can anyone do these calculus problems
Discussion
problem 1.
The expression f(t)=10t^2 x sin(2t) represents the work done in kJ by a mechanism on a canning plant against time. By using the process of integration, determine the power dissipated between 1 and 1.3 seconds after the device starts to function.
By the way, f(t)=10t^2 x sin(2t), with the ^ means to the power of 2. Just cant put the equation in correctly as its an e-mail.
Problem 2.
The function Ln((5t^3 - 6t)^3) represents the position of a lever on a mechanism at some point in time. Differentiate the function to determine its velocity 3 seconds after it starts to move.
thanks
The expression f(t)=10t^2 x sin(2t) represents the work done in kJ by a mechanism on a canning plant against time. By using the process of integration, determine the power dissipated between 1 and 1.3 seconds after the device starts to function.
By the way, f(t)=10t^2 x sin(2t), with the ^ means to the power of 2. Just cant put the equation in correctly as its an e-mail.
Problem 2.
The function Ln((5t^3 - 6t)^3) represents the position of a lever on a mechanism at some point in time. Differentiate the function to determine its velocity 3 seconds after it starts to move.
thanks
moddy said:
problem 1.
The expression f(t)=10t^2 x sin(2t) represents the work done in kJ by a mechanism on a canning plant against time. By using the process of integration, determine the power dissipated between 1 and 1.3 seconds after the device starts to function.
By the way, f(t)=10t^2 x sin(2t), with the ^ means to the power of 2. Just cant put the equation in correctly as its an e-mail.
Problem 2.
The function Ln((5t^3 - 6t)^3) represents the position of a lever on a mechanism at some point in time. Differentiate the function to determine its velocity 3 seconds after it starts to move.
thanks
Problem 1.The expression f(t)=10t^2 x sin(2t) represents the work done in kJ by a mechanism on a canning plant against time. By using the process of integration, determine the power dissipated between 1 and 1.3 seconds after the device starts to function.
By the way, f(t)=10t^2 x sin(2t), with the ^ means to the power of 2. Just cant put the equation in correctly as its an e-mail.
Problem 2.
The function Ln((5t^3 - 6t)^3) represents the position of a lever on a mechanism at some point in time. Differentiate the function to determine its velocity 3 seconds after it starts to move.
thanks
[-5t^2 x cos(2t) - int(-10tcos2t)]
[-5t^2 x cos(2t) - (-5tsin2t - int(-5sin2t))]
[-5t^2 x cos(2t) + 5tsin2t + 2.5cos2t]
therefore sub in 1.3 and 1 you get then take the value from at t=1 from that at t=1.3 you get 2.86kW.
Problem 2.
dx/dt = 9(-2+5t^2)(-6t+5t^3)^2 = (-18+45t^2)(-6t+5t^3)^2
dp/dx= 1/x =1/(5t^3 - 6t)^3
dp/dt=dp/dx*dx/dt = (18-45t^2)/(6t-5t^3)
sub in t=3 v=dp/dt= 3.3m/s
There is something not right about the first question, unless I have missed something obvious. Power is the time derivative of work not the integral, so although denchy's integration of the expression is correct, I don't think it gives a meaningful answer. If the question gave the expression as that for power and asked you for the work done between the times, then denchy's answer would be the way to do it.
Where are the questions from?
Where are the questions from?
tank slapper said:
There is something not right about the first question, unless I have missed something obvious. Power is the time derivative of work not the integral, so although denchy's integration of the expression is correct, I don't think it gives a meaningful answer. If the question gave the expression as that for power and asked you for the work done between the times, then denchy's answer would be the way to do it.
Where are the questions from?
Yep you are correct, didnt read it properly, was late. Question should say expression for Power, intergrate to find the work done.Where are the questions from?
.
. Masters postgraduate, but this stuff is very much dark arts 
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