Problems 87-94 require the following discussion of a secant line. The slope of the secant line containing the two points ( x , f ( x ) ) and ( x + h , f ( x + h ) ) on the graph of a function y = f ( x ) may be given as In calculus, this expression is called the difference quotient of f (a) Express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer. (b) Find m sec for h = 0.5 , 0.1, and 0.01 at x = 1 . What value does m sec approach as h approaches 0? (c) Find an equation for the secant line at x = 1 with h = 0 .01 . (d) Use a graphing utility to graph f and the secant line found in part (c) in the same viewing window. 94. f ( x ) = 1 x 2
Problems 87-94 require the following discussion of a secant line. The slope of the secant line containing the two points ( x , f ( x ) ) and ( x + h , f ( x + h ) ) on the graph of a function y = f ( x ) may be given as In calculus, this expression is called the difference quotient of f (a) Express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer. (b) Find m sec for h = 0.5 , 0.1, and 0.01 at x = 1 . What value does m sec approach as h approaches 0? (c) Find an equation for the secant line at x = 1 with h = 0 .01 . (d) Use a graphing utility to graph f and the secant line found in part (c) in the same viewing window. 94. f ( x ) = 1 x 2
Solution Summary: The author explains how the function f (x) = 1 x 2 requires the discussion of a secant line.
Problems 87-94 require the following discussion of a secant line. The slope of the secant line containing the two points
and
on the graph of a function
may be given as
In calculus, this expression is called thedifference quotient of f
(a) Express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer.
(b) Find msec for
, 0.1, and 0.01 at
. What value does msec approach as h approaches 0?
(c) Find an equation for the secant line at
with
.
(d) Use a graphing utility to graph f and the secant line found in part (c) in the same viewing window.
Good Day,
Would appreciate any assistance with this query.
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This question builds on an earlier problem. The randomized numbers may have changed, but have your work for the previous problem available to help with this one.
A 4-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius 2 cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The
wheel rotates counterclockwise at 1.5 rev/sec. At some point, the rod will be tangent to the circle as shown in the third picture.
A
B
A
B
at some instant, the piston will be tangent to the circle
(a) Express the x and y coordinates of point A as functions of t:
x= 2 cos(3πt)
and y= 2 sin(3t)
(b) Write a formula for the slope of the tangent line to the circle at the point A at time t seconds:
-cot(3πt)
sin(3лt)
(c) Express the x-coordinate of the right end of the rod at point B as a function of t: 2 cos(3πt) +411-
4
-2 sin (3лt)
(d)…
5. [-/1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.5.AE.003.
y
y= ex²
0
Video Example
x
EXAMPLE 3
(a) Use the Midpoint Rule with n = 10 to approximate the integral
कर
L'ex²
dx.
(b) Give an upper bound for the error involved in this approximation.
SOLUTION
8+2
1
L'ex² d
(a) Since a = 0, b = 1, and n = 10, the Midpoint Rule gives the following. (Round your answer to six decimal places.)
dx Ax[f(0.05) + f(0.15) + ... + f(0.85) + f(0.95)]
0.1 [0.0025 +0.0225
+
+ e0.0625 + 0.1225
e0.3025 + e0.4225
+ e0.2025
+
+ e0.5625 €0.7225 +0.9025]
The figure illustrates this approximation.
(b) Since f(x) = ex², we have f'(x)
=
0 ≤ f'(x) =
< 6e.
ASK YOUR TEACHER
and f'(x) =
Also, since 0 ≤ x ≤ 1 we have x² ≤
and so
Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate, we see that an upper bound for the error is as follows. (Round your final
answer to five decimal places.)
6e(1)3
e
24(
=
≈
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