interpreting the meaning of the derivative in context - homework\n1. alvaro is studying the extinction of…

interpreting the meaning of the derivative in context - homework\n1. alvaro is studying the extinction of the bear population of siberia over time. the following\nfunction gives the number of bears t years since alvaro started tracking it.\n\nb(t)=2190e^{-0.3t}\n\na. what is b(3)?\n\nb. interpret b(3) in the context of this problem.\n\n2. the windchill, in fahrenheit, that a human feels is dependent on the air temperature (in\nfahrenheit) as well as the wind velocity, v, (in miles per hour). at a temperature of 32 f, the\nwindchill is given by w(v)=55.6 - 22.1v^{0.16} and is valid for 5 ≤ v ≤ 60.\n\na. find w(20).\n\nb. using correct units, interpret the meaning of w(20) in the context of this problem.\n\n3. johanna is running along a straight path for 40 minutes. her velocity is given by a differentiable\nfunction v(t), where v is in meters and t is in minutes. selected values of v(t) is given in the\ntable below.\n\n\n\n\na. use the table to estimate v(17).\n\nb. using proper units, interpret the meaning of v(17) in the context of this problem.
Answer
1. a.
Explanation:
Step1: Differentiate ( B(t) )
Use the chain rule. If ( y = ae^{bt}), then ( y'=abe^{bt}). For ( B(t)=2190e^{- 0.3t}), ( B'(t)=2190\times(-0.3)e^{-0.3t}=-657e^{-0.3t})
Step2: Substitute ( t = 3 )
Substitute ( t = 3) into ( B'(t)). ( B'(3)=-657e^{-0.3\times3}=-657e^{-0.9}\approx-657\times0.4066=-267.23)
Answer:
( B'(3)\approx - 267.23)
1. b.
Brief Explanations:
The derivative ( B'(t)) represents the rate of change of the bear population. When ( t = 3) (3 years since Alvaro started tracking), ( B'(3)\approx-267.23) means that the bear population is decreasing at a rate of approximately ( 267.23) bears per year at the - third - year mark.
Answer:
The bear population is decreasing at a rate of approximately ( 267.23) bears per year 3 years after Alvaro started tracking.
2. a.
Explanation:
Step1: Differentiate ( W(v))
Use the power rule. If ( y = ax^{n}), then ( y'=nax^{n - 1}). For ( W(v)=55.6-22.1v^{0.16}), ( W'(v)=-22.1\times0.16v^{0.16 - 1}=-3.536v^{-0.84})
Step2: Substitute ( v = 20 )
Substitute ( v = 20) into ( W'(v)). ( W'(20)=-3.536\times20^{-0.84}\approx-3.536\times0.133=-0.47)
Answer:
( W'(20)\approx - 0.47)
2. b.
Brief Explanations:
The derivative ( W'(v)) represents the rate of change of wind - chill with respect to wind velocity. The units of ( W(v)) are degrees Fahrenheit and the units of ( v) are miles per hour. So the units of ( W'(v)) are degrees Fahrenheit per mile per hour. When ( v = 20) (wind velocity of 20 miles per hour), ( W'(20)\approx-0.47) means that the wind - chill is decreasing at a rate of approximately ( 0.47) degrees Fahrenheit per mile per hour when the wind velocity is 20 miles per hour.
Answer:
The wind - chill is decreasing at a rate of approximately ( 0.47) degrees Fahrenheit per mile per hour when the wind velocity is 20 miles per hour.
3. a.
Explanation:
Step1: Use the difference quotient
We use the centered difference quotient ( v'(a)\approx\frac{v(a + h)-v(a - h)}{2h}). To estimate ( v'(17)), we can use ( h = 3) (since ( 17-3 = 14) is not in the table, but we can use the closest values). A better approach is to use the average of the left - hand and right - hand difference quotients. The left - hand difference quotient: ( \frac{v(20)-v(12)}{20 - 12}=\frac{240 - 200}{8}=5). The right - hand difference quotient (using the closest value after 17, which is 24, but it's a bit of a stretch. A more common approach is to use the formula for the average rate of change between two points around ( t = 17)). Another way is to assume a linear approximation between ( t = 12) and ( t = 20). ( v'(17)\approx\frac{v(20)-v(12)}{20 - 12}=\frac{240-200}{8}=5)
Answer:
( v'(17)\approx5)
3. b.
Brief Explanations:
The derivative ( v'(t)) represents the rate of change of velocity (acceleration). The units of ( v(t)) are meters per minute and the units of ( t) are minutes. So the units of ( v'(t)) are meters per minute per minute (or meters per minute²). When ( t = 17) (17 minutes into the run), ( v'(17)\approx5) means that Johanna's acceleration is approximately ( 5) meters per minute² at the 17 - minute mark.
Answer:
Johanna's acceleration is approximately ( 5) meters per minute² 17 minutes into her run.