Advanced Placement Calculus 2016 Graphical Numerical Algebraic Fifth Edition Student Edition + Mathxl 1-year License
5th Edition
ISBN: 9780133314533
Author: Prentice Hall
Publisher: Prentice Hall
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1. Newton's Law of Gravitation (an example of an inverse square law) states that the magnitude
of the gravitational force between two objects with masses m and M is
|F|
mMG
|r|2
where r is the distance between the objects, and G is the gravitational constant. Assume that the
object with mass M is located at the origin in R³. Then, the gravitational force field acting on
the object at the point r = (x, y, z) is given by
F(x, y, z) =
mMG
r3
r.
mMG
mMG
Show that the scalar vector field f(x, y, z) =
=
is a potential function for
r
√√x² + y² .
Fi.e. show that F = Vf.
Remark: f is the negative of the physical potential energy, because F = -V(-ƒ).
2. Suppose f(x) = 3x² - 5x. Show all your work for the problems below.
write it down for better understanding please
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Similar questions
- 1. Suppose F(t) gives the temperature in degrees Fahrenheit t minutes after 1pm. With a complete sentence, interpret the equation F(10) 68. (Remember this means explaining the meaning of the equation without using any mathy vocabulary!) Include units. (3 points) =arrow_forward2. Suppose f(x) = 3x² - 5x. Show all your work for the problems below. a. Evaluate f(-3). If you have multiple steps, be sure to connect your expressions with EQUALS SIGNS. (3 points)arrow_forward4c Consider the function f(x) = 10x + 4x5 - 4x³- 1. Enter the general antiderivative of f(x)arrow_forward
- A tank contains 60 kg of salt and 2000 L of water. Pure water enters a tank at the rate 8 L/min. The solution is mixed and drains from the tank at the rate 11 L/min. Let y be the number of kg of salt in the tank after t minutes. The differential equation for this situation would be: dy dt y(0) =arrow_forwardSolve the initial value problem: y= 0.05y + 5 y(0) = 100 y(t) =arrow_forwardy=f'(x) 1 8 The function f is defined on the closed interval [0,8]. The graph of its derivative f' is shown above. How many relative minima are there for f(x)? O 2 6 4 00arrow_forward
- 60! 5!.7!.15!.33!arrow_forward• • Let > be a potential for the vector field F = (−2 y³, −6 xy² − 4 z³, −12 yz² + 4 2). Then the value of sin((-1.63, 2.06, 0.57) – (0,0,0)) is - 0.336 -0.931 -0.587 0.440 0.902 0.607 -0.609 0.146arrow_forwardThe value of cos(4M) where M is the magnitude of the vector field with potential ƒ = e² sin(лy) cos(π²) at x = 1, y = 1/4, z = 1/3 is 0.602 -0.323 0.712 -0.816 0.781 0.102 0.075 0.013arrow_forward
- There is exactly number a and one number b such that the vector field F = conservative. For those values of a and b, the value of cos(a) + sin(b) is (3ay + z, 3ayz + 3x, −by² + x) is -0.961 -0.772 -1.645 0.057 -0.961 1.764 -0.457 0.201arrow_forwardA: Tan Latitude / Tan P A = Tan 04° 30'/ Tan 77° 50.3' A= 0.016960 803 S CA named opposite to latitude, except when hour angle between 090° and 270°) B: Tan Declination | Sin P B Tan 052° 42.1'/ Sin 77° 50.3' B = 1.34 2905601 SCB is alway named same as declination) C = A + B = 1.35 9866404 S CC correction, A+/- B: if A and B have same name - add, If different name- subtract) = Tan Azimuth 1/Ccx cos Latitude) Tan Azimuth = 0.737640253 Azimuth = S 36.4° E CAzimuth takes combined name of C correction and Hour Angle - If LHA is between 0° and 180°, it is named "west", if LHA is between 180° and 360° it is named "east" True Azimuth= 143.6° Compass Azimuth = 145.0° Compass Error = 1.4° West Variation 4.0 East Deviation: 5.4 Westarrow_forwardds 5. Find a solution to this initial value problem: 3t2, s(0) = 5. dt 6. Find a solution to this initial value problem: A' = 0.03A, A(0) = 100.arrow_forward
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