# Calculus

Topics: Derivative, Velocity, Slope Pages: 11 (2150 words) Published: May 15, 2013
1. ht= -4.9t2+ 450, where t is the time elapsed in seconds and h is the height in metres.

a) Table of Values

t(s)| h(t) (m)|
0| ht= -4.9(0)2+ 450= 450|
1| ht= -4.9(1)2+ 450= 445.1|
2| ht= -4.9(2)2+ 450= 430.4|
3| ht= -4.9(3)2+ 450= 405.9|
4| ht= -4.9(4)2+ 450=371.6|
5| ht= -4.9(5)2+ 450=327.5|
6| ht= -4.9(6)2+ 450= 273.6|
7| ht= -4.9(7)2+ 450= 209.9|
8| ht= -4.9(8)2+ 450= 136.4|
9| ht= -4.9(9)2+ 450=53.1|
10| ht= -4.9(10)2+ 450= -40|

b) Average velocity for the first 2 seconds after the ball was dropped=

h2-h02-0 = 430.4-4502-0 = -19.62 = -9.8 m/s

c) Average velocity for the following time intervals:

i. 1≤t≤4 = h4-h14-1 = 371.6-445.14-1 = -73.53 = -24.5 m/s

ii. 1≤t≤2 = h2-h12-1 = 430.4-445.12-1 = -14.71 = -14.7 m/s

iii. 1≤t≤1.5 = h1.5-h11.5-1 = h1.5=-4.91.52+450-445.11.5-1 = 438.975-445.11.5-1

= -6.1250.5 = -12.25 m/s = -12.3 m/s

d) Instantaneous velocity at t= 1 second = h1-h0.751-0.75 = 445.1- h0.75= -4.90.752+ 4501-0.75 = 445.1-447.21-0.75 = -2.10.25 = -8.4 m/s

FIND GRAPH ON FOLLOWING PAGE

2. M=10.5-0.4t2, where M is the mass in grams and t is the time in seconds.

a) All the sugar has dissolved when M= 0 g
M=10.5-0.4t2
0=10.5-0.4t2
0.4t2=10.5
t2=10.50.4
t2=26.25
t2=√26.25
t=±5.12 s

Since t is time, the negative value cannot be considered and therefore M is 0 g when t= 5.12 s.

b) Average rate of change in the interval 0≤t≤1 = M1-M01-0 = M1=10.5-0.412-M0=10.5-0.4021-0 = 10.1-10.51-0 = -0.41 = -0.4 g/s

c) Table of Values

t(s)| M(t) (g)|
0| M=10.5-0.4(0)2= 10.5|
1| M=10.5-0.4(1)2= 10.1|
2| M=10.5-0.4(2)2= 8.9|
3| M=10.5-0.4(3)2= 6.9|
4| M=10.5-0.4(4)2= 4.1|
5| M=10.5-0.4(5)2= 0.5|
6| M=10.5-0.4(6)2= -3.9|

The instantaneous rate of change at t= 2 seconds = M2-M1.752-1.75 = M2=10.5-0.422-M1.75=10.5-0.41.7522-1.75 = 8.9-9.2752-1.75 = -0.3750.25 = -1.5 g/s FIND GRAPH ON FOLLOWING PAGE

3. dt=2t2, where d is the distance in metres and t is the time in seconds.

a) Table of Values

t(s)| d(t) (m)|
0| d0=2(0)2= 0|
1| d1=2(1)2= 2|
2| d2=2(2)2= 8|
3| d3=2(3)2= 18|
4| d4=2(4)2= 32|
5| d5=2(5)2= 50|
6| d6=2(6)2= 72|
7| d7=2(7)2= 98|
8| d8=2(8)2= 128|
9| d9=2(9)2= 162|

b) The average speed when 4≤t≤7 = d7=272-d4=2427-4

= 98-327-4 = 663
= 22 m/s

c) The instantaneous velocity at t= 4 seconds = d4=242-d3.75=23.7524-3.75 = 32-28.1254-3.75 = 3.8750.25 = 15.5 m/s

FIND GRAPH ON FOLLOWING PAGE

4.
a) The instantaneous rate of change of a function at point (a, fa) can be determined by the formula limh→0fa+h-fah . First, calculate the values of f(a) andf(a+h). Next, substitute the values calculated into the formula and simplify. Once done, let h=0 and substitute into the simplified equation to find the instantaneous rate of change of a function.

Alternatively, the formula fa+h-faa+h-a can be used to determine the instantaneous rate of...