A tug-o-war team produces a tension in a rope described by the rule T=290(8t−0.5t2−1.4) units, where t is the number of seconds after commencing the pull.
1. Sketch a graph of T against t, stating the practical domain.
2. What is the greatest tension produced during a ‘heave’?
At the moment i am clueless and don't know where to start so it would be great if someone could help!
Hi AaronAnton,
For Q1, the practical domain will be t∈ [0,∞). This is because t(i.e time) cannot be negative. As for sketching the graph of the function given, I would just use the CAS calculator or the TI-Nspire calculator to sketch the function. If you want to do it manually, you first need to use the 'completing the square method' to find the turning point. This is shown below:
First make the coefficient of t^2 positive one(i.e +1).
T=290(8t−0.5t^2−1.4)
T=2320t - 145t^2 - 406
T= 145(16t - t^2 - 14/5) ---> T= 145(-t^2+16t-14/5)
T=-145(t^2-16t+14/5)
Now complete the square
T=-145[(t^2-16t+(-16/2)^2)-(-16/2)^2+14/5]
T=-145[(t^2-16t+64)-64+14/5]
T=-145[(t-8)^2-306/5]
T=-145(t-8)^2+8874
Now you know the turning point is at t=8. Also when you sketch the graph, make sure the domain as I said before is t∈ [0,∞).
Q2. This question requires the knowledge of calculus(especially finding the derivative of a function and making it equal to zero to solve for t)
This is how you do it:
T=290(8t−0.5t^2−1.4)
dT/dt=290(8-t)
Let dT/dt=0(we do this because this will find the t value for the turning point, which in most cases will be where the greatest or lowest T value would be)
290(8-t)=0
8-t=0
Therefore, t=8
Now sub t=8 into T=290(8t−0.5t^2−1.4)
Therefore T=8874.
Therefore the greatest tension during the 'heave' is 8874.
I hope my solutions help. Let me know if you are still confused or stuck with any other questions.