Vector Functions of One Variable

Question

Question

A particle moves along the curve of the intersection of the cylinders $y=-x^2$ and $z=x^2$ in the direction in which $x$ increases. (All distances are in cm.) At the instant when the particle is at the point $(1,\,-1,\,1)$ its speed is $9$ cm/s, and the speed is increasing at a rate of $3$ cm/s$^2$. Find the velocity and acceleration of particle at that instant.

Solution (Well at least my attempt at one)

$\mathbf{r}(t) = x\mathbf{i} - x^2\mathbf{j} + x^2\mathbf{k}$

$\mathbf{v}(t) = \left(\mathbf{i} - 2x\mathbf{j} + 2x\mathbf{k}\right)\dfrac{dx}{dt}$

\begin{align*} \implies ||\mathbf{v}|| &=\left(\sqrt{1^2 +4x^2+4x^2}\right) \left|\dfrac{d x}{dt}\right|\\ &=(\sqrt{1+8x^2})\dfrac{dx}{dt} \end{align*}

$|\frac{dx}{dt}|=\frac{dx}{dt}$ since the particle is always increasing

$\implies$ at the point $(1,\,-1,\,1),\,\, ||\mathbf{v}|| = 9$, so $dx/dt = 3$ and the velocity at that point is $\mathbf{v}(t) = 3\mathbf{i} - 6\mathbf{j} + 6\mathbf{k}$

Now acceleration is where I have a bit of trouble

$\mathbf{a}(t) =\dfrac{d^2x}{dt^2}(\mathbf{i} - 2x\mathbf{j} + 2x\mathbf{k}) + (\dfrac{dx}{dt})^2(- 2\mathbf{j} + 2\mathbf{k})$

But I don't exactly know what to do from here any help would be appreciated

Answer 1

Notice that, at the given instant: $x=1$ cm and \begin{align*} \frac{d^2x}{dt^2}&=\frac{d}{dt}\left(\frac{dx}{dt}\right)\\ &=\frac{d}{dt}\left(\frac{v}{\sqrt{1+8x^2}}\right)\qquad\text{where }v=||\mathbf{v}||\\ &=\frac{\sqrt{1+8x^2}\dfrac{dv}{dt}-\frac{8x}{\sqrt{1+8x^2}}\frac{dx}{dt}v}{1+8x^2} \\ &=\frac{\sqrt{1+8(1)^2}(3)-\dfrac{8(1)}{\sqrt{1+8(1)^2}}(3)(9)}{1+8(1)^2}\text{ cm/s}^2\\ &=\frac{9-72}{9}\text{ cm/s}^2\\ \frac{d^2x}{dt^2}&=-7\text{ cm/s}^2 \end{align*} Hence, at the given instant, \begin{align*} \mathbf{a}&=\left(-7 \text{ cm/s}^2\right)(\mathbf{i}-2\mathbf{j}+2\mathbf{k} \text{ cm/s})+(9 \text{ cm/s})^2(-2\mathbf{j}+2\mathbf{k}\text{ cm}^{-1})\\ &=-7\mathbf{i}-4\mathbf{j}+4\mathbf{k}\quad\text{cm/s}^2 \end{align*}

Answer 2

As I mentioned in my comment above, you can figure out $\dfrac{d^2x}{dt^2}$ from your equation in $|\mathbf{v}|$, given that you know that the change of speed is 3cm/s/s.

\begin{eqnarray} |\mathbf{v}| &=& (\sqrt{1+8x^2})\dfrac{dx}{dt}\\ \dfrac{d|\mathbf{v}|}{dt} &=& (\sqrt{1+8^2})\dfrac{d^2x}{dt^2} + \frac{8x}{\sqrt{1+8x^2}}\left(\dfrac{dx}{dt}\right)^2 = 3 \\ \dfrac{d^2x}{dt^2} &=& -7 \end{eqnarray}

Plugging this into your equation for $\mathbf{a}$ gives \begin{equation}\mathbf{a} = -7\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\end{equation} and it can easily be checked that $|\mathbf{a}|=3$