Recent content by opus

Thanks all! I’m about to take an exam so I’ll come back to this when I get home!

So we'd like to find leg C. But we can't use Law of Cosines yet so we will use Law of Sines. We can easily find the length of A and this is ##\sqrt{13}##. With some geometry we can see that ##\angle a = 53.1##. We can now use Law of sines. $$\frac{\sin(a)}{A} = \frac{\sin(b)}{B}$$ We want to...

Ok thank you kindly!

For reference, these points of intersection are all solutions to the equation (in other words, when my two vectors are perpendicular). But am I correct if I say that there is no explicit way to state every single solution here?

If we look at the trigonometric equation in a vacuum, I can see how we can say that the solutions tend to ##n\pi##. I also can see that there are an infinite amount of solutions. But the original question is to find when the vectors are perpendicular (that is, when is cos = 0). So it seems I...

Ok now I definitely can see that. So then would it be satisfactory to say that, with respect to the original problem, ##\cos(\theta) = 0## when t = 0 and when ##t = (n + \frac{1}{2})\pi## where n is an integer?

Im not seeing that at all. For example at n = 1, we should have ##\frac{3\pi}{2}## its a little over ##\frac{\pi}{4}## as I'm looking at Desmos for better accuracy.

Yes indeed, I was trying to graph one function instead of two separate ones not sure why. I've got the sketch now though. Alright I think there are infinite solutions to ##\tan(t) = \sqrt{\frac{1}{t}}## and I see this with the sketch. Now, these values are not pretty so I'm not sure how to...

Ok so spent some time trying to graph this thing and wasnt having much luck. Went to Desmos to have a look and it's an absolute nightmare of a graph. Also tried to express ##\tan(t) = \sqrt{\frac{1}{t}}## a different way but of course that gave me the same thing in Desmos. This is assuming...

Wouldn't I run into problems if I did the following to get tan(t)? : $$t\sin^2(t) = \cos^2(t)$$ $$t\tan^2(t) = 1$$ $$\tan(t) = \sqrt{\frac{1}{t}}$$ The problems I see with this are dividing by ##\cos^2(t)## as it could be zero. Also, t in the denominator on the final line could be zero as well.

Ok wow I guess I should rephrase my question then, since I thought this was going to be a simplish trig equation. This trig equation is a part of a bigger problem where I am trying to find out what values of t make a position vector and an acceleration vector perpendicular. I've attached a pic...

You're talkinga bout the variable t preceding the second term, correct?

I'd like to solve ##0 = \cos^2(t) - t\sin^2(t)## but it's been forever since I've done some trig and I'm real rusty. I've tried rewriting terms using identities such as ##\sin^2(t) = 1 - \cos^2(t)## but not getting anything helpful. Can I get a point in the right direction?

Thanks Mark that is more clear! Going to keep plugging at this to see what I can get

Ok so ##\vec{T}(t)##, ##\vec{N}(t)##, and ##\vec{a}_{v\perp}## arent ever mentioned in any previous problem for this homework, but in the text they are given as: ##\vec{T}(t) = \frac{\vec{v}(t)}{v(t)}## so a unit vector in the direction of the velocity vector. ##\vec{N}(t) =...