Question

Allometric relations often can be modeled by ​f(x)= x^b​, where a and b are constants. One study showed that for a male fiddler crab weighing over 0.75​ gram, the weight of its claws can be estimated by ​f(x)= 0.448 x^1.21. The input x is the weight of the crab in​ grams, and the output​ f(x) is the weight of the claws in grams.

(a) Predict the weight of the claws of a 2-gram crab

(b) Approximate the weight of a crab that has 0.5-gram claws

Answer 1

a)
`f(2)= 0.448 (2)^(1.21)= 1.036 grams`

b)
`0.5 = 0.448 x^(1.21)`

Dividing both sides by 0.448 we got:

`(0.5)/(0.448) = x^(1.21)`

We can appy the exponent
`(1)/(1.21)` in both sides of the equation and we got:

`((0.5)/(0.448))^{(1)/(1.21)} = x= 1.095grams`

Explanation:

For this case we know the following function:

`f(x) = 0.448 x^(1.21)`

The notation is: x is the weight of the crab in​ grams, and the output​ f(x) is the weight of the claws in grams.

Part a

For this case we just need to replace x = 2 gram in the function and we got:

`f(2)= 0.448 (2)^(1.21)= 1.036 grams`

Part b

For this case we know tha value for
`f(x) =0.5` and we want to find the value of x who satisfy this condition:

`0.5 = 0.448 x^(1.21)`

Dividing both sides by 0.448 we got:

`(0.5)/(0.448) = x^(1.21)`

We can appy the exponent
`(1)/(1.21)` in both sides of the equation and we got:

`((0.5)/(0.448))^{(1)/(1.21)} = x= 1.095grams`