The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is
![\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cmathrm%20ds%20%3D%20%5Cint_a%5Eb%20%5Csqrt%7B%5Cleft%28%5Cfrac%7B%5Cmathrm%20dx%7D%7B%5Cmathrm%20dt%7D%5Cright%29%5E2%2B%5Cleft%28%5Cfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D%5Cright%29%5E2%7D%20%5C%2C%5Cmathrm%20dt)
In this case, we have
<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> ) ==> d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )
<em>y(t)</em> = 5 - 2<em>t</em> ==> d<em>y</em>/d<em>t</em> = -2
and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then
![\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E2%20%5Csqrt%7B%5Cleft%28e%5Et-e%5E%7B-t%7D%5Cright%29%5E2%2B%28-2%29%5E2%7D%20%5C%2C%5Cmathrm%20dt%20%3D%20%5Cint_0%5E2%20%5Csqrt%7Be%5E%7B2t%7D-2%2Be%5E%7B-2t%7D%2B4%7D%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%20%5Csqrt%7Be%5E%7B2t%7D%2B2%2Be%5E%7B-2t%7D%7D%20%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Csqrt%7B%5Cleft%28e%5Et%2Be%5E%7B-t%7D%5Cright%29%5E2%7D%20%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Cleft%28e%5Et%2Be%5E%7B-t%7D%5Cright%29%5C%2C%5Cmathrm%20dt)
![=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}](https://tex.z-dn.net/?f=%3D%5Cleft%28e%5Et-e%5E%7B-t%7D%5Cright%29%5Cbigg%7C_0%5E2%20%3D%20%5Cleft%28e%5E2-e%5E%7B-2%7D%5Cright%29-%5Cleft%28e%5E0-e%5E%7B-0%7D%5Cright%29%20%3D%20%5Cboxed%7Be%5E2-%5Cfrac1%7Be%5E2%7D%7D)
They are supplementary angles. Aka equal to 180 degrees.
Answer:
H
Step-by-step explanation:
It appears to be just a slope problem in disguise. Where if you find the slope, you find the density of nickel. In this case, it would be copper. The slope appears to be closer to 9, and if you take some of the points and divide it, like where it seems to cross 4 and 35, you would get an answer close to 9. Same with 40 and 4.5, you get 8.89. The only answer closest to that would be H