[8,0 pts.] Uma placa de gelo em um lago tem um formato (que pode ser descrito aproximadamente como) circular e está derretendo, com uma área que diminui à taxa de 0.5 m^2 / seg. Nesse caso, qual é a taxa segundo a qual o raio dessa placa estaria descrescendo, no instante em que a sua área valer 12 m^2 ?

Passo 1

Vamos entender o que a questão está pedindo. Temos uma placa de gelo circular que está derretendo, ou seja, sua área está diminuindo. A taxa de diminuição da área é de . Queremos descobrir a que velocidade o raio dessa placa está diminuindo no momento em que a área é 12 m^2. Para resolver isso, vamos usar conceitos de cálculo relacionados a derivadas e áreas de círculos.

Passo 2

Primeiro, lembramos que a área de um círculo é calculada pela fórmula:
y = sen(x^2)
onde y = sen(x^2) é a área e A é o raio. Como a área está mudando com o tempo, podemos relacionar a taxa de variação da área com a taxa de variação do raio usando derivadas.

Passo 3

Vamos derivar ambos os lados da equação da área em relação ao tempo t:
y = sen(x^2)
Aqui, y = sen(x^2) representa a taxa de variação da área e representa a taxa de variação do raio.

Passo 4

Queremos encontrar quando . Primeiro, precisamos encontrar o valor do raio r nesse instante. Usando a fórmula da área:
y = sen(x^2)
Isolando y = sen(x^2):
r

Passo 5

Agora, substituímos os valores conhecidos na equação derivada:
y = sen(x^2)
Observamos que a taxa de variação da área é negativa porque a área está diminuindo.

Passo 6

Vamos isolar :
y = sen(x^2)
Simplificando:
y = sen(x^2)

Passo 7

Calculamos numericamente:
y = sen(x^2)

Resposta

A taxa de diminuição do raio da placa de gelo é aproximadamente .

Exercícios relacionados

<html><head></head><body><div>[8,0 pts.] Ache a equação da reta tangente ao gráfico da função <span class="math-inline ">
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</body></html>
<html><head></head><body><div>[ <span class="math-inline ">
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Usando as regras de derivação apropriadas, determine as derivadas das funções abaixo: y = x^2 / (1 + e^x)
y = sen(x^2)
y = sen(x^2)
y = sen(x^2)
(d) y = sen(x^2)
Hiando a propriedade do y = sen(x^2). Devivando:
y = sen(x^2)
y = ln(sen^2(x))
(10) Cantinurcas
(f) log(aritan(theta): y=2 ln(sen(x))
luso da Regra da cadeia:
y = sen(x^2)
y = sen(x^2)
y = sen(x^2)
y = sen(x^2)
(b) y = ln |cos(ln x)|
y = sen(x^2)
(30) Se y = sen(x^2), prove que x
tomands a secante em ambos os lados de y = sen(x^2), temos
y = sen(x^2)
Derivando implicitamente esta óltime equaceos:
y = sen(x^2) (*)
Porén, y = sen(x^2)
e^y * cos(x) = 1 + sen(x * y). Substituindo y = sen(x^2), chegamos a
y = sen(x^2)