Represent any point on the curve by (x, 1-x^2). The distance between (0, 0) and (x, 1-x^2) is
![\sqrt{(x-0)^2+(1-x^2-0)^2}=\sqrt{x^2+(1-x^2)^2}=\sqrt{x^2+1-2x^2+x^4}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28x-0%29%5E2%2B%281-x%5E2-0%29%5E2%7D%3D%5Csqrt%7Bx%5E2%2B%281-x%5E2%29%5E2%7D%3D%5Csqrt%7Bx%5E2%2B1-2x%5E2%2Bx%5E4%7D%20)
To make this easier, let's minimize the SQUARE of this quantity because when the square root is minimal, its square will be minimal.
So minimize
![L=x^4-x^2+1](https://tex.z-dn.net/?f=L%3Dx%5E4-x%5E2%2B1)
Find the derivative of L and set it equal to zero.
![\frac{d}{dx}(L)=4x^3-2x \\ 4x^3-2x=0 \\ 2x(2x^2-1)=0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%28L%29%3D4x%5E3-2x%20%5C%5C%204x%5E3-2x%3D0%20%5C%5C%202x%282x%5E2-1%29%3D0)
This gives you
![x=0](https://tex.z-dn.net/?f=x%3D0)
or
![x^2=\frac{1}{2} \\ x=\pm\sqrt{2}/2](https://tex.z-dn.net/?f=x%5E2%3D%5Cfrac%7B1%7D%7B2%7D%20%5C%5C%20x%3D%5Cpm%5Csqrt%7B2%7D%2F2)
You can use the Second Derivative Test to figure out which value(s) produce the MINIMUM distance.
![\frac{d^2}{dx}=12x^2-2](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2%7D%7Bdx%7D%3D12x%5E2-2)
When x = 0, the second derivative is negative, indicating a relative maximum. When
![x=\pm\frac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=x%3D%5Cpm%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)
, the second derivative is positive, indicating a relative MINIMUM.
The two points on the curve closest to the origin are
The answer is not d it’s b because there is no negative in the 144.
If an airplane flies 920 km in ONE hour then the average speed in km/h is 920 km
Answer:
9 cm
Step-by-step explanation:
Arc length = θ/360 × 2πr
A circle has an arc with a measure of 120 degrees and a lenght of 3π cm. What is the diameter of the circle
Hence,
3π = 120/360 × 2πr
Diameter = 2r
3π = 1/3 × 2πr
Cross Multiply
9π = 2πr
2r = 9π/π
Note that :
Diameter = 2r
Hence,
Diameter = 9 cm
Answer:
yoi look for what you add to the equation to make it a perfect square by dividing 6 by 2 then add 2^2 to both sides.