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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.
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).
(3) If , find
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13) Find the values
of x at which the tangent to the graph
of is
horizontal.
If the tangent is horizontal, its slope
must be 0, so we need to solve
Answer: x
= 3 and x
= 1
(14) An object, moving on
a straight line, is at a distance of 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
feet per second
(b) At what time(s) is the
object's velocity 5 feet per second?
When f '(t)
= 5.
Answer: when t
= 3 seconds and t
= 5 seconds
(15) At which values of x
do the graphs of and have
the same slope?
Answer: x
= 0 and x
= 2/3
(16) Find the slope
of the tangent to the graph of the equation
at the point (3, 2).
(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
We want ?
r and A
obey the equation
(area of a circle). Differentiating both sides with respect to time
t,
10 seconds after the stone is dropped, the ripple's radius is (10 sec)(2 ft/sec)
= 20 feet.
Plugging this in, we get
square feet per second.