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a particle is moving along a curve whose equation is: `(xy^3)/(1+y^2)=8/5 ` assume...
a particle is moving along a curve whose equation is:
assume the x-coordinate is increasing at a rate of 6 units/sec when the particle is at the point (1,2)
a) at what rate is the y-coordinate of the point changing at that instant?
b) is the particle rising or falling at that instant?
1 Answer | add yours
`(xy^3)/(1+y^2) = 8/5`
`5xy^3 = 8(1+y^2)`
Differentiate both sides with respect to time t.
`5(x*3y^2y'+y^3x') = 8xx2yxxy' ---(1)`
It is given that at (1,2) the rate of increasing of x is 6 units per second.
`x = 1`
`y = 2`
`x' = 6`
Applying the above values in (1) will give you;
`5(x*3y^2y'+y^3x') = 8xx2yxxy'`
`5(1xx3xx2^2xxy'+2^3xx6) = 8xx2xx2xxy'`
`15y'+60 = 8y'`
`y' = -60/7`
So the y coordinate is changing at a rate of `60/7` units per second. Since we have a negative value for y' by the usual notation that means y is decreasing or falling.
Posted by jeew-m on July 11, 2013 at 4:29 AM (Answer #1)
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