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Calculus I                                                                      Test #2                                             November 3, 2004

Name____________________                               R.  Hammack                                             Score ______
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(1) Suppose f(x) is a function for which the following limits hold.

Underscript[lim , w0] (f(w) - f(0))/(w - 0) = -1        Underscript[lim , w1] (f(w) - f(1))/(w - 1) = 5         Underscript[lim , w2] (f(w) - f(2))/(w - 2) = 0            Underscript[lim , w5] (f(w) - f(5))/(w - 5) = 3  

Based on this information, answer the following questions.

(a) Is there a value of x at which the tangent line to the graph of y = f(x) is horizontal?  If so, what is x?
The third limit says  f '(2) = 0, so x = 2 is such a value.

(b) Find the slope of the tangent line to the graph of  y = f(x)  at the point where x = 5.
The fourth limit says  f '(5) = 3, so the slope is 3.

(c)  Suppose you also know that f(0) = 5.  Find the equation of the tangent line to the graph of   y = f(x)  at the point where x = 0.
The first limit says  f '(0) = -1, so the tangent line has slope -1 and y-intercept 5. It's equation is thus y = -x +5


(2) The graph of a function g(x) is illustrated below.  Sketch the graph of g '(x).
[Graphics:HTMLFiles/T2F04A_5.gif]


[Graphics:HTMLFiles/T2F04A_6.gif]



(3) If y = tan(x),  find (d^2y)/dx^2

dy/dx = sec^2(x)

(d^2y)/dx^2 = 2sec(x)^1sec(x) tan(x) = 2sec^2(x) tan(x)


(4) d/dx[ 2x^9 - x + 6 ] =  18x^8 - 1


(5) d/dx[ 1/x + 3/x^(1/2)] = d/dx[ x^(-1) + 3x^(-1/2)] = -x^(-2) + 3 (-1/2) x^(-3/2) = -1/x^2 - 3/(2x^(1/2)^3)


(6) d/dx[ x^2cot(x)] = 2x cot(x) + x^2(-csc^2(x)) = 2x cot(x) - x^2csc^2(x)



(7)  d/ds[ (4x + 2)/cos(x)] = (4cos(x) - (4x + 2) (-sin(x)))/cos^2(x) = (4cos(x) + ( 4x + 2) sin(x))/cos^2(x)



(8) d/dx[ sin(1/x)] =cos (1/x) ( -1/x^2) = -cos (1/x)/x^2



(9) d/dx[(x^2 + 9 )^(1/2)] = d/dx[(x^2 + 9)^(1/2) ] = 1/2 (x^2 + 9)^(-1/2) (2x) = x/(x^2 + 9)^(1/2)


(10)  d/dx[  x sec ( π x )   ] =(1) sec(π x) + x sec(π x) tan(π x) π = sec(π x) + π x sec(π x) tan(π x)



(11) d/dx[ sin^2(x) (2x + 1)^3 ] =2sin(x) cos(x) (2x + 1)^3 + sin^2(x) 3 (2x + 1)^22 = 2sin(x) cos(x) (2x + 1)^3 + 6sin^2(x) (2x + 1)^2



(12) d/dx[ tan^3(x^(1/2))] =   d/dx[( tan(x^(1/2)))^3] = 3 ( tan(x^(1/2)))^2d/dx[ tan(x^(1/2))] = 3 tan^2(x^(1/2)) sec^2(x^(1/2)) 1/(2x^(1/2))



(13)   Find the values of x at which the tangent to the graph of   y = 1/(x - 2) + x  is horizontal.

If the tangent is horizontal, its slope  y ' = -1/(x - 2)^2 + 1 must be 0, so we need to solve
0 = -1/(x - 2)^2 + 1  1/(x - 2)^2 = 1  1 = (x - 2)^2  1 = x^2 - 4x + 4  0 = x^2 - 4x + 3  0 = (x + 3) (x + 1)

Answer: x = 3 and x = 1



(14)  An object, moving on a straight line, is at a distance of f(t) = t^3 - 12t^2 + 50 t   feet from a stationary point on the line at time t seconds.

(a)  Find the function for the object's velocity at time t.

Velocity at time t is f ' (t) = 3t^2 - 24t + 50 feet per second


(b)  At what time(s) is the object's velocity 5 feet per second?

When f '(t) = 5.
3t^2 - 24t + 50 = 5  3t^2 - 24t + 45 = 0  3 (t^2 - 8t + 15) = 0  3 (t - 3) (t - 5) = 0

Answer: when t = 3 seconds and t = 5 seconds



(15)  At which values of x do the graphs of  y = x^2  and  y = x^3  have the same slope?

2x = 3x^2  2x - 3x^2 = 0  FormBox[RowBox[{x (2 - 3x), =, RowBox[{0, Cell[]}]}], TraditionalForm]

Answer: x = 0 and x = 2/3


(16)   Find the slope of the tangent to the graph of the equation   2 2 FormBox[RowBox[{ , Cell[TextData[{Cell[BoxData[x - y = x y - 1]], }]]}], TraditionalForm] at the point (3, 2).

d/dx[x^2 - y^2] = d/dx[x y - 1]  2x - 2y y ' = (1) y + x y ' - 0  2x - y = x y ... 371; y ' = (2x - y)/(x + 2y)  y ' | _ ((x, y) = (3, 2)) = (2 (3) - (2))/(3 + 2 (2)) = 4/7



(17)  A stone dropped into a still pond sends out a circular ripple whose radius increases at a rate of 2 feet per second.  How rapidly is the area enclosed by the ripple increasing 10 seconds after the stone is dropped?

Let r be the radius of the ripple, and let A be its area.

We know dr/dt = 2

We want dA/dt =?

r and A obey the equation  A = π r^2 (area of a circle).  Differentiating both sides with respect to time t,
d/dt[A] = d/dt[π r^2]  dA/dt = 2π r dr/dt  dA/dt = 2π r (2)  dA/dt = 4π r
10 seconds after the stone is dropped, the ripple's radius is (10 sec)(2 ft/sec) = 20 feet.
Plugging this in, we get dA/dt = 4π 20 = 80π square feet per second.