Answer:3 units/s
Step-by-step explanation:
Given
![y=\sqrt{x}](https://tex.z-dn.net/?f=y%3D%5Csqrt%7Bx%7D)
Point P lie on this curve so any general point on curve can be written as ![(x,\sqrt{x})](https://tex.z-dn.net/?f=%28x%2C%5Csqrt%7Bx%7D%29)
and ![\frac{\mathrm{d} x}{\mathrm{d} t}=4 units/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20x%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D4%20units%2Fs)
Distance between Point P and (2,0)
![P=\sqrt{(x-2)^2+(\sqrt{x}-0)^2}](https://tex.z-dn.net/?f=P%3D%5Csqrt%7B%28x-2%29%5E2%2B%28%5Csqrt%7Bx%7D-0%29%5E2%7D)
P at x=3 P=2
rate at which distance is changing is
![\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{\mathrm{d} \sqrt{(x-2)^2+(\sqrt{x}-0)^2}}{\mathrm{d} t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20P%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%5Csqrt%7B%28x-2%29%5E2%2B%28%5Csqrt%7Bx%7D-0%29%5E2%7D%7D%7B%5Cmathrm%7Bd%7D%20t%7D)
![\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2x-3}{\sqrt{(x-2)^2+(\sqrt{x}-0)^2}}\times \frac{\mathrm{d} x}{\mathrm{d} t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20P%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D%5Cfrac%7B2x-3%7D%7B%5Csqrt%7B%28x-2%29%5E2%2B%28%5Csqrt%7Bx%7D-0%29%5E2%7D%7D%5Ctimes%20%5Cfrac%7B%5Cmathrm%7Bd%7D%20x%7D%7B%5Cmathrm%7Bd%7D%20t%7D)
![\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2\times 3-3}{2\times 2}\times 4=3 units/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20P%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D%5Cfrac%7B2%5Ctimes%203-3%7D%7B2%5Ctimes%202%7D%5Ctimes%204%3D3%20units%2Fs)
<span>A ball is thrown at a 30 degree angle above the horizontal with a speed of 10 ft/s. After 0.50s the horizontal component of it's velocity will be the same. In a projectile motion the horizontal component of the velocity is said to be constant. Therefore, it will be equal to the initial velocity.</span>
Answer:
FGHE
Step-by-step explanation:
They have the same angles size
Answer:
Solving the given linear system, we get x = 1 and y = -3.
The solution set is: (1,-3)
Step-by-step explanation:
We need to solve the linear system of equations.
![3x-y=6 \\y=x-4](https://tex.z-dn.net/?f=3x-y%3D6%20%5C%5Cy%3Dx-4)
We can write second equation y=x-4 as: ![-x+y=-4](https://tex.z-dn.net/?f=-x%2By%3D-4)
Let:
![3x-y=6--eq(1)\\ -x+y=-4--eq(2)](https://tex.z-dn.net/?f=3x-y%3D6--eq%281%29%5C%5C%20-x%2By%3D-4--eq%282%29)
Now, Adding equation 1 and 2
![3x-y=6\\ -x+y=-4\\--------\\2x+0y=2\\2x=2\\x=\frac{2}{2}\\x=1](https://tex.z-dn.net/?f=3x-y%3D6%5C%5C%20-x%2By%3D-4%5C%5C--------%5C%5C2x%2B0y%3D2%5C%5C2x%3D2%5C%5Cx%3D%5Cfrac%7B2%7D%7B2%7D%5C%5Cx%3D1)
So, we get x = 1
Now, put value of x in second equation to find value of y:
![y=x-4\\Put\:x=1\\y=1-4\\y=-3](https://tex.z-dn.net/?f=y%3Dx-4%5C%5CPut%5C%3Ax%3D1%5C%5Cy%3D1-4%5C%5Cy%3D-3)
So, we get y = -3
Solving the given linear system, we get x = 1 and y = -3.
The solution set is: (1,-3)