Description Usage Format Details Source Examples
The dataset contains measurements of mean arterial pressure (mmHG) and heart rate (b/min) for a baroreflex curve.
1 |
A data frame with 18 observations on the following 2 variables.
pressure
a numeric vector containing measurements of arterial pressure.
rate
a numeric vector containing measurements of heart rate.
The dataset is an example of an asymmetric dose-response curve, that is not
easily handled using the log-logistic or Weibull models (LL.4
, LL.5
,
W1.4
and W2.4
), whereas the baro5
model provides a nice fit.
Ricketts, J. H. and Head, G. A. (1999) A five-parameter logistic equation for investigating asymmetry of curvature in baroreflex studies, Am. J. Physiol. (Regulatory Integrative Comp. Physiol. 46), 277, 441–454.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | ## Fitting the baro5 model
heartrate.m1 <- drm(rate~pressure, data=heartrate, fct=baro5())
plot(heartrate.m1)
coef(heartrate.m1)
#Output:
#b1:(Intercept) b2:(Intercept) c:(Intercept) d:(Intercept) e:(Intercept)
# 11.07984 46.67492 150.33588 351.29613 75.59392
## Inserting the estimated baro5 model function in deriv()
baro5Derivative <- deriv(~ 150.33588 + ((351.29613 - 150.33588)/
(1 + (1/(1 + exp((2 * 11.07984 * 46.67492/(11.07984 + 46.67492)) *
(log(x) - log(75.59392 ))))) * (exp(11.07984 * (log(x) - log(75.59392)))) +
(1 - (1/(1 + exp((2 * 11.07984 * 46.67492/(11.07984 + 46.67492)) *
(log(x) - log(75.59392 )))))) * (exp(46.67492 * (log(x) - log(75.59392 )))))), "x", function(x){})
## Plotting the derivative
#pressureVector <- 50:100
pressureVector <- seq(50, 100, length.out=300)
derivativeVector <- attr(baro5Derivative(pressureVector), "gradient")
plot(pressureVector, derivativeVector, type = "l")
## Finding the minimum
pressureVector[which.min(derivativeVector)]
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