Exercicios Calculo (26)

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33 
 
𝑥2 + (−𝑥)2 = 64 
2𝑥2 = 64 
𝑥2 = 32 
𝑥 = ±4√2 
∴ 𝑦 = ∓4√2 
𝑃𝑜𝑛𝑡𝑜𝑠: 𝐴(4√2, −4√2) 𝑒 𝐵(−4√2,4√2). 
 
𝐶𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑎𝑛𝑔𝑢𝑙𝑎𝑟𝑒𝑠 𝑑𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 à 𝑐𝑢𝑟𝑣𝑎 𝑒𝑚 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠: 
 
𝑃𝑜𝑟 𝑑𝑒𝑟𝑖𝑣𝑎çã𝑜 𝑖𝑚𝑝𝑙í𝑐𝑖𝑡𝑎, 𝑡𝑒𝑚𝑜𝑠: 
𝐶1:𝑥
2 + 𝑦2 = 64 
𝑑
𝑑𝑥
(𝑥2) +
𝑑
𝑑𝑥
(𝑦2) =
𝑑
𝑑𝑥
(64) 
2𝑥 + 2𝑦.
𝑑𝑦
𝑑𝑥
= 0 
𝑑𝑦
𝑑𝑥
= −
𝑥
𝑦
 
𝑑𝑦
𝑑𝑥
|
(4√2,−4√2)
= −
4√2
(−4√2)
= 1 ; 
𝑑𝑦
𝑑𝑥
|
(−4√2,4√2)
= −
(−4√2)
4√2
= 1 
 
𝐶2:𝑥 + 𝑦 = 0 
𝑑
𝑑𝑥
(𝑥)+
𝑑
𝑑𝑥
(𝑦) =
𝑑
𝑑𝑥
(0) 
1 +
𝑑𝑦
𝑑𝑥
= 0 
𝑑𝑦
𝑑𝑥
= −1 
𝑑𝑦
𝑑𝑥
|
(4√2,−4√2)
= −1 ; 
𝑑𝑦
𝑑𝑥
|
(−4√2,4√2)
= −1 
 
𝐶𝑜𝑚𝑜 𝑛𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝐴 𝑒 𝐵 𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑎𝑛𝑔𝑢𝑙𝑎𝑟𝑒𝑠 𝑑𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 à 𝑐𝑢𝑟𝑣𝑎 
𝑥2 + 𝑦2 = 64 𝑠ã𝑜 𝑖𝑔𝑢𝑎𝑖𝑠 𝑎 1 𝑒 à 𝑐𝑢𝑟𝑣𝑎 𝑥+ 𝑦 = 0 𝑖𝑔𝑢𝑎𝑖𝑠 𝑎 − 1 𝑜 𝑝𝑟𝑜𝑑𝑢𝑡𝑜 𝑑𝑒𝑠𝑠𝑒𝑠 
𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟𝑒𝑠 é 𝑖𝑔𝑢𝑎𝑙 à − 1 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 à 𝑒𝑠𝑠𝑎𝑠 
𝑐𝑢𝑟𝑣𝑎𝑠 𝑛𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒çã𝑜 𝑠ã𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟𝑒𝑠 (𝑜𝑟𝑡𝑜𝑔𝑜𝑛𝑎𝑖𝑠). 
 
 
(𝐼𝑙𝑢𝑠𝑡𝑟𝑎çã𝑜 𝑑𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜)