Graphical Vector Problem (What are they asking for?)

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"Graphical" Vector Problem (What are they asking for?)

Homework Statement


Here is the problem statement. I honestly don't understand what it is they are asking for? Do they want actual numbers? Am I supposed to measure the lengths with a ruler or something? I know it's easy once I interpret the question correctly :smile:
Doc-11_14_105_08PM-page-1.jpg


How would you interpret this one?
 
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Hey, that vector A in part (c) is bent. Must have a non-zero curl :-p

Yup, that's what I would do. Draw them carefully, construct the corresponding parallelograms and figure out the alpha and beta by measuring to scale things. "Trace" them certainly suggests that.
 


LCKurtz said:
Hey, that vector A in part (c) is bent. Must have a non-zero curl :-p

Yup, that's what I would do. Draw them carefully, construct the corresponding parallelograms and figure out the alpha and beta by measuring to scale things. "Trace" them certainly suggests that.

Haha! :smile: Thanks LCKurtz! Don't quit your day job (assuming you have one!)!
 


By "graphical means" they mean for you to extend the vectors A and B so that the lengthened vectors form two sides of a parallelogram whose diagonal is C. You'll need to find the lengths of A and B with a ruler so that you can determine the values of \alpha and \beta so that \alphaA + \betaB = C.

For example, if you have to double the lengths of A and B to form the parallelogram, then \alpha = \beta = 2.

Your mileage may vary...
 


I must be an idiot. I cannot see a way to extend or shorten A and B such that C becomes the diagonal of the parallelogram. My brain keeps making B the diagonal since it is more intuitive to me.
 


Don't forget that your coefficients of A and/or B may be negative so you also have to consider the opposite of A or B when making your parallelogram.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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