Calculus Practice Problems with Solutions

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Question 1
a.
c ( t ) = t
16 t2 +10 t+63
(i) Rate of change of nitrate
Using quotient rule;
c ' ( t ) =(16 t¿ ¿2+10 t +63) ( 1 ) −t(32t−10)
¿ ¿ ¿
c ' ( t ) =16 t2 +10 t+63−32 t2−10 t
¿ ¿
c ' ( t ) =−16 t2 +63
¿ ¿
(ii) Concentration after 2 hrs
Replacing in the c’(t)
Concentration will be
c' (2)=−16 x 22 +63
¿ ¿
= -4.63-5
(iii) graphical representation using the following values
Time Concentration (x10-3)
2 -0.0463
4 -1.50
6 -1.05
8 -0.706
10 -0.495
12 -0.362

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b.
Given position equation as s(t) = t3-6t2+ 9t
The equation for velocity is s’(t) =3t2 – 12t
The equation for acceleration is s’’(t) = 6t -12
(i) position, velocity and acceleration at t=0.5s
Position = 0.53-6(0.5)2+9(0.5) = 3.125
Velocity= 3(0.52) -12(0.5) = -5.25
Acceleration = 6(0.5) -12 = -9