heartrate | R Documentation |

The dataset contains measurements of mean arterial pressure (mmHG) and heart rate (b/min) for a baroreflex curve.

`data(heartrate)`

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.

```
## 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)]
```