative is 0.
    2. Use the Mean Value Theorem to show: start square root 1 plus x end square root less than or equal to 1 plus one half x
      • Solution
        Let f of x equal start square root 1 plus x end square root comma a equal 0 comma b equal x
        Using the Mean Value Theorem, there is a c between 0 and x such that f of b minus f of a equals f prime of c times open parenthesis b minus a close parenthesis
        So, start square root 1 plus x end square root minus 1 equals start fraction 1 over 2 times start square root 1 plus c end square root end fraction open parenthesis x minus 0 close parenthesis
        But since c is between 0 and xstart fraction 1 over 2 times start square root 1 plus c end square root end fraction less than or equal to one half.
        Thus start square root 1 plus x end square root minus 1 equals start fraction 1 over 2 times start square root 1 plus c end square root end fraction open parenthesis x minus 0 close parenthesis less than or equal to one half x right arrow start square root 1 plus x end square root less than or equal to 1 plus one half x
    3. Differentiate: tangent open parenthesis x y close parenthesis plus x equals y squared
      • Solution
        Use implicit differentiation to get secant squared open parenthesis x y close parenthesis open parenthesis y plus x start fraction d y over d x end fraction close parenthesis plus 1 equals 2 y start fraction d y over d x end fraction
        So, start fraction d y over d x end fraction equals start fraction negative 1 minus y secant squared open parenthesis x y close parenthesis over x secant squared open parenthesis x y close parenthesis minus 2 y end fraction
    4. Compute: the limit as x approaches infinity of open parenthesis 1 plus start fraction 3 over x end fraction close parenthesis superscript 2x baseline
      • Solution
        Take ln, compute the limit, then take e:
        the limit as x approaches infinity of the natural log of open parenthesis open parenthesis 1 plus 3 over x close parenthesis to the 2 x close parenthesis
        equals the limit as x approaches infinity of 2 x times the natural log of open parenthesis 1 plus 3 over x close parenthesis
        equals 2 times the limit as x approaches infinity of start fraction the natural log of open parenthesis 1 plus 3 over x close parenthesis over start fraction 1 over x end fraction end fraction
        equals 2 times the limit as x approaches infinity of start fraction start fraction 1 over 1 plus start fraction 3 over x end fraction end fraction times negative start fraction 3 over x squared end fraction over negative start fraction 1 over x squared end fraction end fraction
        equals 2 times the limit as x approaches infinity of start fraction 3 over 1 plus start fraction 3 over x end fraction end fraction equals 6
        Taking elimit as x approaches infinity of open parenthesis 1 plus start fraction 3 over x end fraction close parenthesis superscript 2x baseline equals e to the 6
    5. Integrate: integral of start fraction x over start square root 4 minus x to the 4 end square root end fraction d x
      • Solution
        First use u-substitution before using the inverse sine formula.Since x to the fourth equals open parenthesis x squared close parenthesis squared, let u equals x squared right arrow d u equals 2 x d xintegral of start fraction x over start square root 4 minus x to the 4 end square root end fraction d x equals integral of start fraction x over start square root 4 minus open parenthesis x squared close parenthesis squared end square root end fraction d x equals one half integral of start fraction d u over start square root 4 minus u squared end square root end fraction equals one half inverse sine of start fraction u over 2 end fraction plus capital cequals one half inverse sine of start fraction x squared over 2 end fraction plus capital c