Vectors, Calculus and a Particle $P$

The displacement of $P$ at a time $t$ seconds is described by the vector r .

To find the direction that $P$ is travelling in at any given time, we must find the vector describing the velocity of $P$ at time $t$ seconds. Let's call this v .

Now here's the part where calculus comes in. v = $\frac{d}{dt}$ r . To differentiate r , we must differentiate the i , j and k components of r separately.

$\therefore$ v $=(4\phi t+4)$ i $+(7-10t)$ j $+(64-60\mu t^2)$ k

Now let's find the value of $t$ when $t^2+4t-5=0$ .

Factorise and solve for $t$ as follows:

$(t-1)(t+5)=0 \implies t=1, -5$

But $t>=0, \therefore t=1$

Now $t=1 \implies$ v $=(4\phi +4)$ i $+(-3)$ j $+(64-60\mu)$ k

We also know that at $t=1$ , v is parallel to the vector $6$ j $-68$ k (call this a ), so at this time v = $\lambda$ a where $\lambda$ is some non-zero scalar.

To find $\lambda$ , we can write an equation: $(4\phi +4)$ i $+(-3)$ j $+(64-60\mu)$ k $=$ $(6\lambda)$ j $-(68 \lambda)$ k

Comparing coefficients in j , we find $-3=6\lambda \implies \lambda=-\frac{1}{2}$

Applying this to coefficients in k , we can write $64-60\mu=-68(-\frac{1}{2}) \implies 64-60\mu=34 \implies \mu= \frac{1}{2}=0.5$ .

Finally, comparing coefficients in i , we see that $4\phi +4=0 \implies \phi=-1$

$\therefore \mu- \phi =0.5-(-1)=0.5+1=1.5$ .