a) radius of a circle
a=pi*r^2
to write this as a function of r
a("r")=pi*r^2
pug in what r is to find what a equals
b) r(t) = 2.2t
where you plug in what your t is to give you r
try using what I gave you for part a and b to find out the rest. Post if you have questions.
Hey so I need help on a problem from an assignment that my teacher won't help at all.
-WRITE IN COMPLETE SENTENCES-
"An oil strikes a sand bar that rips a hole in the hull of the ship. Oil begins leaking out of a tanker with spilled oil forming an approximate circle around the tanker."
The radius of the circle is increasing at a rate of 2.2ft.
a) write the area of the circle of oil as a function of radius.
b) Write tge radius of the circle as a function of time.
c)Find the radius of the circle of spilled oil after 2 hrs? what's the radius of the circle after 2.5 hrs?
d) use the result in part(c) to determine the area of the circle after 2hrs and 2.5 hrs.
e) write the function that represents area of the circle of spilled oil as a function of time.
f)Use the result of (e) to determine the area of the result of the circle after 2hrs and 2.5 hrs.
g) calculate the average rate of change of the area of the circle from 2hrs to 2.5 hrs.
h)Calculate the average rate of change of the circle from 5hrs to 7.5 hrs.
i)based on the results obtained in parts (g) and (h),what's happening to the average rate of change of the area of the circle as time passes?
j)If the oil tanker is 150yrds from shore,when will the oil spill first reach the shoreline?(1yd=3ft)
k)How long will it be until 6miles of shore line is contaminated with oil?(1mile=580ft).
I know it's a lot. im sorry =(
3 answers
The radius of the circle is increasing at a rate of 2.2ft ??? per second, minute, hour? It is kind of important later, as in f) h) i) J) k)
area=pi*r^2
d area/dt= 2 PI r dr/dt
so depending on the answer to my question above, dr/dt is known.
So d area/dt is then known
for j, dr/dt=dl/dt or Time=L/dr/dt_
where L is the distance to shore
the statement <<The radius of the circle is increasing at a rate of 2.2ft. >> is inadequate. Time has to be stated for the 2.2 ft, otherwise it is not a rate.
area=pi*r^2
d area/dt= 2 PI r dr/dt
so depending on the answer to my question above, dr/dt is known.
So d area/dt is then known
for j, dr/dt=dl/dt or Time=L/dr/dt_
where L is the distance to shore
the statement <<The radius of the circle is increasing at a rate of 2.2ft. >> is inadequate. Time has to be stated for the 2.2 ft, otherwise it is not a rate.
PLEASE PROOF READ - I assume you mean
2.2ft. / HOUR
a) A = pi r^2
b) 2.2 t
c) 2.2 * 2
2.2 * 2.5
d)A(2) = pi * (4.4)*2
A(2.5) = pi (5.5)^2
e) A(t) = pi (2.2 t)^2
f) A(2) = pi (2.2*2)^2
A(2.5) = pi (2.2*2.5)^2
g) [ A(2.5) -A(2) ]/.5
h) [A(7.)-A(5)]/ 2.5
i) you will see that it increases
by the way
dA/dt = 2 pi r dr/dt
the circumference times how fast the radius is increasing (draw a picture)
j) r = 2.2 t from part b
150 yards r = 3 * 150 = 450 feet
2.2 t = 450
t = 450/2.2
k) , you will have to do this trig yourself. Segment of circle is length 6*580
450^2 + (3*580)^2 = r^2 = (2.2 t)^2
2.2ft. / HOUR
a) A = pi r^2
b) 2.2 t
c) 2.2 * 2
2.2 * 2.5
d)A(2) = pi * (4.4)*2
A(2.5) = pi (5.5)^2
e) A(t) = pi (2.2 t)^2
f) A(2) = pi (2.2*2)^2
A(2.5) = pi (2.2*2.5)^2
g) [ A(2.5) -A(2) ]/.5
h) [A(7.)-A(5)]/ 2.5
i) you will see that it increases
by the way
dA/dt = 2 pi r dr/dt
the circumference times how fast the radius is increasing (draw a picture)
j) r = 2.2 t from part b
150 yards r = 3 * 150 = 450 feet
2.2 t = 450
t = 450/2.2
k) , you will have to do this trig yourself. Segment of circle is length 6*580
450^2 + (3*580)^2 = r^2 = (2.2 t)^2
1 answer