Question

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?

Answer 1

Answer:

$\frac{50}{3}$ cm/sec.

Step-by-step explanation:

We have been given that a particle in the first quadrant is moving along a path described by the equation $x^2+xy+2y^2=16$ 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 $x =2$ by substituting $x =2$ in our given equation.

$2^2+2y+2y^2=16$

$4-16+2y+2y^2=16-16$

$2y^2+2y-12=0$

$y^2+y-6=0$

$y^2+3y-2y-6=0$  

$(y+3)(y-2)=0$

$(y+3)=0,(y-2)=0$

$y=-3,y=2$

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

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

$2x\cdot x'+x'y+xy'+4y\cdot y'=0$

We have been given that $y=2$, $x =2$ and $y'=-10$.

$2(2)\cdot x'+(2)x'+2(-10)+4(2)\cdot (-10)=0$

$4\cdot x'+2x'-20-80=0$

$6x'-100=0$

$6x'-100+100=0+100$

$6x'=100$

$\frac{6x'}{6}=\frac{100}{6}$

$x'=\frac{50}{3}$

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