Quiz 2 (Solution)


Date: MATH 1210-4 - Spring 2004

  1. Find $ \int (x^3+3x^2-1) dx$.

    \begin{displaymath}\begin{split}\int (x^3+3x^2-1) dx &= \int x^3 dx + \int 3x^2 ...
...rac{1}{3} x^3 - x +C = \frac{1}{4}x^4 + x^3 - x +C. \end{split}\end{displaymath}    

    We can check the result: $ (\frac{1}{4}x^4 + x^3 - x +C)' =
(\frac{1}{4}x^4)' + (x^3)' - (x)' +(C)' = \frac{1}{4}\cdot 4 x^3 +
3x^2-1+0 = x^3+ 3x^2-1$, as wanted.

  2. Find the area determined by the $ x$-axis and the curve $ y=6x^5-4x$ for $ 2\leq x\leq 5$.

    Since $ 6x^5-4x\geq 0$ for $ 2\leq x\leq 5$ we can compute the area as $ \int_2^5( 6x^5-4x) dx$:

    \begin{displaymath}\begin{split}Area &= \int_2^5( 6x^5-4x) dx = (x^6 - 2x^2)\ver...
...2^6 - 2\cdot 2^2) \\ &= 15625 -50 -(64 -8) = 15519. \end{split}\end{displaymath}    

  3. A ball is thrown straight up from the top of a $ 100 ft$ tall building with an initial velocity of $ 16 ft/s$. The acceleration due to the gravity force is $ -32 ft/s^2$.
    1. Find a formula for the velocity of the ball $ t$ seconds after it is thrown.

      Since $ a=v'$ we have $ v(t) = \int a(t) dt = \int -32 dt = -32 t+
C$. But we know that $ v(0) = 16$, so that $ v(0) = 0+C = 16 \ensuremath{\Rightarrow}\xspace
C=16$ and we have $ v(t) = -32t+16$.

    2. Find a formula for the position (height above the ground) of the ball $ t$ seconds after it is thrown.

      We denote the height above the ground after $ t$ seconds with $ s(t)$. Then, since $ v=s'$ we have $ s(t) = \int v(t) dt = \int
(-32 t+16) dt = -32 \frac{t^2}{2} + 16 t + C$. Additionally, $ s(0)
= 100$, so that $ 100 = s(0) = C$ and the final formula is $ s(t) =
-16t^2 + 16 t + 100$.

    3. What is the maximum height that the ball reaches?

      The time $ t_{max}$ of the maximum height satisfies $ v(t_{max}) =
0$. That is, $ 0 = -32t_{max}+16$, so that $ t_{max}=\frac{1}{2}$.

      The position at $ t_{max}$, the maximum height, is $ s(\frac{1}{2})
= -16(\frac{1}{2})^2 + 16 \cdot \frac{1}{2} + 100 = -4 + 8 + 100 =
104$ ft above the ground.



Javier Fernandez 2004-02-03