The volume of a sphere is increasing at a constant rate of 2952 cubic centimeters per second. At the instant when the radius of the sphere is 1010 centimeters, what is the rate of change of the radius? The volume of a sphere can be found with the equation V=\frac{4}{3}\pi r^3.V= 3
4

πr
3
. Round your answer to three decimal places.

Step-by-step explanation:Let us set up the following variables:

enter image source here

Radius of sphere at time t

(cm)

A

Surface area of sphere at time t

(cm

2

)

V

Volume of sphere at time t

(cm

3

)

t

time

(sec)

Our aim is to find

d

A

d

t

when

V

=

9

π

2

and

d

V

d

t

=

π

3

.

The standard formula for Area & Volume of a sphere are:

V

=

4

3

π

r

3

...

.

[

1

]

A

=

4

π

r

2

...

.

[

2

]

When

V

=

9

π

2

⇒

4

3

π

r

3

=

9

π

2

∴

r

3

=

9

2

⋅

3

4

∴

r

=

3

2

Differentiating [1] and [2] wrt

r

we get;

d

V

d

r

=

4

π

r

2

and

d

A

d

r

=

8

π

r

And from the chain rule we get:

d

A

d

t

=

d

A

d

r

⋅

d

r

d

V

⋅

d

V

d

t

=

8

π

r

⋅

1

4

π

r

2

⋅

d

V

d

t

=

2

r

⋅

d

V

d

t

So when

V

=

9

π

2

,

d

V

d

t

=

π

3

and

r

=

3

2

, then:

d

A

d

t

=

2

3

2

⋅

π

3

=

4

π

9

Answer link

salamat

Feb 20, 2017

4

9

π

cm^2/s

Explanation:

V

=

4

3

π

⋅

r

3

, where V =volume of sphere and r =radius.

given that,

d

V

d

t

=

π

3

and

V

=

9

2

π

4

3

π

⋅

r

3

=

9

2

π

r

3

=

9

2

π

⋅

3

4

π

r

3

=

27

8

,

r

=

3

2

V

=

4

3

π

⋅

r

3

,

then

d

V

d

r

=

4

π

⋅

r

2

when

r

=

3

2

,

d

V

d

r

=

4

π

⋅

(

3

2

)

2

=

9

π

d

r

d

t

=

d

V

d

t

/

(

d

V

d

r

)

=

π

3

9

π

=

1

27

Area pf sphere,

A

=

4

π

r

2

d

A

d

r

=

8

π

r

when

r

=

3

2

,

d

A

d

r

=

8

π

r

=

8

π

(

3

2

)

=

12

π

therefore,

d

A

d

t

=

d

A

d

r

⋅

d

r

d

t

=

12

π

⋅

1

27

=

4

9

π

(

c

m

)

2

s

Answer link

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Let us set up the following variables:

enter image source here

⎧

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎩

r

Radius of sphere at time t

(cm)

A

Surface area of sphere at time t

(cm

2

)

V

Volume of sphere at time t

(cm

3

)

t

time

(sec)

Our aim is to find

d

A

d

t

when

V

=

9

π

2

and

d

V

d

t

=

π

3

.

The standard formula for Area & Volume of a sphere are:

V

=

4

3

π

r

3

...

.

[

1

]

A

=

4

π

r

2

...

.

[

2

]

When

V

=

9

π

2

⇒

4

3

π

r

3

=

9

π

2

∴

r

3

=

9

2

⋅

3

4

∴

r

=

3

2

Differentiating [1] and [2] wrt

r

we get;

d

V

d

r

=

4

π

r

2

and

d

A

d

r

=

8

π

r

And from the chain rule we get:

d

A

d

t

=

d

A

d

r

⋅

d

r

d

V

⋅

d

V

d

t

=

8

π

r

⋅

1

4

π

r

2

⋅

d

V

d

t

=

2

r

⋅

d

V

d

t

So when

V

=

9

π

2

,

d

V

d

t

=

π

3

and

r

=

3

2

, then:

d

A

d

t

=

2

3

2

⋅

π

3

=

4

π

9

Answer link

salamat

Feb 20, 2017

4

9

π

cm^2/s

Explanation:

V

=

4

3

π

⋅

r

3

, where V =volume of sphere and r =radius.

given that,

d

V

d

t

=

π

3

and

V

=

9

2

π

4

3

π

⋅

r

3

=

9

2

π

r

3

=

9

2

π

⋅

3

4

π

r

3

=

27

8

,

r

=

3

2

V

=

4

3

π

⋅

r

3

,

then

d

V

d

r

=

4

π

⋅

r

2

when

r

=

3

2

,

d

V

d

r

=

4

π

⋅

(

3

2

)

2

=

9

π

d

r

d

t

=

d

V

d

t

/

(

d

V

d

r

)

=

π

3

9

π

=

1

27

Area pf sphere,

A

=

4

π

r

2

d

A

d

r

=

8

π

r

when

r

=

3

2

,

d

A

d

r

=

8

π

r

=

8

π

(

3

2

)

=

12

π

therefore,

d

A

d

t

=

d

A

d

r

⋅

d

r

d

t

=

12

π

⋅

1

27

=

4

9

π

(

c

m

)

2

s

Answer link

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How do you determine the rate of change of a function?

What are partial derivatives?

How do you find a function f(x), which, when multiplied by its derivative, gives you

x

3

, and...

How do you graph the derivative of a function when you are given the graph of the function?

What is the purpose of a derivative?

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See all questions in Rate of Change of a Function

Impact of this question

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You can reuse this answer

Creative Commons License

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation:

70.30

step-by-step explanation: