The length (in centimeters) of a typical Pacific halibut t years old is approximately
f(t) = 190(1 − 0.955e−0.25t).
(b) How fast is the length of a typical 9-year-old Pacific halibut increasing?
cm/yr
(c) What is the maximum length a typical Pacific halibut can attain?
cm
You sure are having quite a time with these exponential functions!
y = 190 [ 1- .955 e^-(.25 t) ]
If t = 9
y = 190 [ 1 - .955(.105) ]
= 190 [ 1 - .1 ]
= 171 cm
In other words it has reached 90% of max length in 9 years
now
dy/dt = 190(-.955) (-.25)(.105)
= 4.76 cm/year
c) when t gets very large e^-.25 t = 1/e^oo = 0
so
max length = (190)(1) = 190 cm
To find the rate at which the length of a typical 9-year-old Pacific halibut is increasing, we need to find the derivative of the function f(t) with respect to t and then evaluate it at t=9.
The derivative of f(t) with respect to t can be found using the chain rule of differentiation. The chain rule states that if we have a function g(h(t)), then the derivative of g(h(t)) with respect to t is given by g'(h(t))*h'(t).
For our function f(t) = 190(1 − 0.955e−0.25t), let's go step by step to find its derivative:
Step 1: Take the derivative of the outer function:
g(t) = 190*(1 − 0.955e^(-0.25t))
g'(t) = 190*(-0.955)(e^(-0.25t))
Step 2: Take the derivative of the inner function:
h(t) = e^(-0.25t)
h'(t) = (-0.25)(e^(-0.25t))
Step 3: Apply the chain rule:
f'(t) = g'(h(t))*h'(t)
= 190*(-0.955)(e^(-0.25t))*(-0.25)(e^(-0.25t))
= 47.75(e^(-0.25t))^2
Now, to find the rate at which the length of a typical 9-year-old Pacific halibut is increasing, we substitute t=9 into the derivative:
f'(9) = 47.75(e^(-0.25*9))^2
= 47.75(e^(-2.25))^2
≈ 9.854 cm/yr
Therefore, the length of a typical 9-year-old Pacific halibut is increasing at a rate of approximately 9.854 cm/yr.
To find the maximum length a typical Pacific halibut can attain, we need to find the limit of the function f(t) as t approaches infinity.
Taking the limit of f(t) as t approaches infinity:
lim(t→∞) f(t) = lim(t→∞) 190(1 − 0.955e^(-0.25t))
As t approaches infinity, the term e^(-0.25t) becomes very close to zero, hence approaching 0. Therefore:
lim(t→∞) 0.955e^(-0.25t) = 0
So, the maximum length a typical Pacific halibut can attain is given by:
lim(t→∞) f(t) = lim(t→∞) 190(1 − 0) = 190 cm