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QuestãoColaborador
- Mensagens : 329
Inscrição : 08/09/2009
Área de formação : Pedagogia
Localização : Rio de Janeiro
![[EN/2006] Área lateral de um tanque com a forma de um cilindro circular reto Empty](https://2img.net/i/empty.gif)
[EN/2006] Área lateral de um tanque com a forma de um cilindro circular reto
Dom 06 Fev 2022, 14:39
Um tanque de combustível tem a forma de um cilindro circular reto e sua altura mede três metros. O raio da base do cilindro vale, em metros, o dobro da soma dos cubos dos inversos das raízes da equação: x4 + 4x3 + 8x2 + 8x + 4 = 0. A área lateral do tanque, em m², mede
A) 6pi
B) 12pi
C) 18pi
D) 36pi
E) 48pi
A) 6pi
B) 12pi
C) 18pi
D) 36pi
E) 48pi
- Gabarito:
- B
DanielFerreiraModerador
- Mensagens : 336
Inscrição : 29/08/2009
Idade : 38
Área de formação : Licenciatura em Matemática
Localização : Itacuruçá / Mangaratiba (RJ)
Re: [EN/2006] Área lateral de um tanque com a forma de um cilindro circular reto
Dom 06 Fev 2022, 19:44
A área lateral do cilindro em questão é dada por:
&space;%5Ccdot&space;h%7D%7D "\displaystyle \boxed{\mathtt{A_l = (2\pi r) \cdot h}}")
Onde r é o raio e h a altura do cilindro.
De acordo com o enunciado,
e
%7D%7D "\displaystyle \boxed{\mathtt{r = 2 \cdot \left ( \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3} + \frac{1}{x_4^3} \right )}}")
Onde
são raízes da equação 
.
Posto isto, determinemos a medida do raio. Segue,
![svg.image?\\&space;\displaystyle&space;\mathtt{x^4&space;+&space;4x^3&space;+&space;8x^2&space;+&space;8x&space;+&space;4&space;=&space;0}&space;\\&space;\mathtt{x^4&space;+&space;(4x^3&space;+&space;4)&space;+&space;(8x^2&space;+&space;8x)&space;=&space;0}&space;\\&space;\mathtt{x^4&space;+&space;4(x^3&space;+&space;1)&space;+&space;8x(x&space;+&space;1)&space;=&space;0}&space;\\&space;\mathtt{x^4&space;+&space;4(x&space;+&space;1)(x^2&space;-&space;x&space;+&space;1)&space;+&space;8x(x&space;+&space;1)&space;=&space;0}&space;\\&space;\mathtt{x^4&space;+&space;4(x&space;+&space;1)&space;\cdot&space;\left&space;[&space;(x^2&space;-&space;x&space;+&space;1)&space;+&space;2x&space;\right&space;]&space;=&space;0}&space;\\&space;\mathtt{x^4&space;+&space;4(x&space;+&space;1)&space;\cdot&space;(x^2&space;+&space;x&space;+&space;1)&space;=&space;0}](https://latex.codecogs.com/svg.image?%5C%5C&space;%5Cdisplaystyle&space;%5Cmathtt%7Bx%5E4&space;+&space;4x%5E3&space;+&space;8x%5E2&space;+&space;8x&space;+&space;4&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7Bx%5E4&space;+&space;(4x%5E3&space;+&space;4)&space;+&space;(8x%5E2&space;+&space;8x)&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7Bx%5E4&space;+&space;4(x%5E3&space;+&space;1)&space;+&space;8x(x&space;+&space;1)&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7Bx%5E4&space;+&space;4(x&space;+&space;1)(x%5E2&space;-&space;x&space;+&space;1)&space;+&space;8x(x&space;+&space;1)&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7Bx%5E4&space;+&space;4(x&space;+&space;1)&space;%5Ccdot&space;%5Cleft&space;%5B&space;(x%5E2&space;-&space;x&space;+&space;1)&space;+&space;2x&space;%5Cright&space;%5D&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7Bx%5E4&space;+&space;4(x&space;+&space;1)&space;%5Ccdot&space;(x%5E2&space;+&space;x&space;+&space;1)&space;=&space;0%7D "\\ \displaystyle \mathtt{x^4 + 4x^3 + 8x^2 + 8x + 4 = 0} \\ \mathtt{x^4 + (4x^3 + 4) + (8x^2 + 8x) = 0} \\ \mathtt{x^4 + 4(x^3 + 1) + 8x(x + 1) = 0} \\ \mathtt{x^4 + 4(x + 1)(x^2 - x + 1) + 8x(x + 1) = 0} \\ \mathtt{x^4 + 4(x + 1) \cdot \left [ (x^2 - x + 1) + 2x \right ] = 0} \\ \mathtt{x^4 + 4(x + 1) \cdot (x^2 + x + 1) = 0}")
Por conseguinte, tomemos
.
