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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.
