A college requires all incoming freshmen to live on campus. the college has two freshman dormitories, east and west. for an upco
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Unfortunately the portion of the graph shown does not show the locations of points A and B, nor does it show sufficient parts of the curve.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.
Answer:
Step-by-step explanation:
ALRIGHT ARE YOU READY FOR EPIC MATH TIME?
because i know i am. so lets get this bread.
WATCH: we have
interesting... that first character is called pi. maybe i should have said, "lets get this pi".
so now lets do some algebra. i am going to work from right to left. lets multiply 6r by 2. that gives us 12r. now we have...
WE ARE MAKING EPIC PROGRESS, AREN'T WE?
lets multiply 12r by 3r. then we have....
notice how i multiplied 12 by 3 and also r by r (which gives 36 and respectively)
so your simplified expression is: