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APPLICATIONS OF CALCULUS 1
APPLICATIONS OF CALCULUS (SOLUTIONS)
1 The Point O is the intersection of two roads that cross at right angles as shown. One car travels towards O from the north at 20ms−1 while the second travels due east towards O also at 20ms−1 .
(a) Show that after t seconds their distance apart, d, is given by
d = 100− 20t( )2 + 80− 20t( )2
(b) Show that this simplifies to
d2 = 400 5 − t( )2 + 4 − t( )2⎡⎣ ⎤⎦
𝑑
𝑑! = 20(5− 𝑡) ! + 20(4− 𝑡) ! = 400 5− 𝑡 ! + 400 4− 𝑡 ! = 400[ 5− 𝑡 ! + 4− 𝑡 !]
O 100 𝑚
20 𝑚𝑠!!
20 𝑚𝑠!!
80 𝑚
𝑑
100 − 20𝑡
80−20𝑡
Find distance of each from O at time t then use Pythagoras’ for distance between them.
APPLICATIONS OF CALCULUS 2
(c) Show, using calculus that the minimum distance between the two cars is 𝟏𝟎 𝟐 𝒎. 𝐷! = 400[ 5− 𝑡 ! + 4− 𝑡 !]
Minimum distance at !(!
!)!"
= 0 !(!
!)!"
= 400[2 5− 𝑡 −1 + 2 4− 𝑡 −1 ] 400[2 5− 𝑡 −1 + 2 4− 𝑡 −1 ] = 0 800𝑡 − 4000− 3200+ 800𝑡 = 0 1600𝑡 = 7200 𝑡 = !"##
!"##= !
!
Minimum distance between them is at time 𝑡 = 4.5 𝑠𝑒𝑐𝑠 𝐷! = 400[ 0.5 ! + (−0.5)! ] = 400(0.25+ 0.25) = 200 𝑫 = 𝟏𝟎 𝟐 𝒎 (d) Now show, without using calculus, that the minimum distance between the two cars is 𝟏𝟎 𝟐 𝒎 (Using relative velocity)
APPLICATIONS OF CALCULUS 3
2 Two straight roads cross at right angles at O. John is running along one of the roads towards O at 𝟕 𝒎𝒔!𝟏 . Mary is cycling along the other road at 𝟐𝟒 𝒎𝒔!𝟏 . When John is 100 𝒎 from O, Mary is at O. (𝒊) Find an expression for the distance, D, between John and Mary. Distance of John from junction is 100− 7𝑡 Distance of Mary from junction is 24𝑡 𝐷! = 100− 7𝑡 ! + (24𝑡)! 𝒊𝒊 Hence find the shortest distance between John and Mary at any time 𝒕. Shortest distance between them is at !(!
!)!"
= 0 !(!
!)!"
= 2 100− 7𝑡 −7 + (2)576𝑡 2 100− 7𝑡 −7 + 1152𝑡 = 0 98𝑡 − 1400+ 1152𝑡 = 0 1250𝑡 = 1400
𝒕 =𝟐𝟖𝟐𝟓
𝐷! = 100− 7𝑡 ! + (24𝑡)!
= !"#$!"
!+ !"#
!!
!
𝑫 = 𝟗𝟔 𝒎
𝑀𝑎𝑟𝑦
𝐽𝑜ℎ𝑛 O 100 𝑚
24 𝑚𝑠!!
7 𝑚𝑠!!
APPLICATIONS OF CALCULUS 4
1987 QUESTION 2b: AN ALTERNATIVE TO USING 𝑽𝑨 = 𝑽𝑩 A car starts from a point O with an initial speed of 8 𝒎𝒔!𝟏 and then travels with a uniform acceleration of 4 𝒎𝒔!𝟐. Two seconds later a second car Q starts with an initial velocity of 30 𝒎𝒔!𝟏 and then moves with a uniform acceleration of 𝟑 𝒎𝒔!𝟐. Show that after passing P, Q will never be ahead by more than 74m. 𝐢 Find an expression for the distance travelled by car P at time 𝐭 seconds 𝑆! = 𝑢 𝑡 + !
!(𝑎)𝑡! 𝑢 = 8 𝑎 = 4 𝑡 = 𝑡 + 2 𝑠 = 𝑠!
𝑆! = 8 𝑡 + 2 + !
