A particle is moving along the curve whose equation is $\frac{8}{5} = \frac{x y^{3}}{1 + y^{2}}$ $8/5=(xy^3)/(1+y^2)$ . Assume that the x-coordinate is increasing at the rate of 6 units/sec when the particle is at the point (1,2). At what rate is y-coordinate of the point changing at that instant?

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A particle is moving along the curve whose equation is $\frac{8}{5} = \frac{x y^{3}}{1 + y^{2}}$ $8/5=(xy^3)/(1+y^2)$ . Assume that the x-coordinate is increasing at the rate of 6 units/sec when the particle is at the point (1,2). At what rate is y-coordinate of the point changing at that instant?

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Answer 1 · 2018-06-06T04:10:41

$- \frac{60}{7}$ $- 60/7$ units sec^-1

Explanation:

$\frac{8}{5} = \frac{x y^{3}}{1 + y^{2}}$ $8/5=[xy^3]/[1+y^2]$ Cross multiply, $5 x y^{3} = 8 [1 + y^{2}]$ $5xy^3=8[1+y^2]$ ........ $[1]$ $[1]$

Differentiating.... $[1]$ $[1]$ wrt $x$ $x$ using the product rule and the general power rule, implicitly. i.e, $d [u v] = v d u + u d v$ $d[uv]=vdu+udv$ [where $u$ $u$ and $v$ $v$ are both functions of $x$ $x$ ]

$\frac{d}{d x} [5 x y^{3}] = \frac{d}{d x} [8 + 8 y^{2}]$ $d/dx[5xy^3]=d/dx[8+8y^2]$ therefore, $5 y^{3} + 5 x [3 y^{2}] \frac{d y}{d x} = 16 y \frac{d y}{d x}$ $5y^3+5x[3y^2]dy/dx=16ydy/dx$ , so , $[5 y^{3} + 15 x y^{2} \frac{d y}{d x} = 16 y \frac{d y}{d x}]$ $[5y^3+15xy^2dy/dx=16ydy/dx]$

Solving for $\frac{d y}{d x}$ $dy/dx$ , ... $\frac{d y}{d x} = [\frac{5 y^{3}}{16 y - 15 x y^{2}}]$ $dy/dx=[[5y^3]/[16y-15xy^2]]$

We are told that the rate of change of $x$ $x$ wrt t [ time]= $6$ $6$ units ${sec}^{- 1}$ $sec^-1$ ,i.e, $\frac{d x}{d t} = 6$ $dx/dt=6$ .

#dx/dt.dy/dx= dy/dt# thus, $\frac{d y}{d t} = \frac{6 [5 y^{3}]}{16 y - 15 x y^{2}}$ $dy/dt = [6[5y^3]]/[16y-15xy^2]$

so $\frac{d y}{d t}$ $dy/dt$ [rate of change of $y$ $y$ wrt t ] = $\frac{30 y^{3}}{16 y - 15 x y^{2}}$ $[30y^3]/[16y-15xy^2]$ and at the point $[1, 2]$ $[1,2]$ , $\frac{d y}{d t} = [30] \frac{2^{3}}{16 [2] - 15 [1] [2^{2}]}$ $dy/dt = [30][2^3]/[16[2]-15[1][2^2]$ = - $\frac{60}{7}$ $60/7$ $u n i t s {sec}^{- 1}$ $units sec^-1$

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A particle is moving along the curve whose equation is $\frac{8}{5} = \frac{x y^{3}}{1 + y^{2}}$ $8/5=(xy^3)/(1+y^2)$ . Assume that the x-coordinate is increasing at the rate of 6 units/sec when the particle is at the point (1,2). At what rate is y-coordinate of the point changing at that instant?

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1 Answer

Explanation:

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A particle is moving along the curve whose equation is $\frac{8}{5} = \frac{x y^{3}}{1 + y^{2}}$ $8/5=(xy^3)/(1+y^2)$ . Assume that the x-coordinate is increasing at the rate of 6 units/sec when the particle is at the point (1,2). At what rate is y-coordinate of the point changing at that instant?