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MA 114 Exam 1 Fall 07 Solution

The statements are abridged. See your paper for the exact statement of the problems.

  1. Fill in the blanks.
    1. (2 points) `int cos(x) dx = sin(x) + C`
    2. (2 points) `int 1/x dx = ln(|x|) + C`
    3. (2 points) `int 1/(sqrt(1-x^2))dx = sin^(-1)(x) + C`
    4. (2 points) `int sec^2(x) dx = tan(x) + C`
    5. (2 points) `int 1/(1+x^2) dx = tan^(-1)(x) + C`
  2. SHOW ALL WORK on the following.
    1. (5 points) Suppose `A` is the area under the graph of `f(t) = 1/t`between `t=10` and the unknown number `Z`. If `e^A = 100`, then `Z=1000`.

      Note that there was a figure which indicated `w` instead of `Z`.

      The area under the graph between the two points is

      `A = int_10^Z 1/t dt = {:ln(t) |_10^Z = ln(Z) - ln(10) = ln(Z/10)`
      So `100 = e^A = e^(ln(Z/10)) = Z/10` because the exponential function is the inverse of the natural logarithm function. Solving for `Z` gives `Z = 1000`.

    2. (5 points) If `16*2^(3N+5) + 7 = 71`, then N = -1

      Solving for `2^(3N+5)` gives

      `2^(3N+5) = (71 - 7)/16 = 64/16 = 4
      Take the natural logarithm of both sides to get:
      `ln(4) = ln(2^(3N+5)) = (3N + 5) ln(2)`
      Solving for N, one gets
      ` (ln(4))/(ln(2)) = 3N + 5`
      `N = ((ln(4))/(ln(2)) - 5)/3 = ((ln(2^2))/(ln(2)) - 5)/3 = ((2ln(2))/(ln(2)) - 5)/3 = (2 - 5)/3 = -1`

  3. Calculate the following. SHOW ALL WORK.
    1. (5 points) `int sin(x) e^(7 cos(x)) dx = -1/7 e^(7 cos(x)) + C`

      Using the substitution `u = 7 cos(x)`, one has `du = -7 sin(x)dx` and so

      `int sin(x)e^(7cos(x)) dx = int (-1/7) e^u du = -1/7 e^u + C = -1/7 e^(7 cos(x)) + C`

    2. (5 points) `int (x+1)/(4x^2+1) dx = ln(5)/8 + tan^(-1)(2)`

      Use the substitution `u = 2x` which gives `du = 2 dx` and

      `int_0^1 (x+1)/(4x^2+1) dx = int_0^2 1/2((u/2) + 1)/(u^2+1) du = int_0^2 1/4 u/(u^2+1) du + int_0^2 1/2 1/(u^2+1)`
      where the limits of integration are calculated via `u = 2x`.

      For the first integral, use the substitution `v=u^2+1` with `dv = 2du` to get

      `int_0^2 1/4 u/(u^2+1) = 1/8 int_1^5 (dv)/v = {:1/8 ln(|v|)|_1^5 = 1/8(ln(5) - ln(1)) = (ln(5))/8`
      since `ln(1) = 0`.

      For the second integral, `int_0^2 1/2 1/(u^2+1) du = {:1/2 tan^(-1)(u)|_0^2 = tan^(-1)(2) - tan^(-1)(0) = tan^(-1)(2)` because `tan^(-1)(0) = 0`. Combining results, one has the answer ` ln(5)/8 + tan^(-1)(2)`.

  4. (10 points) If `f(x) = x^2 + x + 2` for `x > 0`, what is the equation of the tangent line to the graph of `f^(-1)(x)` at the point `(4, 1)`? SHOW ALL WORK!

    To find the equation of a line one needs the slope and a point. The point is `(4,1)`. The slope is given by the derivative

    ` f^(-1) '(4) = 1/(f'(f^(-1)(4))) = 1/(2x + 1) |_(x= 1) = 1/3`
    where we used the formula for the derivative of the inverse function. We know that there actually is an inverse function because `f'(x) = 2x + 1 > 0` for `x > 0` and so `f` is an increasing function and all increasing functions are one-to-one.

    Using the point slope form of the equation of the line:

    `y = m(x - x_0) + y_0 = 1/3(x - 4) + 1 = 1/3(x-1)`
    The answer is `y = 1/3(x - 1)`.

  5. Suppose `f(x) = (2x+1)/(3x+5)`.
    1. (5 points Calculate `f^(-1)(x)`

      The function is `y = (2x + 1)/(3x + 5)`. Swap the variable names to get `x = (2y+1)/(3y+5)` and solve for `y` in terms of `x`: `3xy + 5x = 2y + 1`, `3xy - 2y = 1 - 5x`, `(3x-2)y = 1- 5x`, and so `y = (1-5x)/(3x-2)`. The answer is `f^(-1)(x) = (1 - 5x)/(3x-2)`.

    2. (5 points For your function `f^(-1)(x)` in part (a), show `f^(-1)(f(x)) = x`.

      Substitute the expression for `f(x)` into that for `f^(-1)(x)` and simplify:

      `f^(-1)(f(x)) = (1 - 5((2x+1)/(3x+5)))/(3((2x+1)/(3x+5))-2)`
      `=((3x+5) - 5(2x+1))/(3(2x+1)-2(3x+5))`
      `=(-7x)/(-7) = x`
      as was to be shown.