!(4) 𝑡 + 2 !
= 8𝑡 + 16+ 2𝑡! + 8𝑡 + 8 𝑆! = 2𝑡! + 16𝑡 + 24 𝐢𝐢 Find an expression for the distance travelled by car Q at time 𝐭 seconds 𝑆! = 𝑢 𝑡 + !
!(𝑎)𝑡! 𝑢 = 30 𝑎 = 3 𝑡 = 𝑡 𝑠 = 𝑠!
𝑆! = 30 𝑡 + !
!(3) 𝑡 !
𝑆! = 30 𝑡 + !
!𝑡 !
(𝐢𝐢𝐢) Find an expression for the distance between the two cars Let 𝐷 = 𝑆! − 𝑆! 𝑆! − 𝑆! = 30 𝑡 + !
!𝑡 ! − 2𝑡! + 16𝑡 + 24
𝐷 = − !
!𝑡 ! + 14𝑡 − 24
𝐢𝐯 Find the time when distance between them is minimum Min distance between them at !"
!"= 0 !"
!"= −𝑡 + 14
𝑀𝑖𝑛𝑖𝑚𝑖𝑢𝑚 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑡 − 𝑡 + 14 = 0 𝑡 = 14 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
APPLICATIONS OF CALCULUS 5
(𝒗) Find minimum distance between them 𝐷 = − !
!𝑡 ! + 14𝑡 − 24
𝑎𝑡 𝑡 = 14 𝐷 = − !
!14 ! + 14(14)− 24
= −98+ 196− 24 = 𝟕𝟒 𝒎
APPLICATIONS OF CALCULUS 6
1992 QUESTION 1b Two particles P and Q are moving in the same direction along parallel straight lines. Their accelerations are 5 m/s2 and 4 m/s2, respectively. At a certain instant P has a velocity 1 m/s and Q is 25.5 m behind P moving with velocity 11 m/s. (i) Prove that Q will overtake P and that P will in turn overtake Q. (ii) When Q is in front of P find the greatest distance between the particles. Find expression for distance travelled by P 𝑠 = 𝑢𝑡 + !
!𝑎𝑡! 𝑢 = 1 𝑎 = 5
𝑆! = (1)𝑡 + !!(5)𝑡!
Find expression for distance travelled by Q 𝑠 = 𝑢𝑡 + !
!𝑎𝑡! 𝑢 = 11 𝑎 = 4
𝑆! = (11)𝑡 + !!(4)𝑡!
Find an expression for the distance between the particles when Q is in front of P
Let distance between them equal D 𝐷 = 𝑆! − 𝑆! − 25.5 𝐷 = 11𝑡 + 2𝑡! − 𝑡 − 2.5 𝑡! − 25.5 = −0.5𝑡! + 10𝑡 − 25.5 Find the maximum distance between them Maximum distance at !"
!"= 0
!"
!"= −𝑡 + 10
−𝑡 + 10 = 0 𝒕 = 𝟏𝟎 Maximum distance 𝐷 = −0.5 𝟏𝟎 ! + 10 𝟏𝟎 − 25.5 = 𝟐𝟒.𝟓
Q P 25.5 𝑚 𝑆!
𝑆!
𝑫 = 𝑺𝑸 − 𝑺𝑷 − 𝟐𝟓.𝟓
APPLICATIONS OF CALCULUS 7
1997 QUESTION 3b A particle is projected from a point p with initial speed 15m/s, down a plane inclined at an angle of 300 to the horizontal. The direction of projection is at right angles to the inclined plane. (The plane of projection is vertical and contains the line of greatest slope). Find (i) the perpendicular height of the particle above the plane after t seconds and hence, or otherwise, show that the vertical height ℎ of the particle above the plane after t seconds is 10 3 t – 4.9t2 (ii) the greatest vertical height it attains above the plane (i.e. the maximum value of h) correct to two places of decimals.
ℎ = 10 3𝑡 − 4.9𝑡! Maximum height at !!
!"= 0
!!
!"= 10 3− 9.8𝑡
10 3− 9.8𝑡 = 0 𝑡 = !" !
!.!
Maximum height = 10 3 !" !!.!
− 4.9 !" !!.!
!
= !""
!.!− !"#
!.!
= 𝟏𝟓.𝟑 𝒎
APPLICATIONS OF CALCULUS 8
