In problem 79-90, find the difference quotient of f: that is find f ( x + h ) − f ( x ) h , h ≠ 0 for each function. Be sure to simplify. f ( x ) = x − 2 [ Hint: Rationalize the numerator.]
In problem 79-90, find the difference quotient of f: that is find f ( x + h ) − f ( x ) h , h ≠ 0 for each function. Be sure to simplify. f ( x ) = x − 2 [ Hint: Rationalize the numerator.]
Solution Summary: The author explains how to calculate the difference quotient of f.
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(
=
≈
2. [-/1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.5.015.
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.)
ASK YOUR TEACHER
3
1
3 +
dy, n = 6
(a) the Trapezoidal Rule
(b) the Midpoint Rule
(c) Simpson's Rule
Need Help? Read It
Watch It
Chapter 1 Solutions
Pearson eText for Precalculus: Concepts Through Functions, A Right Triangle Approach to Trigonometry -- Instant Access (Pearson+)
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY