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What is the solution of the derivative of square root of t divided by the square root of pie?
One expression of the answer is(square root of pi) / (2*(square root of t)*(pi))To solve this problem, you will need to apply two differential calculus rules:1. The constant times a function rule2. The power rule.Because it is so hard to write out these rules within the limits of the answers.com display system, you should look up these rules on the internet if you are not familiar with them.Let SQRT() be the square root function. We are to find the derivative ofSQRT(t)/SQRT(pi)We can write this as(1/SQRT(pi)) * SQRT(t), where "*" represents multiplication.The first factor(1/SQRT(PI))is just a constant. However, we often need forms where we remove the SQRT from the denominator. To do so, we multiply the numerator and denominator by SQRT(pi), which gives usSQRT(pi)/piThe above expression is the constant we will use when we apply rule 1 above. That constant will remain a factor of the answer, according to the rule.Next we apply the second rule toSQRT(t). In order to apply this rule to the expression, we convert the expression to power format. Let ^ represent the exponentiation operator. ThenSQRT(t) = t^(1/2)Applying the power rule to the second factor of our original expression, we get(1/2) * (t^(-1/2)) =(1/2) * (1/(t^(1/2) =(1/2) * (1/(SQRT(t)) =1 / (2 *SQRT(t))Combining the factors from the application of both rules we have( SQRT(pi)/pi ) * (1 / (2 *SQRT(t)) ) =SQRT(pi) / (2 * SQRT(t) * pi) = (the answer given above)SQRT(pi) / (2 * pi * SQRT(t)) = an equivalent answer(SQRT(pi) * SQRT(t)) / (2 * pi * t) = another equivalent answer(SQRT(pi * t)) / (2 * pi * t) = another equivalent answer.=================================Answer #2:Wow! If that kind of problem had required that kind of work,I'm afraid I would have dropped Calculus, and missed out onthe fun of Math and Engineering completely.How about 1/sqrt(pi * t) ?
What is the time period of simple pendulum if its length is 100 cm?
T = 2*pi*sqrt(L/g) = 2.006 seconds (approx). T = 2*pi*sqrt(L/g) = 2.006 seconds (approx). T = 2*pi*sqrt(L/g) = 2.006 seconds (approx). T = 2*pi*sqrt(L/g) = 2.006 seconds (approx).
The frequency and period of a wave are related?
The frequency is proportional to the reciprocal of the period and vice versa. Generally this proportion is 2*pi*f = 1/t and t = 1/(2*pi*f) where the frequency is f and the period is t.
What are all the exact values for which tan t equals sqrt3?
sin(60) or sin(PI/3) = sqrt(3)/2 cos(60) or cos(PI/3)=1/2 tan(60) or tan(PI/3) = sin(60)/cos(60)=sqrt(3) But we want tan for -sqrt(3). Tangent is negative in quadrant II and IV. In Quadrant IV, we compute 360-60=300 or 2PI-PI/3 =5PI/3 tan(5PI/3) = -sqrt(3) Tangent is also negative in the second quadrant, so we compute PI-PI/3=2PI/3 or 120 degrees. tan(t)=-sqrt(3) t=5PI/3 or 2PI/3 The period of tan is PI The general solution is t = 5PI/3+ n PI, where n is any integer t = 2PI/3+ n PI, where n is any integer
If you have Cos theta equals 0.92568 How do you find theta in radians?
cos(t) = 0.92568 therefore t = cos-1(0.92568) = 0.3880. If the answer comes out as 22.23, the calculator is set to degrees. Simply multiply that result by pi/180 to convert to radians (or reset the calculator to work in radians). Excel, for example, works in radians. From that primary value you get t = 0.3880 + 2*k*pi and t = 2*k*pi - 0.3880 for all integer values of k.
Why prefer nominal pi circuit over nominal T circuit in transmission line modeling?
Nominal Pi is used for the transmission line modeling as it makes the Y matrix representation more accurately and in actual sense we get a conditioned matrix.
What is the period of xt equals 3.4 Cos5 Pi t - Pi?
Assuming the question refers to x(t) = 3.4*cos(5*pi*t) - pi, the periodicity of the cos function is 2*pi So 5*pi*t = 2*pi or 5*t = 2 or t = 0.4
What has the author P T Cox written?
P. T. Cox has written: 'Pi-Pi resonance poles'
What has the author H T Reynolds written?
H. T. Reynolds has written: 'Analysis of nominal data'
What is the solution of the derivative of square root of t divided by the square root of pie?
One expression of the answer is(square root of pi) / (2*(square root of t)*(pi))To solve this problem, you will need to apply two differential calculus rules:1. The constant times a function rule2. The power rule.Because it is so hard to write out these rules within the limits of the answers.com display system, you should look up these rules on the internet if you are not familiar with them.Let SQRT() be the square root function. We are to find the derivative ofSQRT(t)/SQRT(pi)We can write this as(1/SQRT(pi)) * SQRT(t), where "*" represents multiplication.The first factor(1/SQRT(PI))is just a constant. However, we often need forms where we remove the SQRT from the denominator. To do so, we multiply the numerator and denominator by SQRT(pi), which gives usSQRT(pi)/piThe above expression is the constant we will use when we apply rule 1 above. That constant will remain a factor of the answer, according to the rule.Next we apply the second rule toSQRT(t). In order to apply this rule to the expression, we convert the expression to power format. Let ^ represent the exponentiation operator. ThenSQRT(t) = t^(1/2)Applying the power rule to the second factor of our original expression, we get(1/2) * (t^(-1/2)) =(1/2) * (1/(t^(1/2) =(1/2) * (1/(SQRT(t)) =1 / (2 *SQRT(t))Combining the factors from the application of both rules we have( SQRT(pi)/pi ) * (1 / (2 *SQRT(t)) ) =SQRT(pi) / (2 * SQRT(t) * pi) = (the answer given above)SQRT(pi) / (2 * pi * SQRT(t)) = an equivalent answer(SQRT(pi) * SQRT(t)) / (2 * pi * t) = another equivalent answer(SQRT(pi * t)) / (2 * pi * t) = another equivalent answer.=================================Answer #2:Wow! If that kind of problem had required that kind of work,I'm afraid I would have dropped Calculus, and missed out onthe fun of Math and Engineering completely.How about 1/sqrt(pi * t) ?
What is the time period of simple pendulum if its length is 100 cm?
T = 2*pi*sqrt(L/g) = 2.006 seconds (approx). T = 2*pi*sqrt(L/g) = 2.006 seconds (approx). T = 2*pi*sqrt(L/g) = 2.006 seconds (approx). T = 2*pi*sqrt(L/g) = 2.006 seconds (approx).
The frequency and period of a wave are related?
The frequency is proportional to the reciprocal of the period and vice versa. Generally this proportion is 2*pi*f = 1/t and t = 1/(2*pi*f) where the frequency is f and the period is t.
If a maximum acceleration is 1G for a pure sin wave at 50Hz what is the velocity and displacement?
Y = A sin(100 pi t)Velocity = y' = 100 pi A cos (100 pi t)Acceleration = V' = y'' = -10,000 pi2 A sin(100 pi t) = 9.8 m/sec2Max acceleration = 10,000 pi2 A = 9.8 m/sec2Max Displacement = Amplitude = 9.8 / (10,000 pi2) = 0.0000993 = 0.0993 millimeterMax Velocity = 100 pi A = 9.8 / (100 pi) = 0.0312 = 3.12 centimeter/second
What is the integration OF a minus bcos theta WHEN a AND b ARE constants andheta IS minus pi by 2 TO pi by 2?
let p=pi and t=theta so you're looking at: /2p / | | | b-a*Cos[t] dt=b*t-a*Sin[t] as t goes from -2p to 2p | | / / -2p now evaluate at 2p. 2pb-a*Sin[2p]=2pb do the same for -2p. -2pb-a*Sin[-2p]=-2pb the you take b-acos[t] evaluated at 2p minus it evaluated at -2p [2pb]-[-2pb]=4pb so 4 pi times b
What is a nominal leader?
It is a temporary account which records revenues gains & losses etc. In other words it is a t shape account
What is the equation for a sine wave operating at 375 Hz?
It is f(t) = a*sin(2*pi*375*t) where a is the amplitude and t is the time in seconds.
What are the formulas for Inverted pendulum model of walking?
T=2 pi srq (L/g) and omega= (2 pi /T) is simple pendulum Vw = 2 L/T natural speed of walking Fr=V^2 /(gL) Froude number
