A person’s blood pressure, P(t), in millimeters of mercury (mm Hg), is modeled by the function P(t)=100+20cos(8π/3t), where t is the time in seconds. Estimate the instantaneous rate of change...

A person’s blood pressure, P(t), in millimeters of mercury (mm Hg), is modeled by the function P(t)=100+20cos(8π/3t), where t is the time in seconds.
Estimate the instantaneous rate of change in a person’s blood pressure at t=0.5. Do not use a derivative. Use the formula : 
( f(a-0.001)-f(a) ) / 0.001

Asked on by lskajf

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llltkl | College Teacher | (Level 3) Valedictorian

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The instantaneous rate of change at any point where x=a, can be numerically obtained by determining the slope of a typical secant to the curve y=f(x) centered around the point x=a. It is obtainable by determining the values of function f(x) in two closely spaced instants (a+h) and (a-h) and dividing by the interval, i.e. 2h. The formula is:

Rate of instantaneous change=(f(a+h)~f(a+h))/(2h)

The smaller the value of h, more accurate is the estimate of the slope and hence of the instantaneous rate of change.

If you take f(a-h) and f(a) instead, you are likely to get the instantaneous rate of change around the point (a-h/2) and not a. That is the reason I am going to use the earlier formula for this calculation.

A person’s blood pressure P(t), in mm of Hg, is modelled by the function:

`P(t)= 100+20cos((8pi)/(3t))`

The instantaneous rate of change of the person’s blood pressure at time t=0.5 s has to be determined.

`P(0.500+0.001)= 100+20cos((8pi)/(3*(0.500+0.001)))=(100-10.5736)`

`P(0.500-0.001)= 100+20cos((8pi)/(3*(0.500-0.001)))=(100-9.41289)`

The instantaneous rate of change of the person’s blood pressure at time t=0.5 s:

`= (P(0.500-0.001)- P(0.500+0.001))/(2*0.001)=580.33`

Make the interval finer by one decimal point and instantaneous rate of change of the person’s blood pressure becomes 580.415 which is very close to the actual derivative (dy/dx) value of 580.416.

Sources:

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