Com efeito,%5E2&space;%5C&space;e&space;%5C&space;x%5E4&space;=&space;(%5Clambda&space;-&space;1)%5E4%7D "\\ \displaystyle \mathtt{x^2 = (\lambda - 1)^2 \ e \ x^4 = (\lambda - 1)^4}")
.
Daí,
![svg.image?\\&space;\displaystyle&space;\mathtt{x^4&space;+&space;4(x&space;+&space;1)&space;\cdot&space;(x^2&space;+&space;x&space;+&space;1)&space;=&space;0}&space;\\&space;\mathtt{(\lambda&space;-&space;1)^4&space;+&space;4&space;\lambda&space;\left&space;[&space;(\lambda&space;-&space;1)^2&space;+&space;\lambda&space;\right&space;]&space;=&space;0}&space;\\&space;\mathtt{(\lambda&space;-&space;1)^4&space;+&space;4\lambda(\lambda&space;-&space;1)^2&space;+&space;4\lambda^2&space;=&space;0}&space;\\&space;\mathtt{\left&space;[&space;(\lambda&space;-&space;1)^2&space;+&space;2\lambda&space;\right&space;]^2&space;=&space;0}&space;\\&space;\mathtt{(\lambda^2&space;+&space;1)^2&space;=&space;0}](https://latex.codecogs.com/svg.image?%5C%5C&space;%5Cdisplaystyle&space;%5Cmathtt%7Bx%5E4&space;+&space;4(x&space;+&space;1)&space;%5Ccdot&space;(x%5E2&space;+&space;x&space;+&space;1)&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7B(%5Clambda&space;-&space;1)%5E4&space;+&space;4&space;%5Clambda&space;%5Cleft&space;%5B&space;(%5Clambda&space;-&space;1)%5E2&space;+&space;%5Clambda&space;%5Cright&space;%5D&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7B(%5Clambda&space;-&space;1)%5E4&space;+&space;4%5Clambda(%5Clambda&space;-&space;1)%5E2&space;+&space;4%5Clambda%5E2&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7B%5Cleft&space;%5B&space;(%5Clambda&space;-&space;1)%5E2&space;+&space;2%5Clambda&space;%5Cright&space;%5D%5E2&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7B(%5Clambda%5E2&space;+&space;1)%5E2&space;=&space;0%7D "\\ \displaystyle \mathtt{x^4 + 4(x + 1) \cdot (x^2 + x + 1) = 0} \\ \mathtt{(\lambda - 1)^4 + 4 \lambda \left [ (\lambda - 1)^2 + \lambda \right ] = 0} \\ \mathtt{(\lambda - 1)^4 + 4\lambda(\lambda - 1)^2 + 4\lambda^2 = 0} \\ \mathtt{\left [ (\lambda - 1)^2 + 2\lambda \right ]^2 = 0} \\ \mathtt{(\lambda^2 + 1)^2 = 0}")
![svg.image?\\&space;\displaystyle&space;\mathtt{(\lambda^2&space;-&space;i^2)^2&space;=&space;0}&space;\\&space;\mathtt{\left&space;[&space;(\lambda&space;+&space;i)&space;\cdot&space;(\lambda&space;-&space;i)&space;\right&space;]^2&space;=&space;0}&space;\\&space;\mathtt{(\lambda&space;+&space;i)^2&space;\cdot&space;(\lambda&space;-&space;i)^2&space;=&space;0}&space;\\&space;\boxed{\mathtt{S_\lambda&space;=&space;\left\{&space;\pm&space;i&space;\right\}}}](https://latex.codecogs.com/svg.image?