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Dmitrij [34]
3 years ago
9

Find the derivative of y = sin(ln(5x2 − 2x))

Engineering
1 answer:
pickupchik [31]3 years ago
3 0

Answer:

y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)

Explanation:

Let y = \sin[\ln(5\cdot x^{2}-2\cdot x)] and we proceed to find the derivative by the following steps:

1) y = \sin[\ln(5\cdot x^{2}-2\cdot x)] Given

2) y = \sin [\ln[x\cdot (5\cdot x - 2)]] Distributive property

3) y = \sin[\ln x + \ln (5\cdot x - 2 )] \ln (a\cdot b) = \ln a + \ln b

4) y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)  \frac{d}{dx} (\sin x) = \cos x/\frac{d}{dx}(\ln x) = \frac{1}{x}/\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}/Rule of chain/Result

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A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum
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Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

  • Density = 61 veh/km.
  • Speed = 59 km/hr.
  • Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

q_{max}=\frac{V_f \times K_i}{4}

<u>Where:</u>

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In order to calculate the free flow speed, we would use this formula:

V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr

Substituting the parameters into the model, we have:

q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

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What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4
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Answer:

maximum stress is 2872.28 MPa

Explanation:

given data

radius of curvature = 3 × 10^{-4} mm

crack length = 5.5 × 10^{-2} mm

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to find out

maximum stress

solution

we know that  maximum stress formula that is express as

\sigma m = 2 ( \sigma o ) \sqrt{\frac{a}{\delta t}}     ......................1

here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack

so put here all value in equation 1 we get

\sigma m = 2 ( \sigma o) \sqrt{\frac{a}{\delta t}}  

\sigma m = 2(150) \sqrt{ \frac{\frac{5.5*10^{-2}}{2}}{3*10^{-4}}}  

σm = 2872.28 MPa

so maximum stress is 2872.28 MPa

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