In a candy factory, nutty chocolate bars contain 22.0% by mass pecans. If 4.5 kg of pecans were used for candy last Tuesday, how
1 answer:
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Answer:
A ( t ) = -4t + 51
B ( t ) = -2t + 29
t < 11 hours ... [ 0 , 11 ]
Step-by-step explanation:
Given:-
- The height of snowman A, Ao = 51 in
- The height of snowman B, Bo = 29 in
Solution:-
- The day Mason made two snowmans ( A and B ) with their respective heights ( A(t) and B(t) ) will be considered as the initial value of the following ordinary differential equation.
- To construct two first order Linear ODEs we will consider the rate of change in heights of each snowman from the following day.
- The rate of change of snowman A's height ( A ) is:
- The rate of change of snowman B's height ( B ) is:
Where,
t: The time in hours from the start of melting process.
- We will separate the variables and integrate both of the ODEs as follows:
- Evaluate the constant of integration ( c ) for each solution to ODE using the initial values given: A ( 0 ) = Ao = 51 in and B ( 0 ) = Bo = 29 in:
- The solution to the differential equations are as follows:
A ( t ) = -4t + 51
B ( t ) = -2t + 29
- To determine the time domain over which the snowman A height A ( t ) is greater than snowman B height B ( t ). We will set up an inequality as follows:
A ( t ) > B ( t )
-4t + 51 > -2t + 29
2t < 22
t < 11 hours
- The time domain over which snowman A' height is greater than snowman B' height is given by the following notation:
Answer: [ 0 , 11 ]