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

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: 

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