A particle in the first quadrant is moving along a path described by the equation LaTeX: x^2+xy+2y^2=16x 2 + x y + 2 y 2 = 16 such that at the moment its x-coordinate is 2, its y-coordinate is decreasing at a rate of 10 cm/sec. At what rate is its x-coordinate changing at that time?

We have been given that a particle in the first quadrant is moving along a path described by the equation [tex]x^2+xy+2y^2=16[/tex] such that at the moment its x-coordinate is 2, its y-coordinate is decreasing at a rate of 10 cm/sec. We are asked to find the rate at which x-coordinate is changing at that time.

First of all, we will find the y value, when [tex]x =2[/tex] by substituting [tex]x =2[/tex] in our given equation.

[tex]2^2+2y+2y^2=16[/tex]

[tex]4-16+2y+2y^2=16-16[/tex]

[tex]2y^2+2y-12=0[/tex]

[tex]y^2+y-6=0[/tex]

[tex]y^2+3y-2y-6=0[/tex]

[tex](y+3)(y-2)=0[/tex]

[tex](y+3)=0,(y-2)=0[/tex]

[tex]y=-3,y=2[/tex]

Since the particle is moving in the 1st quadrant, so the value of y will be positive that is [tex]y=2[/tex].

Now, we will find the derivative of our given equation.

[tex]2x\cdot x'+x'y+xy'+4y\cdot y'=0[/tex]

We have been given that [tex]y=2[/tex], [tex]x =2[/tex] and [tex]y'=-10[/tex].

[tex]2(2)\cdot x'+(2)x'+2(-10)+4(2)\cdot (-10)=0[/tex]

[tex]4\cdot x'+2x'-20-80=0[/tex]

[tex]6x'-100=0[/tex]

[tex]6x'-100+100=0+100[/tex]

[tex]6x'=100[/tex]

[tex]\frac{6x'}{6}=\frac{100}{6}[/tex]

[tex]x'=\frac{50}{3}[/tex]

Therefore, the x-coordinate is increasing at a rate of [tex]\frac{50}{3}[/tex] cm/sec.