%5C%5C&space;%5Cdisplaystyle&space;%5Cmathtt%7B(%5Clambda%5E2&space;-&space;i%5E2)%5E2&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7B%5Cleft&space;%5B&space;(%5Clambda&space;+&space;i)&space;%5Ccdot&space;(%5Clambda&space;-&space;i)&space;%5Cright&space;%5D%5E2&space;=&space;0%7D&space;%5C%5C&space;%5Cmathtt%7B(%5Clambda&space;+&space;i)%5E2&space;%5Ccdot&space;(%5Clambda&space;-&space;i)%5E2&space;=&space;0%7D&space;%5C%5C&space;%5Cboxed%7B%5Cmathtt%7BS_%5Clambda&space;=&space;%5Cleft%5C%7B&space;%5Cpm&space;i&space;%5Cright%5C%7D%7D%7D "\\ \displaystyle \mathtt{(\lambda^2 - i^2)^2 = 0} \\ \mathtt{\left [ (\lambda + i) \cdot (\lambda - i) \right ]^2 = 0} \\ \mathtt{(\lambda + i)^2 \cdot (\lambda - i)^2 = 0} \\ \boxed{\mathtt{S_\lambda = \left\{ \pm i \right\}}}")
Tendo em vista que mudamos a incógnita e estamos interessados nos valores de x, fazemos:
Isto é,
e 
Portanto,
%7D&space;%5C%5C%5C%5C&space;%5Cmathtt%7Br&space;=&space;2&space;%5Ccdot&space;%5Cleft&space;(&space;%5Cfrac%7B1%7D%7Bx_1%5E3%7D&space;+&space;%5Cfrac%7B1%7D%7Bx_2%5E3%7D&space;+&space;%5Cfrac%7B1%7D%7Bx_1%5E3%7D&space;+&space;%5Cfrac%7B1%7D%7Bx_2%5E3%7D&space;%5Cright&space;)%7D&space;%5C%5C%5C%5C&space;%5Cmathtt%7Br&space;=&space;2&space;%5Ccdot&space;%5Cleft&space;(&space;%5Cfrac%7B2%7D%7Bx_1%5E3%7D&space;+&space;%5Cfrac%7B2%7D%7Bx_2%5E3%7D&space;%5Cright&space;)%7D "\\ \displaystyle \mathtt{r = 2 \cdot \left ( \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3} + \frac{1}{x_4^3} \right )} \\\\ \mathtt{r = 2 \cdot \left ( \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_1^3} + \frac{1}{x_2^3} \right )} \\\\ \mathtt{r = 2 \cdot \left ( \frac{2}{x_1^3} + \frac{2}{x_2^3} \right )}")
%7D&space;%5C%5C%5C%5C&space;%5Cmathtt%7Br&space;=&space;4&space;%5Ccdot&space;%5Cfrac%7B(x_1%5E3&space;+&space;x_2%5E3)%7D%7Bx_1%5E3&space;%5Ccdot&space;x_2%5E3%7D%7D&space;%5C%5C%5C%5C&space;%5Cmathtt%7Br&space;=&space;4&space;%5Ccdot&space;%5Cfrac%7B(x_1&space;+&space;x_2)&space;%5Ccdot&space;(x_1%5E2&space;-&space;x_1&space;%5Ccdot&space;x_2&space;+&space;x_2%5E2)%7D%7B(x_1&space;%5Ccdot&space;x_2)%5E3%7D%7D "\\ \displaystyle \mathtt{r = 4 \cdot \left ( \frac{1}{x_1^3} + \frac{1}{x_2^3} \right )} \\\\ \mathtt{r = 4 \cdot \frac{(x_1^3 + x_2^3)}{x_1^3 \cdot x_2^3}} \\\\ \mathtt{r = 4 \cdot \frac{(x_1 + x_2) \cdot (x_1^2 - x_1 \cdot x_2 + x_2^2)}{(x_1 \cdot x_2)^3}}")
![svg.image?\\&space;\displaystyle&space;\mathtt{r&space;=&space;4&space;\cdot&space;\frac{(-&space;1&space;-&space;i&space;-&space;1&space;+&space;i)&space;\cdot&space;[(-&space;1&space;-&space;i)^2&space;-&space;(-&space;1&space;-&space;i)(-&space;1&space;+&space;i)&space;+&space;(-&space;1&space;+&space;i)^2]}{[(-&space;1&space;-&space;i)(-&space;1&space;+&space;i)]^3}}&space;\\\\&space;\mathtt{r&space;=&space;4&space;\cdot&space;\frac{(-&space;2)(1&space;+&space;2i&space;+&space;i^2&space;-&space;(1&space;-&space;i^2)&space;+&space;1&space;-&space;2i&space;+&space;i^2)}{(1&space;-&space;i^2)^3}}](https://latex.codecogs.com/svg.image?