Differentiating vectors

In a nutshell

Calculus can be used with vectors to solve problems involving 2D motion with variable acceleration. The general form of a displacement vector is $\bold r = f(t) \bold i + g(t) \bold j$ , and you can differentiate both functions separately.

Equations

Description

Equations

Expression of the velocity given that $\bold r = x\bold i + y \bold j$ .

$\bold v = \cfrac{d \bold r}{dt} = \bold{\dot r} = { \dot x}\bold i + {\dot y}\bold j$

Expression of the acceleration given that $\bold r = x \bold i + y \bold j$ .

$\bold a = \cfrac{d^2 \bold r}{dt^2} = \bold{\ddot r} = { \ddot x}\bold i + {\ddot y}\bold j$

$\bold a = \cfrac{d \bold v}{dt} = \bold{\dot v} = { \dot v_x}\bold i + {\dot v_y}\bold j$

Differentiating vectors

Example 1

A body has a position vector $\bold r = \left[(3t^3 -11 t^2) \bold i + (-4t + 10 t^2)\bold j\right] \ m$ with respect to a fixed origin.

i) Find the speed at $t = 10 \ s$ to two decimal places.

ii) Calculate the acceleration at that time.

i) Find the speed at $t= 10 \ s$ to two decimal places.

Using $\bold v = \cfrac{d\bold r}{dt} = \bold{\dot r}$ gives:

$\begin{aligned}\bold v &= \cfrac{d\bold r}{dt} =\\&= \cfrac{d}{dt}\left[(3t^3 - 11 t^2) \bold i + (-4t +10t^2)\bold j \right]=\\&=(9t^2 - 22 t)\bold i + (-4+20t)\bold j \end{aligned}$ 

Substituting $t= 10 \ s$ :

$\bold v = (680 \bold i + 196 \bold j) \ ms^{-1}$ 

Now, the speed is given by the formula $Speed = \sqrt{v_x^2 + v_y ^2}$ . Therefore:

$Speed = \sqrt{680^2 + 196 ^2} = \underline{707.68 \ ms^{-1} \ (2 \ d.p.) }$ 

ii) Calculate the acceleration.

Using $\bold a = \cfrac{d\bold v}{dt} = \bold{\dot v}$ you can write:

$\begin{aligned}\bold a &= \cfrac{d\bold v}{dt} =\\&= \cfrac{d}{dt}\left[(9t^2 - 22 t) \bold i + (-4+20t) \bold j \right]=\\&= (18 t -22) \bold i + 20 \bold j \ ms^{-2}\end{aligned}$ 

Therefore, at that time ( $t=10\ s$ ) you have:

$\underline{\bold a = (158 \bold i + 20 \bold j) \ ms^{-2}}$

Example 2

A body with a mass $m = 10 \ kg$ is acted on by a force $P \ N$ . Relative to the origin, its position vector is $\bold r = \left(4t^2 \bold i + 10 t^{\frac32} \bold j\right) \ m.$ 

i) Find the speed of this body when $t = 10 \ s$ to one decimal place.

ii) Find the acceleration when $t=3\ s.$

iii) Calculate the force $P \ N$ acting on the body at that time.

i) Find the speed of this body when $t=10\ s.$

First find the velocity:

$\begin{aligned}\bold v &= \cfrac{d\bold r}{dt} =\\&= \cfrac{d}{dt}\left(4t^2 \bold i + 10 t^{\frac32} \bold j\right)=\\&=8t \bold i +15 \sqrt t \bold j \end{aligned}$

Substitute in $t=10 \ s$ :

$\bold v = (80 \bold i + 15 \sqrt{10} \bold j) m.s^{-1}$ 

Therefore,use $Speed = \sqrt{v_x^2 + v_y^2}$ to give:

$Speed = \sqrt{80^2 +\left(15\sqrt{10}\right)^2}= \underline{93.0\ ms^{-1} \ (1 \ d.p.)}$ 

ii) Find the acceleration when $t=3\ s.$ 

Use $\bold a = \cfrac{d \bold v}{dt} = \bold {\dot v}$ to give:

$\begin{aligned}\bold a &= \cfrac{d \bold v}{dt} =\\&= \cfrac{d}{dt}(8t \bold i +15t^\frac12 \bold j)=\\& = \left(8 \bold i + \cfrac{15}{2\sqrt{t}} \ \ \bold j\right) \ ms^{-2}\end{aligned}$ 

Substituting $t = 3 \ s .$ 

$\underline{\bold a = \left( 8 \bold i + \cfrac{15}{2\sqrt3}\ \ \bold j\right) \ ms^{-2} }$ 

iii) Calculate the force acting on the body at that time.

Use $\bold F = m \times \bold a$ to give:

$\bold F = m \times \bold a = \underline{\left(80 \bold i +25 \sqrt{3} \ \bold j\right) \ N }$ 