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jeyben [28]
2 years ago
13

Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25

Engineering
1 answer:
Sedaia [141]2 years ago
4 0

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:

y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x

y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}

dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx

dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)

dy = \frac{1}{2\pi}

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.

To learn more on differentials: brainly.com/question/24062595

#SPJ1

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two pints of ethyl ether evaporate over a period of 1,5 hours. what is the air flow necessary to remain at 10% or less of ethyl
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The air flow necessary to remain at the lower explosive level is 4515. 04cfm

<h3>How to solve for the rate of air flow</h3>

First we have to find the rate of emission. This is solved as

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We have the following details

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Read more on the rate of air flow on brainly.com/question/13289839

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Two blocks of rubber with a modulus of rigidity G =10 MPa are bonded to rigid supports and to a plate AB. Knowing that b = 200 m
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4 years ago
An enrichment plant has a throughput of 32,000 kg U/day and produces 26,000 kg U as tails. What is the enrichment of the product
torisob [31]

Answer:

1.10 %

Explanation:

The enrichment of the product can be calculated using the following equation:

\frac{W}{P} = \frac{x_{p} - x_{f}}{x_{f} - x_{w}}   (1)

<em>where W: is waste or tails = 26000, P: is the product = 32000, xp: is the wt. fraction of ²³⁵U in product, xf: is the wt. fraction of ²³⁵U in feed = 0.72% and xw: is the wt. fraction of ²³⁵U in waste or tails = 0.25%.      </em>

By solving equation (1) for xp, we can find the enrichment of the product:

x_{p} = \frac {W}{P} \cdot (x_{f} - x_{w}) + x_{f}        

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Therefore, the enrichment of the product is 1.10 %.

I hope it helps you!

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