%5C%5C&space;%5Cdisplaystyle&space;%5Cmathtt%7Br&space;=&space;4&space;%5Ccdot&space;%5Cfrac%7B(-&space;1&space;-&space;i&space;-&space;1&space;+&space;i)&space;%5Ccdot&space;%5B(-&space;1&space;-&space;i)%5E2&space;-&space;(-&space;1&space;-&space;i)(-&space;1&space;+&space;i)&space;+&space;(-&space;1&space;+&space;i)%5E2%5D%7D%7B%5B(-&space;1&space;-&space;i)(-&space;1&space;+&space;i)%5D%5E3%7D%7D&space;%5C%5C%5C%5C&space;%5Cmathtt%7Br&space;=&space;4&space;%5Ccdot&space;%5Cfrac%7B(-&space;2)(1&space;+&space;2i&space;+&space;i%5E2&space;-&space;(1&space;-&space;i%5E2)&space;+&space;1&space;-&space;2i&space;+&space;i%5E2)%7D%7B(1&space;-&space;i%5E2)%5E3%7D%7D "\\ \displaystyle \mathtt{r = 4 \cdot \frac{(- 1 - i - 1 + i) \cdot [(- 1 - i)^2 - (- 1 - i)(- 1 + i) + (- 1 + i)^2]}{[(- 1 - i)(- 1 + i)]^3}} \\\\ \mathtt{r = 4 \cdot \frac{(- 2)(1 + 2i + i^2 - (1 - i^2) + 1 - 2i + i^2)}{(1 - i^2)^3}}")
(1&space;+&space;2i&space;+&space;i%5E2&space;-&space;(1&space;-&space;i%5E2)&space;+&space;1&space;-&space;2i&space;+&space;i%5E2)%7D%7B(1&space;-&space;i%5E2)%5E3%7D%7D&space;%5C%5C%5C%5C&space;%5Cmathtt%7Br&space;=&space;-&space;8&space;%5Ccdot&space;%5Cfrac%7B(-&space;1&space;+&space;i%5E2)%7D%7B(1&space;+&space;1)%5E3%7D%7D&space;%5C%5C%5C%5C&space;%5Cmathtt%7Br&space;=&space;-&space;8&space;%5Ccdot&space;%5Cfrac%7B(-&space;2)%7D%7B8%7D%7D&space;%5C%5C%5C%5C&space;%5Cboxed%7B%5Cboxed%7B%5Cmathtt%7Br&space;=&space;2&space;%5C,&space;m%7D%7D "\\ \displaystyle \mathtt{r = 4 \cdot \frac{(- 2)(1 + 2i + i^2 - (1 - i^2) + 1 - 2i + i^2)}{(1 - i^2)^3}} \\\\ \mathtt{r = - 8 \cdot \frac{(- 1 + i^2)}{(1 + 1)^3}} \\\\ \mathtt{r = - 8 \cdot \frac{(- 2)}{8}} \\\\ \boxed{\boxed{\mathtt{r = 2 \, m}}")
Por fim,
&space;%5Ccdot&space;h%7D&space;%5C%5C&space;%5Cmathtt%7BA_l&space;=&space;2%5Cpi&space;%5Ccdot&space;2&space;%5Ccdot&space;3%7D&space;%5C%5C&space;%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cmathtt%7BA_l&space;=&space;12%5Cpi&space;%5C,&space;m%5E2%7D%7D%7D%7D "\\ \displaystyle \mathtt{A_l = (2\pi r) \cdot h} \\ \mathtt{A_l = 2\pi \cdot 2 \cdot 3} \\ \boxed{\boxed{\boxed{\mathtt{A_l = 12\pi \, m^2}}}}")
Onde r é o raio e h a altura do cilindro.
De acordo com o enunciado,
Onde
Posto isto, determinemos a medida do raio. Segue,
Por conseguinte, tomemos
Com efeito,
Daí,
Tendo em vista que mudamos a incógnita e estamos interessados nos valores de x, fazemos:
Isto é,
Portanto,
Por fim,
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