I don't know of any automatic model fitting, but it seems like in your case the relationship between and is . Thus, you could calculate each and for each in order to find the 'general eqausion'
So per your data set
a <- data.frame(c =c(0.01,0.02,0.03,0.05,0.07,0.10,0.15,0.20,0.30,0.50,0.70,1),
t1 =c(5,5.55,5.94,6.53,6.98,7.54,8.29,8.90,9.92,11.51,12.80,14.46),
t2 = c(4.35,4.84,5.19,5.72,6.12,6.62,7.29,7.84,8.74,10.16,11.33,12.80),
t3 = c(3.70, 4.13,4.44,4.89,5.25,5.68,6.26,6.73,7.52,8.74, 9.74,10.99),
t4 = c(3.08,3.45,3.70,4.09,4.39,4.75,5.23,5.63,6.28,7.28,8.09,9.11),
t5 = c(2.51,2.80,3.01,3.33,3.57,3.86,4.25,4.56,5.07,5.85,6.54,7.19))
A function that calculates estimates and s.e for each
ResFunc <- function(x) {
temp <- lm(reformulate("c", response = x), log(a))
c(exp(coef(temp)[[1]]), coef(temp)[[2]], exp(summary(temp)$coefficients[1,2]), summary(temp)$coefficients[2,2])
}
Running the function
temp <- as.data.frame(t(sapply(setdiff(names(a),"c"), ResFunc)))
colnames(temp) <- c("a", "b", "S.E (a)", "S.E (b)")
temp
# a b S.E (a) S.E (b)
#t1 13.422867 0.2314997 1.024901 0.009622679
#t2 11.888155 0.2353401 1.024803 0.009585284
#t3 10.237551 0.2375002 1.024013 0.009283321
#t4 8.523443 0.2366266 1.022568 0.008730912
#t5 6.831186 0.2321247 1.020344 0.007879240
Now you could potentially estimate each line. For example, estimating and comparing
b <- data.frame(c = a$c, t1 = a$t1, t1.est = temp[1,1]*(a$c^temp[1,2]))
test <- melt(b, "c")
ggplot(test, aes(x = c, y = value, color = variable)) + geom_line()