Problem statement:
Find the arc length of the curve defined by x = √t and y = 3t -1 on the interval 0 < t < 1
attempted solution:
dx/dt = 1/2t-1/2 , dy/dt = 3 and dx = dt/ 2√t
dy/dx = 6√t
length = ∫01 √(1 + (6√t)2) .dt/ 2√t
= ∫01 √1 + 36t) dt/2√t
now I'm stuck with a product that is very...
I'm struggling to solve the following integral
∫ x/(√27-6x-x2)
my attempt is as follows:
∫x/(√36 - (x+3)2)
= ∫1/ √(36 - (x+3)2) + ∫x+1/ √(36 - (x+3)2)
= arcsin (x + 3)/6 + this is where I got stuck.
1. Please see the attachments for my attempts at answering the question.
Please note that this is not a homework question as the answers are provided. I just want to know how to get there. Thanks in advance.
Correct me of I'm wrong but if the second litre is heated immediately after the first, 150g of the system is theoretically at 90 degrees correct? You'll have to pardon my reasoning, I'm a first year electrical engineering student.
1. The element of a heater kettle has a constant resistance of 105 ohms and applied voltage of 240V.
1. Calculate the time taken to raise the temperature of water from 15 degrees Celsius to 90 assuming that. 80% of the power input to the kettle is usefully employed. I worked this out and got 11...