Subido por Julio Julca Castillo

Ing-Materiales-Lista-01

List 1
List of Proposed Problems
Mathematical Analysis
1. Increases. Find the increase in the independent variable for the given points.
a) x1 = 2; x2 = 3
c) t1 = 1.0; t2 = 1.25
b) x1 = 4.5; x2 = 6.3
d) s1 = −2.4; s2 = −2.37
2. Increases. Find the increments of the following functions for the given intervals
a) f (x) = 2x + 7; x = 3, ∆x = 0.2
b) f (x) = 3x2 + 3x − 5; x = 2, ∆x = 0.1
x2 − 4
c) f (x) =
; x = 1, ∆x = 0.2
x−2
4x2 − 3x + 1
; x = 3, ∆x = 0.3
x−2
500
e) p(t) = 2000 +
; t = 2, ∆t = 1
1 + t2
d) f (x) =
f ) h(x) = ax2 + bx + c,
x
to x + ∆x
5
,
t
to t + ∆t
1+t
3. Slope of the secant line. Find the slope of the secant line passing through the
points (x1, y1) y (x2, y2) where y1 = f (x1) and y2 = f (x2) for the function and
given points.
g) G(t) = 300 +
a) f (x) = 3x − 2, x1 = 2.5, x2 = 2.4
b) f (x) = x2 + 2x, x1 = −1.4, x2 = −2.1
c) f (x) = 1 − 3x2, x1 = 0.4, x2 = 1.2
1
d) f (x) = , x1 = 1, x2 = 1.4
x
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4. Average speed. Find the average velocity for the position function h(t) = −4.9t2 +
50t + 40, meters in the given time intervals.
a) t1 = 1 to t2 = 1.5 s.
b) t1 = 1 to t2 = 1.1 s.
c) t1 = 1 to t2 = 1.01 s.
d) t1 = 1 to t2 = 1.001 s.
e) t1 = 1 a t2 = 1 + h seconds, where h is the increment of time.
f ) what happens if we make h → 0 (the result obtained is the instant speed flame
in t = 1 seconds)
5. Average exchange rate. Find the average exchange rate of each function in the
given interval.
a) f (x) = x2 − 2x + 3; x1 = 2, x2 = 2.5
x
b) f (x) =
; x1 = 1.1, x2 = 1.4
1 + x2
6. Concept questions refer to the following figure. Let P (2, f (2)) and Q(2+h, f (2+
h)) be points on the graph of a function f .
a) Find an expression for the slope of the secant line passing through P and Q.
b) Find an expression for the slope of the tangent line passing trough P .
7. Interpretation. (a) Give a geometric and a physical interpretation of the expression
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f (x + h) − f (x)
h
(b) Give a geometric and a physical interpretation of the expression
f (x + h) − f (x)
h→0
h
lı́m
8. Slope. Let f (x) = x2
a) Compute the slope of the secant line joining the points on the graph of f whose
x coordinates are x = −2 and x = −1.9
b) Use calculus to compute the slope of the line that is tangent to the graph of f
when x = −2 and compare with the slope found in part (a)
x
9. Slope. Let f (x) =
x−1
a) Compute the slope of the secant line joining the points on the graph of f whose
x coordinates are x = −1 and x = −0.5
b) Use calculus to compute the slope of the line that is tangent to the graph of f
when x = −1 and compare with the slope found in part (a)
10. Derivative. Find the derivative of the given function using the following limit
′
f (x + h) − f (x)
h→0
h
f (x) = lı́m
if the limit exists, for
a) f (x) = x2
b) f (x) = x2 − 3x
c) f (x) = 4x − 3
d) f (x) = 1 − x2
2
e) f (x) =
x+1
x
f ) f (x) =
x−1
2x + 3
g) f (x) =
x+4
h) f (x) =
√
2x + 1
i) f (x) = sin x
j) f (x) = cos x
11. Derivative. Find the derivative of the given function using the following limit
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f (x) − f (x0)
0
x − x0
′
f (x0) = x→x
lı́m
if the limit exists, for
d) f (x) = x2 − 3x − 1, x0 = −1
√
e) f (x) = x, x0 = 4
a) f (x) = 10x2 − 3, x0 = 1
b) f (x) = x3 − 4x2, x0 = −1
c) f (x) =
4
, x0 = 2
3−x
12. Slope. Find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
√
d) f (x) = x + 1, (8, 3)
a) f (x) = x2 + 1, (2, 5)
x
x−1
b) f (x) =
, (3, 3)
e)
f
(x)
=
, (0, −1)
x−2
x+1
8
c) f (x) = 2 , (2, 2)
x
13. Slope and tangent line. Let f (x) =
′
a) Find the derivative f of f
1
x−1
b) Find an equation of the tangent line to the curve at the point (−1, − 12 )
c) Sketch the graph of f and the tangent line to the curve at (−1, − 12 )
14. Average rate. Let y = x2 + x
a) Find the average rate of change of y with respect to x in the interval from
x = 2 to x = 3, from x = 2 to x = 2.5, and from x = 2 to x = 2.1
b) Find the instantaneous rate of change of y at x = 2
c) Compare the results obtained in part (a) with that of part (b)
15. Velocity of a car. Suppose the distance s (in feet) covered by a car moving along a
straight road after t sec is given by the function s = 2t2 + 48t
a) Calculate the average velocity of the car over the time intervals [20, 21], [20, 20.1]
and [20, 20.01]
b) Calculate the instantaneous velocity of the car when t = 20
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c) Compare the results of part (a) with the part (b)
16. Velocity of a ball thrown into the air. A ball is thrown straight up with an initial
velocity of 128 f t/sec, so that its height (in feet) after t sec is given by s(t) =
128t − 16t2.
a) What is the average velocity of the ball over the time intervals [2, 3], [2, 2.5],
and [2, 2.1] ?
b) What is the instantaneous velocity at time t = 2 ?
c) What is the instantaneous velocity at time t = 5 ?. Is the ball rising of falling
at this time?
d) When will the ball hit the ground?
17. Hot-air balloon rises. A hot-air balloon rises vertically from the ground so that its
1
1
height after t sec is h = t2 + t f t 0 ≤ t ≤ 60.
2
2
a) What is the height of the balloon at the end of 40 sec ?
b) What is the average velocity of the balloon between t = 0 and t = 40 ?
c) What is the velocity of the balloon a the end of 40 sec?
18. Temperature. At a temperature of 20oC, the volume V (in liters) of 1.33 g of O2 is
1
related to its pressure p (in atmospheres) by the formula V =
p
a) What is the average rate of change of V with respect to p as p increases from
p = 2 to p = 3?
b) What is the rate of change of V with respect to p when p = 2?
19. Tangent lines with specified slopes.
a) At what points do the graphs of the functions have horizontal tangents? (a)
f (x) = x2 + 4x − 1 and (b) g(x) = x3 − 3x
b) Find equations of all lines having slope −1 that are tangent to the curve
1
y=
x−1
c) Find an equation
of the straight line having slope 1/4 that is tangent to the
√
curve y = x
20. Vertical tangents. Graph the curves in the exercises (a) Where do the graphs appear to have vertical tangents ?. (b) Confirm your findings in part (a) with limit
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calculations.
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a) y = x2/5
d) y = x3/5
g) y = x2/3 − (x − 1)1/3
b) y = x4/5
e) y = 4x2/5 − 2x
h) y = x1/3 + (x − 1)1/3
c) y = x1/5
f ) y = x5/3 − 5x2/3
i) y = |4 − x|
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