# The problems is stated like this:

A particle moves along the curve of intersection of the cylinders y=-x2 and z=x2 in the direction in which x increases. At the instant when the particle is at the point (1,-1,1), its speed is 9cm/s and that speed is increasing at a rate of 3 cm/s2. Find the velocity and acceleration of the particle at that instant.

What I did:
I said, let x=u to parametrize the curve
thus I got r(t)=ui+(-u2)j+(u2)k
as x=u
y=-x2=-u2
z=x2=u2

ds/dt=9cm/s
d/dt(ds/dt)=3cm/s2

thus,
v(t)=1i-2uj+2uk

however when I take the magnitude of v and input P (1,-1,1), I obtain something that is note equal to the speed given in the problem. I am not sure what I am doing wrong.

Thank you!

Your vector $v(t)$ describes what the velocity of the particle would be (in cm/s) if its $x$-coordinate were increasing at the rate of $1$ cm/s.

The problem statement never says how fast the $x$-coordinate of the particle should be increasing as the particle passes the point $(1,-1,1).$ The particle's $x$-coordinate may be increasing slower or faster than $1$ cm/s.

What will be true of the particle in the question is that its velocity vector will be in the same direction as your vector $v(t),$ since the particle is required to follow the same curve regardless of how fast its $x$-coordinate might increase. So you now have the direction (from your calculation) and magnitude (from the problem statement) of the velocity vector of the particle. From those you should be able to find correct components.

Update: Let's try to look at this a slightly different way.

When you decided to let $x=u,$ you defined $u$ as just another name for $x.$ Therefore you don't need $u.$ You already have a perfectly good name for $x,$ namely the symbol $x$ itself. So use it.

Using the symbol $x$ when you mean $x,$ you found that $$\newcommand{i}{\mathbf i}\newcommand{j}{\mathbf j}\newcommand{k}{\mathbf k} \newcommand{r}{\mathbf r} \r(t) = x \i - x^2 \j + x^2 \k.$$

If you differentiate with respect to $x$ (which is what you did), you find that $$\frac{d}{dx} \r(t) = \i - 2x \j + 2x \k.$$

If you evaluate this at $x=1,$ and take the magnitude, you should get $3,$ because that is the correct value of $\left\lVert \frac{d}{dx} \r(t)\right\rVert$ when $x=1.$ (At least, I hope that is what you got--you did not say.) So now you have \begin{align} \left\lVert \frac{d}{dx} \r(t)\right\rVert &= 3 && \text{what you calculated}\\ \left\lVert \frac{d}{dt} \r(t)\right\rVert &= 9 && \text{given in the problem statement}\\ \frac{d}{dx} \r(t) &\neq \frac{d}{dt} \r(t) && \text{the logical conclusion} \end{align}

The last fact should be no surprise at all, because you never had any information to indicate that $x$ and $t$ are just different names for the same variable, and therefore there was never any reason to believe that taking a derivative $\frac{d}{dx}$ would give you the same result as taking a derivative $\frac{d}{dt}.$

One way (but not the only way!) to get the result you need is to start all over again, but this time take the derivative with respect to $t$ (which is what you need) instead of $x.$ For example, you do not just want to merely do this with the $\i$ component: $$\frac{d}{dx} (x\i) = \left(\frac{dx}{dx}\right) \i = \i.$$

Instead you want $$\frac{d}{dt} (x\i) = \left(\frac{dx}{dt}\right) \i.$$

For the $\j$ component you want $$\frac{d}{dt} (-x^2\j) = \left(\frac{d}{dt} (-x^2)\right) \j = \left(-2x\frac{dx}{dt}\right) \j.$$

You might recognize the use of the Chain Rule in the previous equation: $$\frac{df(x)}{dt} = \frac{df(x)}{dx} \cdot \frac{dx}{dt},$$ with $f(x) = -x^2$ in this case.

It is useful to know that the Chain Rule also applies when $f(t)$ is a vector:

\begin{align} \frac{d}{dt} \mathbf f(x) &= \frac{d}{dt} (f_1(x) \i + f_2(x) \j + f_3(x) \k) \\ &= \frac{df_1(x)}{dt} \i + \frac{df_2(x)}{dt} \j + \frac{df_3(x)}{dt} \k \\ &= \left(\frac{df_1(x)}{dx} \frac{dx}{dt}\right)\i + \left(\frac{df_2(x)}{dx} \frac{dx}{dt}\right)\j + \left(\frac{df_3(x)}{dx} \frac{dx}{dt}\right)\k \\ &= \left(\frac{df_1(x)}{dx} \i + \frac{df_2(x)}{dx} \j + \frac{df_3(x)}{dx} \k\right) \frac{dx}{dt}\\ &= \left(\frac{d}{dx}(f_1(x)\i + f_2(x)\j + f_3(x)\k)\right) \frac{dx}{dt}\\ &= \left(\frac{d}{dx}\mathbf f(x)\right) \frac{dx}{dt}. \end{align}

Therefore the velocity of your particle is $$\frac{d}{dt} \r(t)= \left( \frac{d}{dx} \r(t)\right) \frac{dx}{dt}.$$

This gives you a second way to find the velocity of the particle. This way does not require you to start all over from the beginning. Take the magnitude of both sides, and remember that $dx/dt$ is a scalar: $$\left\lVert \frac{d}{dt} \r(t)\right\rVert = \left\lVert \frac{d}{dx} \r(t)\right\rVert \frac{dx}{dt}.$$

You already know everything in that equation (at the instant when $x=1$) except $dx/dt,$ so you can solve it for $dx/dt.$ Then use the previous equation with all the information you have to find the velocity of the particle.

If we also have to work out the acceleration at this level of detail, I think we will need a new question just to do that.

• I understand that, but shouldn't I find what u is equal to? Or is it only equal to one in the equation for r(u)? Do I have to find u for velocity and acceleration separately? I am very lost :(
– lola
Apr 10, 2017 at 3:08
• You already set $u=x.$ At the point $(1,-1,1),$ clearly $x=1.$ You can't fiddle with $u$ to get an answer. You can try to determine what $dx/dt$ is at that point. You might also want to use a symbol different from $v(t)$ for your constructed vector, because people usually write $v(t)$ to denote the velocity at time $t$ and this particular vector is something different from that. Apr 10, 2017 at 3:11
• So I would have v(u)=1i-2uj+2uk which would be v(1)=1i+2j+2k, but if I take the magnitude of this I don't obtain 9, that's why I am confused.
– lola
Apr 10, 2017 at 3:29