The answer would be de = 14x but i’m not fully confident in the answer
X: 2/3.
Y: -1
Those are the values
Answer:
a) The sequence converges to 0
b) The lenght of the curve is ![\frac{1}{54}(217^{3/2}-37^{3/2})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B54%7D%28217%5E%7B3%2F2%7D-37%5E%7B3%2F2%7D%29)
Step-by-step explanation:
Consider the sequence ![a_n = \frac{-6n^6 + \sin^2(7n)}{n^7+11}](https://tex.z-dn.net/?f=a_n%20%3D%20%5Cfrac%7B-6n%5E6%20%2B%20%5Csin%5E2%287n%29%7D%7Bn%5E7%2B11%7D)
a) We will prove it using the sandwich lemma. Note that for all n
, then
![\frac{-6n^6 -1}{n^7+11}\leq\frac{-6n^6 + \sin^2(7n)}{n^7+11}\leq \frac{-6n^6 + 1}{n^7+11}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-6n%5E6%20-1%7D%7Bn%5E7%2B11%7D%5Cleq%5Cfrac%7B-6n%5E6%20%2B%20%5Csin%5E2%287n%29%7D%7Bn%5E7%2B11%7D%5Cleq%20%5Cfrac%7B-6n%5E6%20%2B%201%7D%7Bn%5E7%2B11%7D)
Note that the expressions on the left and the right hand side have a greater degree on the denominator than the one on the numerator. Then, by takint the limit n goes to infinty on both sides, we have that
![0 \leq\frac{-6n^6 + \sin^2(7n)}{n^7+11} \leq 0](https://tex.z-dn.net/?f=0%20%5Cleq%5Cfrac%7B-6n%5E6%20%2B%20%5Csin%5E2%287n%29%7D%7Bn%5E7%2B11%7D%20%5Cleq%200)
So, the sequence converges to 0.
b) The function
the formula of curve lenght is given by
![s = \int_a^b \sqrt[]{1+(f'(x))^2}dx](https://tex.z-dn.net/?f=s%20%3D%20%5Cint_a%5Eb%20%5Csqrt%5B%5D%7B1%2B%28f%27%28x%29%29%5E2%7Ddx)
in this case, a=1, b=6
Note that
. Then
. Take u = 1+36x. Then du= 36dx (i.e du/36 = dx). If x = 1, then u = 37 and if x = 6 then u = 217. So,
10%= .10
.10(11,800)= 1,180
1,180(8)= 9,440
11,800- 9,440= 2,360
= $2,360
I hope this helped.