Answer:
[tex]51.72\text{ cells per hour}[/tex]
Step-by-step explanation:
So, the function, P(t), represents the number of cells after t hours.
This means that the derivative, P'(t), represents the instantaneous rate of change (in cells per hour) at a certain point t.
C)
So, we are given that the quadratic curve of the trend is the function:
[tex]P(t)=6.10t^2-9.28t+16.43[/tex]
To find the instanteous rate of growth at t=5 hours, we must first differentiate the function. So, differentiate with respect to t:
[tex]\frac{d}{dt}[P(t)]=\frac{d}{dt}[6.10t^2-9.28t+16.43][/tex]
Expand:
[tex]P'(t)=\frac{d}{dt}[6.10t^2]+\frac{d}{dt}[-9.28t]+\frac{d}{dt}[16.43][/tex]
Move the constant to the front using the constant multiple rule. The derivative of a constant is 0. So:
[tex]P'(t)=6.10\frac{d}{dt}[t^2]-9.28\frac{d}{dt}[t][/tex]
Differentiate. Use the power rule:
[tex]P'(t)=6.10(2t)-9.28(1)[/tex]
Simplify:
[tex]P'(t)=12.20t-9.28[/tex]
So, to find the instantaneous rate of growth at t=5, substitute 5 into our differentiated function:
[tex]P'(5)=12.20(5)-9.28[/tex]
Multiply:
[tex]P'(5)=61-9.28[/tex]
Subtract:
[tex]P'(5)=51.72[/tex]
This tells us that at exactly t=5, the rate of growth is 51.72 cells per hour.
And we're done!
