Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 5 down vote accepted

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.

share|cite|improve this answer

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

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 $\sqrt {13}$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.