The red squirrel population, p, in a neighborhood park can be modelled by the function p(t)= srqt(210t+44t^2), where t is time in year.
a) Determine the rate of growth of the squirrel population at t= 2years
Ans: find the 1st derivative and substitute 2 in the function.
b) When will the population reach 60 sqirrels
Ans: set p(t) = 60
c) What is the instantaneous rate of change of the population at the time in part b)?
Ans: the time obtained in b, substitute that into the 1st derivative.
d) When is the instantaneous rate of the squirrel population approximately 7squirrels per year?
Ans: set the first derivative p'(t) = 7
your answers are correct
a) Ah, the rate of growth of those red squirrels at t=2 years, huh? Well, first we need to find the first derivative of the squirrel population function. But I have a joke for you while we're at it:
Why don't scientists trust atoms?
Because they make up everything!
Alright, back to business. The first derivative of p(t) = sqrt(210t+44t^2) is p'(t) = (210+88t) / (2*sqrt(210t+44t^2)). Now, plug in t=2 and calculate. But don't overthink it, just like a squirrel in a tree: just do it acorn-cording to plan!
b) So, when will those red squirrels reach a population of 60? Let's set the population function equal to 60 and solve for t. But before we do that, a squirrel pun for you:
Why did the squirrel take up knitting?
Because it heard it was good at making sweaters!
Alright, let's get back on track. We have p(t) = sqrt(210t+44t^2) = 60. Solve that equation and you'll find the squirrel-astic answer!
c) Now, let's find the instantaneous rate of change of the squirrel population at the time obtained in part b, shall we? Just plug that time into the first derivative we found earlier (p'(t) = (210+88t) / (2*sqrt(210t+44t^2))) and calculate. Easy-peasy, squirrel-squeezy!
d) Ah, the elusive instantaneous rate of change of approximately 7 squirrels per year! Set the first derivative, which is p'(t) = (210+88t) / (2*sqrt(210t+44t^2)), equal to 7 and solve for t. But here's something to lighten the mood while you ponder that:
Why don't spiders go to school?
Because they already know their webs-ite!
Happy squirrel math-ing!
a) To determine the rate of growth of the squirrel population at t = 2 years, we need to find the first derivative of the function p(t) = sqrt(210t + 44t^2) and substitute t = 2.
First, find the derivative of p(t):
p'(t) = (1/2)*(210 + 88t) / sqrt(210t + 44t^2)
Now substitute t = 2:
p'(2) = (1/2)*(210 + 88(2)) / sqrt(210(2) + 44(2^2))
= (1/2)*(210 + 176) / sqrt(420 + 176)
= (1/2)*(386) / sqrt(596)
= 193 / sqrt(596)
≈ 7.88
Therefore, the rate of growth of the squirrel population at t = 2 years is approximately 7.88 squirrels per year.
b) To determine when the population will reach 60 squirrels, we need to set p(t) = 60 and solve for t.
So, we have:
sqrt(210t + 44t^2) = 60
Squaring both sides of the equation, we get:
210t + 44t^2 = 3600
Rearranging the equation, we have:
44t^2 + 210t - 3600 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula to find the values of t when p(t) = 60.
c) To determine the instantaneous rate of change of the population at the time obtained from part b, substitute that value of t into the first derivative p'(t).
d) To find when the instantaneous rate of the squirrel population is approximately 7 squirrels per year, set the first derivative p'(t) = 7 and solve for t.
To answer part a) of the question, we need to find the rate of growth of the squirrel population at t=2 years. For this, we will take the first derivative of the function p(t). The first derivative represents the rate of change or the slope of the function.
To find the first derivative of p(t), we differentiate the function p(t) = sqrt(210t + 44t^2) with respect to t.
p'(t) = (1/2)*(210t + 44t^2)^(-1/2) * (210 + 88t)
Now, substitute t=2 into the first derivative equation p'(t) to get the rate of growth at t=2 years.
To answer part b) of the question, we need to find when the population will reach 60 squirrels. For this, we set p(t) equal to 60 and solve for t.
p(t) = sqrt(210t + 44t^2) = 60
To determine the time when the population reaches 60 squirrels, you need to solve this equation for t.
To answer part c) of the question, we find the instantaneous rate of change of the population at the time obtained in part b). This can be done by substituting the value of t obtained in part b) into the first derivative equation p'(t).
To answer part d) of the question, we need to find the time when the instantaneous rate of the squirrel population is approximately 7 squirrels per year. For this, we set the first derivative p'(t) equal to 7 and solve for t.