Now I want to apologize in advance for not knowing how to use mathematical expressions here and would appreciate if someone is willing to edit this.

A point is moving on a circle with radius of 2 meters and angular velocity of $\frac12 \text{ rad}\cdot s^{-1}$, with starting angle of $\frac{\pi}{4}$.

In which moment $t \in [0;\frac{\pi}{2}]$ cosine of angle between tangent on $(x(t),y(t))$ and line $y= \Biggl( \frac{-1}{3^{1/2}}\Biggr) x + 7$ will be biggest (maximum).

Now coordinates for the point can be found with formulas and we get $$x(t)=2\cos \biggl(\bigl(\frac12\bigr)t + \frac\pi4\biggr)$$ and $$y(t)=2\sin\biggl(\bigl(\frac12\bigr)t + \frac \pi4\biggr).$$

I went and found the derivate of both $x(t)$ and $y(t)$ and we find the tangent $$y'(t)= \cot \biggl(\bigl(\frac12\bigr)t + \frac \pi4\biggr).$$ Now I'm stuck and I don't really know how to set up the function in which we can find maximum of and angle. Thank you in advance for any kind of answer or hint.


First of all, some corrections. The derivative that you found in the end of your post is NOT $y'(t)$ — it's $y'(x)=\frac{dy}{dx}$. And it's slightly wrong — a negative sign is missing. Do you see why?

Now, what is the maximal possible value of cosine? And what value of an angle, $\theta=\ldots$, has this maximal possible value of cosine? That's the angle you want between the two lines (the given line and the tangent line). Then think geometrically: two lines have this angle of $\theta=\ldots$ between them is the same as saying that the lines are $\ldots$ relative to each other.

Your know the slope of the given line, viz. $m_1=-\frac{1}{\sqrt{3}}$. The slope of the tangent line is $m_2=\frac{dy}{dt}=\cot\left(\frac{1}{2}t+\frac{\pi}{4}\right)$. According to the conclusion in the last paragraph, these slopes must be $\ldots$ — that's how you can set up an equation to solve for $t$.


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