N= 75 (let me know if you need more help)
Answer:
4 sin (2x - 4).
Step-by-step explanation:
The amplitude of sin x = 1 so our curve with amplitude 4 will have 4 sin x as part of the general form. To alter the period from 2pi to pi we multiply the x by 2, giving 4 sin 2x, Finally to get a phase shift of 2 to the right, we replace the x by x - 2.
Our formula is 4 sin (2(x - 2))
= 4 sin (2x - 4).
Answer:
The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.
Step-by-step explanation:
This is a problem of optimization.
We have to minimize the time it takes for the lifeguard to reach the child.
The time can be calculated by dividing the distance by the speed for each section.
The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.
Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:
![d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}](https://tex.z-dn.net/?f=d_s%5E2%3D%2860-x%29%5E2%2B40%5E2%3D60%5E2-120x%2Bx%5E2%2B40%5E2%3Dx%5E2-120x%2B5200%5C%5C%5C%5Cd_s%3D%5Csqrt%7Bx%5E2-120x%2B5200%7D)
Then, the time (speed divided by distance) is:
![t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1](https://tex.z-dn.net/?f=t%3Dd_b%2Fv_b%2Bd_s%2Fv_s%5C%5C%5C%5Ct%3Dx%2F4%2B%5Csqrt%7Bx%5E2-120x%2B5200%7D%2F1.1)
To optimize this function we have to derive and equal to zero:
![\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\](https://tex.z-dn.net/?f=%5Cdfrac%7Bdt%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B4%7D%2B%5Cdfrac%7B1%7D%7B1.1%7D%28%5Cdfrac%7B1%7D%7B2%7D%29%5Cdfrac%7B2x-120%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%5C%5C%5C%5C%5C%5C%5Cdfrac%7Bdt%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B4%7D%20%2B%5Cdfrac%7B1%7D%7B1.1%7D%20%5Cdfrac%7Bx-60%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%3D0%5C%5C%5C%5C%5C%5C%20%20%5Cdfrac%7Bx-60%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%3D%5Cdfrac%7B1.1%7D%7B4%7D%3D%5Cdfrac%7B2%7D%7B7%7D%5C%5C%5C%5C%5C%5C%20x-60%3D%5Cdfrac%7B2%7D%7B7%7D%5Csqrt%7Bx%5E2-120x%2B5200%7D%5C%5C%5C%5C%5C%5C%28x-60%29%5E2%3D%5Cdfrac%7B2%5E2%7D%7B7%5E2%7D%28x%5E2-120x%2B5200%29%5C%5C%5C%5C%5C%5C%28x-60%29%5E2%3D%5Cdfrac%7B4%7D%7B49%7D%5B%28x-60%29%5E2%2B40%5E2%5D%5C%5C%5C%5C%5C%5C%281-4%2F49%29%28x-60%29%5E2%3D4%2A40%5E2%2F49%3D6400%2F49%5C%5C%5C%5C%2845%2F49%29%28x-60%29%5E2%3D6400%2F49%5C%5C%5C%5C45%28x-60%29%5E2%3D6400%5C%5C%5C%5C)
![x](https://tex.z-dn.net/?f=x%3C60%5C%5C%5C%5C%2860-x%29%3D%5Csqrt%7B6400%2F45%7D%3D11.92%5C%5C%5C%5Cx%3D60-11.926%3D48.074)
As
, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.
Answer is in the attachment below.
-1<f(x)<1
The Answer Is No Solution!