I was studying vector velocity, and was looking at the following example. There's one part that doesn't turn out right and I'm pretty sure it's my calculus game. The problem is:

c) The position of another sail boat in function of time, for $t>20.0\ \mathrm{s}$, is given by

$$\begin{align} b_x(t) &= b_1 + b_2 t \\ y(t) &= c_1 + \frac{c_2}{t} \end{align}$$ where

  • $b_1 = 100\ \mathrm{m}$
  • $b_2 = 0.500\ \mathrm{m/s}$
  • $c_1 = 200\ \mathrm{m}$
  • $c_2 = 360\ \mathrm{m}$

Determine velocity in function of time for $t > 20\ \mathrm{s}$.

So I reconstructed the proof of vectorial velocity, where the vectorial velocity is equal to $$\frac{\mathrm{d}x}{\mathrm{d}t}\hat{i} + \frac{\mathrm{d}y}{\mathrm{d}t}\hat{j}$$ with $\hat{i}$, $\hat{j}$ being the unit vectors. I know that $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $\frac{\mathrm{d}y}{\mathrm{d}t}$ are the derivatives of the position, so it's equal to the velocity.

I have the velocity of $\frac{\mathrm{d}x}{\mathrm{d}t}$ which is $0.500\ \mathrm{m}$, but the other one in the solution is $-c_2 t^{-2}$. I tried doing it, but since you have to do the derivatives of numbers, shouldn't it be all 0?

The given solution is: $$v = b_2\hat{i} - c_2 t^{-2}\hat{j} = 0.500\mathrm{\frac{m}{s}}\hat{i} - \frac{360\ \mathrm{m\,s}}{t^2}\hat{j}$$

  • $\begingroup$ Hi InfoB; what do you mean by "since you have to do the derivatives of numbers"? Could you expand on that? $\endgroup$ – David Z Nov 23 '17 at 23:40

If you think that way then $\frac{d}{dx} x^2$ would be also 0 because $x^2$ is any number. Your thoughts are wrong because you misunderstood the issue.

The derivative of a function is the rate of instantenous change of that funciton along the variable. For example, $d/dt$ is the rate of change in time of the function.

If the function depends on "t", the derivative is not 0 (in general). Just calculate it:

$y'(t)=0+ c_2 \frac{d}{dt} \left(\frac{1}{t}\right)= c_2 ·\left( \frac{-1}{t^2}\right)$

Then, you might want to EVALUATE tis derivative function for some concrete values of $t$, but the derivative is also a function of the same variables. You cannot think of variables as just numbers, because they're variying along.

  • $\begingroup$ Woah, true that! I was thinking that all became 0 because i substituted the number in there! Many thanks! $\endgroup$ – InfoB Nov 23 '17 at 23:53

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