a. If a child were to step off from the edge of a spinning merry-go-round wearing rollerblades, she would move in the direction of the line tangent to the path of the merry-go-round at the point where she steps off. Suppose a merry-go-round of radius 10 is centered at the origin (x^2 + y^2 = 100) and the child steps off from the point (6,8). Calculate the value of the derivative to the curve at x=6. Use that value to find the slope of the line along which she will move (considering her horizontal motion only)
b. Suppose a man was riding "The Quadratic Expression", a train from (-2, 4) to (2,4) along the track that followed the curve y= x^2. If he happened to be riding on top of the train and stepped off just as he was passing the origin, along what line would he move? What if he stepped off at (-1, 1)? (Horizontal motion considered only)
2007-01-16 07:47:43 · 8 answers · asked by aimsnapfall 2 in Science & Mathematics ➔ Mathematics
Calculus is actually fun; it's just that I don't get it at times.
Nope, exam review.
Riighttt. 2 points for you guys. Hehe.
2007-01-16 07:53:55 · update #1
The answer for part b are y=0 and y= -2x -17
:( This is just hard.
2007-01-16 08:11:31 · update #2
My bad.
For part b, the equation is y= -2x-1
2007-01-16 08:20:26 · update #3
/a/ So differentiate 2x +2yy’ =0, y’ =-x/y, where x=6 and y=8; thus y’=-3/4 =k, where k is the slope of tangent to be found: y=k·x+b; I mean b to be found! So y= -3x/4 +b; 3x+4y =b; normalize the line, that is divide by sqtr(3*3 +4*4) =5; thence (3/5)x +(4/5)y =b/5; b/5 is distance of tangent line, which is = to giddy-go R=10, hence b/5 =10 and b=50; thus y=-3x/4 +50;
\b\ thus, if there is a girl she has a fun! If there is a man he is dead for sure! What a weird school you are in? now business! k= y’ =2x, where k is the slope of tangent y=kx+b he is moving along to break his neck. Since it is origin, i.e. x=0 and y=0, then k=0 and 0=0*0 +b, hence b=0; thus y=0 or x-axis!
/b1/ P=(-1,1), y’(P) = k = 2*(-1), k=-2; and P belongs to tangent, hence 1=k*(-1) +b, b= 1+k = -1; this time y=-2x-1 is the tangent he is moving along to break his neck. Upbeat ending.
2007-01-16 08:59:01 · answer #1 · answered by Anonymous · 0⤊ 0⤋
a. first, solve for y: y=(100-x^2)^(1/2)
then substitute something for the 100-x^2=U: y=U^(1/2)
then derivatize that: .5du*U^(-1/2)
and the 100-x^2: -2x=dU
then put U and dU in the equation: dy/dx=.5(-2x)(100-x^2)^(-1/2)
put in 6 for x and the slope is -3/4
b. even easier, derivatise the equation: dy/dx=2x
put in -1 for x to get the slope: -2
and make a line with a slope of -2 that goes through -1,1
the equation for the line is: y=-2x-1
2007-01-16 16:09:14 · answer #2 · answered by Ben B 4 · 0⤊ 0⤋
a.
Differentiate both sides of x^2 + y^2 = 100 with respect to x,
2x+2yy' = 0
Solve for y',
y' = -x/y
Plug in x = 6 and y = 8,
y' = -6/8 = -3/4
b.
Can someone else try?
2007-01-16 16:04:57 · answer #3 · answered by sahsjing 7 · 0⤊ 1⤋
try asking your professor for help... I'm not just trying to get two points. I'm tryin to help u solve ur problem lol...
2007-01-16 15:55:51 · answer #4 · answered by ... 3 · 0⤊ 1⤋
sorry for this time,meet u in ur next question
2007-01-24 05:04:48 · answer #5 · answered by vamsi 2 · 0⤊ 0⤋
Calculus sucks, sorry but I don't have a clue.
2007-01-16 15:49:56 · answer #6 · answered by The Ultimate Answerer 3 · 0⤊ 2⤋
Homework?
....
2007-01-16 15:51:25 · answer #7 · answered by god knows and sees else Yahoo 6 · 0⤊ 1⤋
oh
2007-01-24 15:00:57 · answer #8 · answered by Anonymous · 0⤊ 0⤋