Derivatives with Related Rates

up vote
1
down vote

favorite

The radius of a circle is increasing at a constant rate of $7 cm/min$. When the area of the circle is 16$pi$ $cm^2$, what is the rate of change of the area? Round your answer to three decimal places.

This is what I have so far:

$A=pi r^2$ $to$ dA/dr= 2$pi$r.

So, 16$pi$ $cm^2$ = $pi r^2$ -> r=4

$dr/dt$= 7 cm/min [Given]

Find $dA/dt$.

I know $dA/dt = (dA/dr)cdot(dr/dt)$.

So, I have $dA/dt$ = (2$pi$r)$cdot (7 cm/min $), that is $ dA/dt = 175.929$

I will appreciate any assistance with this. Thanks in advance!

edited Nov 20 at 14:29

asked Nov 20 at 14:15

Lola

156

add a comment |

up vote
1
down vote

favorite

The radius of a circle is increasing at a constant rate of $7 cm/min$. When the area of the circle is 16$pi$ $cm^2$, what is the rate of change of the area? Round your answer to three decimal places.

This is what I have so far:

$A=pi r^2$ $to$ dA/dr= 2$pi$r.

So, 16$pi$ $cm^2$ = $pi r^2$ -> r=4

$dr/dt$= 7 cm/min [Given]

Find $dA/dt$.

I know $dA/dt = (dA/dr)cdot(dr/dt)$.

So, I have $dA/dt$ = (2$pi$r)$cdot (7 cm/min $), that is $ dA/dt = 175.929$

I will appreciate any assistance with this. Thanks in advance!

edited Nov 20 at 14:29

asked Nov 20 at 14:15

Lola

156

add a comment |

up vote
1
down vote

favorite

The radius of a circle is increasing at a constant rate of $7 cm/min$. When the area of the circle is 16$pi$ $cm^2$, what is the rate of change of the area? Round your answer to three decimal places.

This is what I have so far:

$A=pi r^2$ $to$ dA/dr= 2$pi$r.

So, 16$pi$ $cm^2$ = $pi r^2$ -> r=4

$dr/dt$= 7 cm/min [Given]

Find $dA/dt$.

I know $dA/dt = (dA/dr)cdot(dr/dt)$.

So, I have $dA/dt$ = (2$pi$r)$cdot (7 cm/min $), that is $ dA/dt = 175.929$

I will appreciate any assistance with this. Thanks in advance!

edited Nov 20 at 14:29

asked Nov 20 at 14:15

Lola

156

The radius of a circle is increasing at a constant rate of $7 cm/min$. When the area of the circle is 16$pi$ $cm^2$, what is the rate of change of the area? Round your answer to three decimal places.

This is what I have so far:

$A=pi r^2$ $to$ dA/dr= 2$pi$r.

So, 16$pi$ $cm^2$ = $pi r^2$ -> r=4

$dr/dt$= 7 cm/min [Given]

Find $dA/dt$.

I know $dA/dt = (dA/dr)cdot(dr/dt)$.

So, I have $dA/dt$ = (2$pi$r)$cdot (7 cm/min $), that is $ dA/dt = 175.929$

I will appreciate any assistance with this. Thanks in advance!

calculus

edited Nov 20 at 14:29

asked Nov 20 at 14:15

Lola

156

edited Nov 20 at 14:29

asked Nov 20 at 14:15

Lola

156

edited Nov 20 at 14:29

asked Nov 20 at 14:15

Lola

156

asked Nov 20 at 14:15

Lola

156

asked Nov 20 at 14:15

Lola

156

add a comment |

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

When the area of the circle is $16pi, cm^2$ then the radius is $r$ is equal to $4, cm$.
Hence, according to your work,
$$frac{dA}{dt}=2pi rcdot frac{dr}{dt}=?; cm^2/min$$
What is the correct result?

edited Nov 20 at 14:24

answered Nov 20 at 14:18

Robert Z

91.3k1058129

Would it be 2π(4) ⋅ 7 = 56π = 175.929? Thank you very much for your help @Robert Z !!
– Lola
Nov 20 at 14:24

@Lola Yes, and don't forget the units $cm^2/min$.
– Robert Z
Nov 20 at 14:25

@Lola BTW, if you are new here, please take a few second for a tour: math.stackexchange.com/tour
– Robert Z
Nov 20 at 14:27

thanks for the link! It was a nice mini tour and covered the important details. I appreciate it. :)
– Lola
Nov 20 at 14:32

add a comment |

up vote
0
down vote

We have $r(t)=7t$, hence $A(t)= 49 pi t^2$, thus $A'(t)=98 pi t$.

Let $t_0$ so that $A(t_0)=16 pi$. This gives $t_0=4/7$. Hence $A'(t_0)=A'(4/7)=56$.

answered Nov 20 at 14:27

Fred

42.9k1643

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006357%2fderivatives-with-related-rates%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

When the area of the circle is $16pi, cm^2$ then the radius is $r$ is equal to $4, cm$.
Hence, according to your work,
$$frac{dA}{dt}=2pi rcdot frac{dr}{dt}=?; cm^2/min$$
What is the correct result?

edited Nov 20 at 14:24

answered Nov 20 at 14:18

Robert Z

91.3k1058129

Would it be 2π(4) ⋅ 7 = 56π = 175.929? Thank you very much for your help @Robert Z !!
– Lola
Nov 20 at 14:24

@Lola Yes, and don't forget the units $cm^2/min$.
– Robert Z
Nov 20 at 14:25

@Lola BTW, if you are new here, please take a few second for a tour: math.stackexchange.com/tour
– Robert Z
Nov 20 at 14:27

thanks for the link! It was a nice mini tour and covered the important details. I appreciate it. :)
– Lola
Nov 20 at 14:32

add a comment |

up vote
2
down vote

accepted

When the area of the circle is $16pi, cm^2$ then the radius is $r$ is equal to $4, cm$.
Hence, according to your work,
$$frac{dA}{dt}=2pi rcdot frac{dr}{dt}=?; cm^2/min$$
What is the correct result?

edited Nov 20 at 14:24

answered Nov 20 at 14:18

Robert Z

91.3k1058129

Would it be 2π(4) ⋅ 7 = 56π = 175.929? Thank you very much for your help @Robert Z !!
– Lola
Nov 20 at 14:24

@Lola Yes, and don't forget the units $cm^2/min$.
– Robert Z
Nov 20 at 14:25

@Lola BTW, if you are new here, please take a few second for a tour: math.stackexchange.com/tour
– Robert Z
Nov 20 at 14:27

thanks for the link! It was a nice mini tour and covered the important details. I appreciate it. :)
– Lola
Nov 20 at 14:32

add a comment |

up vote
2
down vote

accepted

When the area of the circle is $16pi, cm^2$ then the radius is $r$ is equal to $4, cm$.
Hence, according to your work,
$$frac{dA}{dt}=2pi rcdot frac{dr}{dt}=?; cm^2/min$$
What is the correct result?

edited Nov 20 at 14:24

answered Nov 20 at 14:18

Robert Z

91.3k1058129

When the area of the circle is $16pi, cm^2$ then the radius is $r$ is equal to $4, cm$.
Hence, according to your work,
$$frac{dA}{dt}=2pi rcdot frac{dr}{dt}=?; cm^2/min$$
What is the correct result?

edited Nov 20 at 14:24

answered Nov 20 at 14:18

Robert Z

91.3k1058129

edited Nov 20 at 14:24

answered Nov 20 at 14:18

Robert Z

91.3k1058129

answered Nov 20 at 14:18

Robert Z

91.3k1058129

answered Nov 20 at 14:18

Robert Z

91.3k1058129

Would it be 2π(4) ⋅ 7 = 56π = 175.929? Thank you very much for your help @Robert Z !!
– Lola
Nov 20 at 14:24

@Lola Yes, and don't forget the units $cm^2/min$.
– Robert Z
Nov 20 at 14:25

@Lola BTW, if you are new here, please take a few second for a tour: math.stackexchange.com/tour
– Robert Z
Nov 20 at 14:27

thanks for the link! It was a nice mini tour and covered the important details. I appreciate it. :)
– Lola
Nov 20 at 14:32

add a comment |

Would it be 2π(4) ⋅ 7 = 56π = 175.929? Thank you very much for your help @Robert Z !!
– Lola
Nov 20 at 14:24

@Lola Yes, and don't forget the units $cm^2/min$.
– Robert Z
Nov 20 at 14:25

@Lola BTW, if you are new here, please take a few second for a tour: math.stackexchange.com/tour
– Robert Z
Nov 20 at 14:27

thanks for the link! It was a nice mini tour and covered the important details. I appreciate it. :)
– Lola
Nov 20 at 14:32

Would it be 2π(4) ⋅ 7 = 56π = 175.929? Thank you very much for your help @Robert Z !!
– Lola
Nov 20 at 14:24

@Lola Yes, and don't forget the units $cm^2/min$.
– Robert Z
Nov 20 at 14:25

@Lola BTW, if you are new here, please take a few second for a tour: math.stackexchange.com/tour
– Robert Z
Nov 20 at 14:27

thanks for the link! It was a nice mini tour and covered the important details. I appreciate it. :)
– Lola
Nov 20 at 14:32

add a comment |

up vote
0
down vote

We have $r(t)=7t$, hence $A(t)= 49 pi t^2$, thus $A'(t)=98 pi t$.

Let $t_0$ so that $A(t_0)=16 pi$. This gives $t_0=4/7$. Hence $A'(t_0)=A'(4/7)=56$.

answered Nov 20 at 14:27

Fred

42.9k1643

add a comment |

up vote
0
down vote

We have $r(t)=7t$, hence $A(t)= 49 pi t^2$, thus $A'(t)=98 pi t$.

Let $t_0$ so that $A(t_0)=16 pi$. This gives $t_0=4/7$. Hence $A'(t_0)=A'(4/7)=56$.

answered Nov 20 at 14:27

Fred

42.9k1643

add a comment |

up vote
0
down vote

We have $r(t)=7t$, hence $A(t)= 49 pi t^2$, thus $A'(t)=98 pi t$.

Let $t_0$ so that $A(t_0)=16 pi$. This gives $t_0=4/7$. Hence $A'(t_0)=A'(4/7)=56$.

answered Nov 20 at 14:27

Fred

42.9k1643

We have $r(t)=7t$, hence $A(t)= 49 pi t^2$, thus $A'(t)=98 pi t$.

Let $t_0$ so that $A(t_0)=16 pi$. This gives $t_0=4/7$. Hence $A'(t_0)=A'(4/7)=56$.

answered Nov 20 at 14:27

Fred

42.9k1643

answered Nov 20 at 14:27

Fred

42.9k1643

answered Nov 20 at 14:27

Fred

42.9k1643

answered Nov 20 at 14:27

Fred

42.9k1643

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Mathematics Stack Exchange!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Vrftsjtryk