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Calculus I                                                                      Test #2                                             October 29, 2003

Name____________________                               R.  Hammack                                             Score ______
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(1) Use the limit definition of the derivative to find the derivative of the function f(x) = x^2.


Underscript[lim , wx] (f(w) - f(x))/(w - x) = Underscript[lim , wx] (w^2 - x^2 ... ipt[lim , wx] (w - x) (w + x)/(w - x) = Underscript[lim , wx] (w + x) = x + x = 2x


(2) The graph of a function g(x) is a straight line that is inclined at an angle of 30 degrees, as illustrated below.  Find g ' (0).
[Graphics:HTMLFiles/T2F03Dsol_5.gif]
g ' (x) = slope of the line = -1/3^(1/2)


(3) Sketch the graph of a function  f  for which f(0) = -1, f ' (0) = 0,  f ' (2) = 0, f ' (x) ≥0 when x≥0, and  f ' (x) ≤0 when x≤0.

[Graphics:HTMLFiles/T2F03Dsol_15.gif]

(4) Suppose that the cost of drilling x feet for an oil well is C = f(x) dollars, and suppose that  f ' (300) = 1000. Explain, in non-mathematical terms, what the statement  f ' (300) = 1000 means.

When the well is 300 feet deep, it will cost an extra $1000 to dig the next foot.
(Because rate of change of cost is $1000 per foot when x = 300 feet.)

(5) d/dx[ 3x^4 + x + 2] = 12x^3 + 1

(6) d/dx[ -3/x^8 + 2x^(1/2)] = d/dx[ -3x^(-8) + 2x^(1/2)] = 24x^(-9) + x^(-1/2) = 24/x^9 + 1/x^(1/2)


(7)  d/ds[ 3/(2s + 1)] =d/ds[ 3 (2s + 1)^(-1)] = -3 (2s + 1)^(-2) 2 = -6/(2s + 1)^2


(8) d/dx[ sec(x) tan(x) ] =sec(x) tan(x) tan(x) + sec(x) sec^2(x) = sec(x) tan^2(x) + sec^3(x)


(9) d/dx[sin(x) /x^2] =(cos(x) x^2 - sin(x) 2x)/x^4


(10)  If y = cos(x), find (d^2y)/dx^2.
dy/dx = -sin(x)

(d^2y)/dx^2 = -cos(x)


(11) d/dx[ (x^3 + 3x)^10] =10 (x^3 + 3x)^9 (3x^2 + 3)


(12) d/dx[ sin(x^3) ] =cos(x^3) 3x^2


(13)  d/dx[ sin(x^3)^(1/2)] = d/dx[ (sin(x^3))^(1/2)] = 1/2 (sin(x^3))^(-1/2) d/dx[ sin(x^3) ] = (cos(x^3) 3x^2)/(2sin(x^3)^(1/2))


(14) d/dx[ x^5sec(1/x)] = 5x^4sec(1/x) + x^5sec(1/x) tan(1/x) (-1/x^2) = 5x^4sec(1/x) - x^3sec(1/x) tan(1/x)


(15)  Suppose f  is a function for which f(4) = 3 and f ' (4) = -5.  If g(x) = x f(x), find g ' (4).

g ' (x) = (1) f(x) + x f ' (x)
g ' (4) = (1) f(4) + 4 f ' (4) = 3 + (4) (-5) = -17


(16)  Find all points on the graph of  y = 1 - x^2  at which the tangent line passes through the point (0, 2).

A typical point on the graph will have form (x, 1 - x^2) , and the line passing through this point and (0, 2) will have slope (2 - (1 - x^2))/(0 - x) = -(1 + x^2)/x.
We are interested in when this line is a tangent line, i.e. when its slope is dy/dx = -2x.
Thus we need to solve the equation
-2x = -(1 + x^2)/x

2x^2 = 1 + x^2

x^2 - 1 = 0

(x - 1) (x + 1) = 0

Thus x must be 1 or -1, so the points on the graph are (1,f(1)) = (1, 0) and (-1,f(-1)) = (-1, 0).



(17)  Find the equation of the tangent line to the graph of y = tan(x) at the point where x = π/4.

The slope at x  is dy/dx = sec^2(x).
We are interested in the point where  x = π/4, and the slope is then FormBox[RowBox[{sec^2(π/4), =, 2.}], TraditionalForm]
A point on the tangent line is (π/4, tan(π/4)) = (π/4, 1).

Using the point-slope formula,
y - 1 = 2 (x - π/4)
y - 1 = 2x - π/2
y = 2x + 1 - π/2

y = 2x + (2 - π)/2



(18)  A search light is trained on a tall building. As the light rotates, the spot it illuminates moves up  the side of the building. That is, the distance D between the ground and the illuminated spot is a function of the angle θ formed by the light beam and the horizontal ground. If the search light is located 50 meters from the building, find the function giving the rate of change of D with respect to θ.

tan(θ) = OPP/ADJ = D/50

Thus D = 50tan(θ)

Rate of change is D ' = 50sec^2(θ) meters per radian


(19)  Use implicit differentiation to find dy/dx:     cos(x y) = 1/2.

  d/dx[cos(x y)] = d/dx[1/2]
  
-sin(xy) (y + xdy/dx) = 0

(y + xdy/dx) = 0/-sin(xy)

y + xdy/dx = 0

dy/dx = -y/x




(20)  A conical water tank has a height of 24 feet and a radius of 12 feet, as illustrated. If water flows into the tank at a constant rate of 20 cubic feet per minute, how fast is the depth h of the water changing when h = 2?
(Hint: You may be interested to know that the volume V of a cone of height h and radius r is V = π/3r^2h.)

Let V be the volume of the water in the tank.

Know dV/dt = 20

Want dh/dt

By similar triangles, you get 12/24 = r/h so r = h/2.

By the volume formula,
V = π/3r^2h

V = π/3 (h/2)^2h

V = π/12h^3

d/dt[V] = d/dt[π/12h^3]

dV/dt = π/123h^2dh/dt

20 = π/42^2dh/dt

dh/dt = 20/πfeet per minute