# Maximum speed of a particle given velocity function in terms of vectors

Find the maximum speed of a particle whose velocity, $\mathbf v$ m/s at time t seconds is given by: $$v=2\mathbf isin(t)+\mathbf jcos(t)+3\mathbf k, t\ge0$$

How do I solve this? I tried differentiating and equating to zero but I don't know if it's a valid approach here and if it is, how to take it from there.

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What do you mean by $\sin(t\hat i)$? – GoodDeeds Feb 27 at 10:59
There's no such thing as $\sin$ or $\cos$ of a vector. I guess that there's a typo. – Ahmed Hussein Feb 27 at 10:59
Isn't it : $v=2sin(t)\mathbf i+cos(t)\mathbf j+3\mathbf k$ ? – nicomezi Feb 27 at 10:59
They didn't use brackets in the textbook so I assumed that the sine and cosine are of the whole expression i.e. $sin(t\mathbf i)$ when I guess now it should be $sin(t)\mathbf i$. – Richard Smith Feb 27 at 11:02
@nicomezi - Yeah, the notation was a bit ambiguous. – Richard Smith Feb 27 at 11:03

Hint: Speed is the magnitude of the velocity. So if the velocity is $$v(t) =(2\sin t,\cos t, 3)$$ then the speed is $$f(t)=|v(t)|=\sqrt{4\sin^2 t +\cos^2 t + 9}=\sqrt{10+3\sin^2 t}$$ using the facts that $|(a,b,c)|^2=a^2+b^2+c^2$ and $\sin^2 t+\cos^2 t=1$.

Now you have a real-valued function to maximize on $[0,\infty)$, which is a Calculus I problem.

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$$v=2\sin(t)\mathbf i+\cos(t)\mathbf j+3\mathbf k, t\ge0$$

First find the speed, which is the norm of $v$:

$|v(t)| = \sqrt{4\sin^2 t + \cos^2 t + 9} = \sqrt{3\sin^2 t + 10}$

You now need to maximise that. It's not that difficult. First note that the square root is a monotonic, strictly increasing function. Then note that $\sin^2 t$ has a maximum value of $1$. This immediately gives the max. speed as $\sqrt {13}$

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No, you have messed up the derivative.

First of all, you should have written the velocity as $$v=2\sin(t)\mathbf {\hat i}+\cos(t)\mathbf {\hat j}+3\mathbf {\hat k}$$ according to the conventional notation and not $$v=2\sin(t\mathbf i)+\cos(t\mathbf j)+3\mathbf k$$ where you seem to take the $\sin$ or $\cos$ of a vector.

Secondly, the derivative of a vector of the form $$\vec r=a\mathbf {\hat i}+b\mathbf {\hat j}+c\mathbf {\hat k}$$ is generally expressed as $$\frac{d\vec r}{dt}=\frac{da}{dt}\mathbf {\hat i}+\frac{db}{dt}\mathbf {\hat j}+\frac{dc}{dt}\mathbf {\hat k}$$ A vector is not like a constant, you just do not differentiate it.

So as you can now calculate, $$\frac{dv}{dt}=2\cos (t)\mathbf {\hat i}-\sin (t)\mathbf {\hat j}+0\cdot \mathbf {\hat k}$$ or $$\frac{dv}{dt}=2\cos (t)\mathbf {\hat i}-\sin (t)\mathbf {\hat j}$$

Last but not the least, the question asks for the maximum speed and not the maximum velocity.

And speed $= |v| = \sqrt{4\sin^2(t)+\cos^2(t)+3^2} = \sqrt{3\sin^2t+10}$

Now maximise $|v|$ by calculating and equating $\frac{d|v|}{dt}=0$.

Hope this helps you.

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