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 r = f ( t ) i + g ( t ) j \bold r = f(t) \bold i + g(t) \bold j r = f ( t ) i + g ( t ) j
Equations Description Equations Expression of the velocity given that r = x i + y j \bold r = x\bold i + y \bold j r = x i + y j
v = d r d t = r ˙ = x ˙ i + y ˙ j \bold v = \cfrac{d \bold r}{dt} = \bold{\dot r} = { \dot x}\bold i + {\dot y}\bold j v = d t d r = r ˙ = x ˙ i + y ˙ j
Expression of the acceleration given that r = x i + y j \bold r = x \bold i + y \bold j r = x i + y j
a = d 2 r d t 2 = r ¨ = x ¨ i + y ¨ j \bold a = \cfrac{d^2 \bold r}{dt^2} = \bold{\ddot r} = { \ddot x}\bold i + {\ddot y}\bold j a = d t 2 d 2 r = r ¨ = x ¨ i + y ¨ j
a = d v d t = v ˙ = v ˙ x i + v ˙ y j \bold a = \cfrac{d \bold v}{dt} = \bold{\dot v} = { \dot v_x}\bold i + {\dot v_y}\bold j a = d t d v = v ˙ = v ˙ x i + v ˙ y j
Differentiating vectors Example 1 A body has a position vector r = [ ( 3 t 3 − 11 t 2 ) i + ( − 4 t + 10 t 2 ) j ] m \bold r = \left[(3t^3 -11 t^2) \bold i + (-4t + 10 t^2)\bold j\right] \ m r = [ ( 3 t 3 − 11 t 2 ) i + ( − 4 t + 10 t 2 ) j ] m
i) Find the speed at t = 10 s t = 10 \ s t = 10 s
ii) Calculate the acceleration at that time.
i) Find the speed at t = 10 s t= 10 \ s t = 10 s
Using v = d r d t = r ˙ \bold v = \cfrac{d\bold r}{dt} = \bold{\dot r} v = d t d r = r ˙
v = d r d t = = d d t [ ( 3 t 3 − 11 t 2 ) i + ( − 4 t + 10 t 2 ) j ] = = ( 9 t 2 − 22 t ) i + ( − 4 + 20 t ) j \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} v = d t d r = = d t d [ ( 3 t 3 − 11 t 2 ) i + ( − 4 t + 10 t 2 ) j ] = = ( 9 t 2 − 22 t ) i + ( − 4 + 20 t ) j
Substituting t = 10 s t= 10 \ s t = 10 s
v = ( 680 i + 196 j ) m s − 1 \bold v = (680 \bold i + 196 \bold j) \ ms^{-1} v = ( 680 i + 196 j ) m s − 1
Now, the speed is given by the formula S p e e d = v x 2 + v y 2 Speed = \sqrt{v_x^2 + v_y ^2} Sp ee d = v x 2 + v y 2
S p e e d = 68 0 2 + 19 6 2 = 707.68 m s − 1 ( 2 d . p . ) ‾ Speed = \sqrt{680^2 + 196 ^2} = \underline{707.68 \ ms^{-1} \ (2 \ d.p.) } Sp ee d = 68 0 2 + 19 6 2 = 707.68 m s − 1 ( 2 d . p . )
ii) Calculate the acceleration.
Using a = d v d t = v ˙ \bold a = \cfrac{d\bold v}{dt} = \bold{\dot v} a = d t d v = v ˙
a = d v d t = = d d t [ ( 9 t 2 − 22 t ) i + ( − 4 + 20 t ) j ] = = ( 18 t − 22 ) i + 20 j m s − 2 \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} a = d t d v = = d t d [ ( 9 t 2 − 22 t ) i + ( − 4 + 20 t ) j ] = = ( 18 t − 22 ) i + 20 j m s − 2
Therefore, at that time (t = 10 s t=10\ s t = 10 s
a = ( 158 i + 20 j ) m s − 2 ‾ \underline{\bold a = (158 \bold i + 20 \bold j) \ ms^{-2}} a = ( 158 i + 20 j ) m s − 2
Example 2 A body with a mass m = 10 k g m = 10 \ kg m = 10 k g P N P \ N P N r = ( 4 t 2 i + 10 t 3 2 j ) m . \bold r = \left(4t^2 \bold i + 10 t^{\frac32} \bold j\right) \ m. r = ( 4 t 2 i + 10 t 2 3 j ) m .
i) Find the speed of this body when t = 10 s t = 10 \ s t = 10 s to one decimal place.
ii) Find the acceleration when t = 3 s . t=3\ s. t = 3 s .
iii) Calculate the force P N P \ N P N
i) Find the speed of this body when t = 10 s . t=10\ s. t = 10 s .
First find the velocity:
v = d r d t = = d d t ( 4 t 2 i + 10 t 3 2 j ) = = 8 t i + 15 t j \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} v = d t d r = = d t d ( 4 t 2 i + 10 t 2 3 j ) = = 8 t i + 15 t j
Substitute in t = 10 s t=10 \ s t = 10 s
v = ( 80 i + 15 10 j ) m . s − 1 \bold v = (80 \bold i + 15 \sqrt{10} \bold j) m.s^{-1} v = ( 80 i + 15 10 j ) m . s − 1
Therefore,use S p e e d = v x 2 + v y 2 Speed = \sqrt{v_x^2 + v_y^2} Sp ee d = v x 2 + v y 2
S p e e d = 8 0 2 + ( 15 10 ) 2 = 93.0 m s − 1 ( 1 d . p . ) ‾ Speed = \sqrt{80^2 +\left(15\sqrt{10}\right)^2}= \underline{93.0\ ms^{-1} \ (1 \ d.p.)} Sp ee d = 8 0 2 + ( 15 10 ) 2 = 93.0 m s − 1 ( 1 d . p . )
ii) Find the acceleration when t = 3 s . t=3\ s. t = 3 s .
Use a = d v d t = v ˙ \bold a = \cfrac{d \bold v}{dt} = \bold {\dot v} a = d t d v = v ˙ to give:
a = d v d t = = d d t ( 8 t i + 15 t 1 2 j ) = = ( 8 i + 15 2 t j ) m s − 2 \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} a = d t d v = = d t d ( 8 t i + 15 t 2 1 j ) = = ( 8 i + 2 t 15 j ) m s − 2
Substituting t = 3 s . t = 3 \ s . t = 3 s .
a = ( 8 i + 15 2 3 j ) m s − 2 ‾ \underline{\bold a = \left( 8 \bold i + \cfrac{15}{2\sqrt3}\ \ \bold j\right) \ ms^{-2} } a = ( 8 i + 2 3 15 j ) m s − 2
iii) Calculate the force acting on the body at that time.
Use F = m × a \bold F = m \times \bold a F = m × a
F = m × a = ( 80 i + 25 3 j ) N ‾ \bold F = m \times \bold a = \underline{\left(80 \bold i +25 \sqrt{3} \ \bold j\right) \ N } F = m × a = ( 80 i + 25 3 j ) N