  6. SHOW ALL WORK on the following. Suppose `g(x)` is a function defined for all `x` and such that
    1. `g(7) = 2`
    2. `g'(7) = -5`
    Let `F(x) = 3^(g(x))`.
    1. (2 points) `F(7) = 3^(g(7)) = 3^2 = 9`
    2. (2 points) `F'(7) = -45 ln(3)`

      One calculates the derivative `F'(7) = (3^(g(x))' = ln(3) 3^(g(x)) g'(x)` using the chain rule. Substituting `x=7` gives `F'(7) = ln(3) 3^(g(7)) g'(7) = ln(3)3^2 (-5) = -45 ln(3)`.

    3. (3 points) The equation for the tangent line to the graph of `F(x)` at `x=7` is `-45ln(3)(x - 7) + 9`

      Use the point slope form of the line to write

      `y = m(x - x_0) + y_0 = g'(7)(x - 7) + g(7) = -45ln(3)(x - 7) + 9`

    4. (3 points) The equation for the tangent line to the graph of `F^(-1)(x)` at `(F(7), 7)` is `x = -45ln(3)(y-7) + 9`.

      The tangent line to the inverse is the reflection of the original tangent line across the 45 degree line `y = x`. So, to obtain its equation, just swap the variables `x` and `y`. If you want, you can also solve for `y` giving `y = -(x-9)/(45 ln(3)) + 7`.

  7. SHOW ALL WORK on the following.
    (10 points) An observer is at ground level 100 feet from the base of a tower 300 feet high. She watches as an exterior elevator ascends from ground level. At time `t` seconds the elevator is at `4t` feet above the ground.

    If `T(t)` is the angle (in radians) formed by her line of sight and the horizontal at time `t` seconds, then:

    1. `T(t) = tan^(-1) ((4t)/100)` because `tan(T(t)) = (4t)/100`.
    2. `T'(t) = 25/(625+t^2)

      Differentiate using the chain rule:

      `T'(t) = 1/(1+ (t/25)^2) 1/25 = 25/(625+t^2)`

    3. `T'(7) = 25/(625 + 7^2) = 25/674`
  8. SHOW ALL WORK on the following.
      (5 points) `lim_(x->oo) (4e^(2x)-5x^2+11)/(5e^(2x)-4x^2+11) =4/5`

      The limit of the numerator and denominator is infinity and so one could try to apply L'Hospital's rule repeatedly to get the answer. However, it is quicker to rewrite the expression by dividing numerator and denominator by `e^(2x)`. This gives

      `lim_(x->oo) (4e^(2x)-5x^2+11)/(5e^(2x)-4x^2+11) = lim_(x->oo) (4 - 5x^2e^(-2x) + 11 e^(-2x))/(5 -4x^2e^(-2x) + 11e^(-2x)) = (lim_(x->oo) 4 - 5x^2e^(-2x) + 11 e^(-2x))/(lim_(x->oo)5 -4x^2e^(-2x) + 11e^(-2x)) = 4/5`

    1. (5 points) `lim_(x->0) (sin(3x^2))/(cos(5x)-1) = -6/25`

      One applies L'Hopital's rule. The limit of the numerator and denominator are both zero. So,

      `lim_(x->0 )(sin(3x^2))/(cos(5x)-1) = lim_(x->0)(6xcos(3x^2))/(-5 sin(5x))`
      Again, the limit of the numerator and denominator are both zero, so one can use L'Hospital's rule to get
      `lim_(x->0)(6xcos(3x^2))/(-5 sin(5x))= lim_(x->0)(6 cos(3x^2) +-36x^2 sin(3x^2))/(-25 cos(5x)) = -6/25`

  9. Calculate the following and SHOW ALL WORK
    1. (3 points) `int x cos(x) dx = x sin(x) + cos(x) + C`

      Use integration by parts with `u=x` and `v' = cos(x)`. Then `u' = 1` and `v = sin(x)`. So

      `int x cos(x) dx = int udv = uv - int v du = x sin(x) - int sin(x)dx = x sin(x) + cos(x) + C`

    2. (3 points) `int x ln(x) dx = 1/2 x^2 ln(x) - x^2/4 + C`

      Use integration by parts with `u = ln(x)` and `v' = x` so that `u' = 1/x` and `v = x^2/2`. One has

      `int x ln(x) dx = 1/2 x^2 ln(x) - int (x^2)/2 1/x dx = 1/2 x^2 ln(x) - x^2/4 + C`

    3. (4 points) `int sin(x) e^x dx =int sin(x) e^x dx = 1/2(sin(x) e^x - cos(x)e^x )`

      Use integration by parts with `u = sin(x)` and `v' = e^x` so that `u' = cos(x)` and `v = e^x`. This gives

      `int sin(x) e^x dx = sin(x) e^x - int cos(x) e^x dx`
      Now use integration by parts again with `u = cos(x)` and `v = e^x` so that `v' - sin(x)` and `v' = e^x`. One gets
      `sin(x)e^x - (cos(x)e^x - int -sin(x)e^x dx)`
      Solve for the integral to get
      `int sin(x) e^x dx = 1/2(sin(x) e^x - cos(x)e^x )`

modified: Monday, February 04, 2008
 

Modified 1/18/2008 12:55


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