By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
<h3>How to determine the differential of a one-variable function</h3>
Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:
dy = y'(x) · dx (1)
If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:





By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
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Answer:
The tension in the rope at the lowest point is 270 N
Explanation:
Given;
weight of the ball, W = 150 N
length of the rope, r = 4 m
velocity of the ball, v = 5.6 m/s
When the ball passes through the lowest point, the tension on the rope is the sum of weight of the ball and centripetal force.
T = W + F
Centripetal force, F = mv²/r
where;
m is the mass of the ball
m = W/g
m = 150 / 9.8 = 15.306 kg
Centripetal force, F = mv²/r
F = (15.306 x 5.6²)/4
F = 120 N
T = W + F
T = 150 + 120
T = 270 N
Therefore, the tension in the rope at the lowest point is 270 N
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Answer:
Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:
Aug=[A eye(size(A))]
Above command is tested in MATLAB command window and is attached in figure below