How can self-study lead to long-term learning compared to classroom instruction?

[harpazo]

As I study calculus 3, I often revisit calculus 1 and 2. The following application is from single variable calculus, partucularly calculus 1, called RELATED RATES. I have not seen a related rates problem since the 2015. I am a bit rusty with the set up. Can someone help me set it up? I can take it from there.

A balloon, initially of radius 1 cm, is being inflated in such a way that its radius at time t seconds is (1+2t) cm. What is the rate of increase of its volume when t = 2?

 

[MarkFL]

As I study calculus 3, I often revisit calculus 1 and 2. The following application is from single variable calculus, partucularly calculus 1, called RELATED RATES. I have not seen a related rates problem since the 2015. I am a bit rusty with the set up. Can someone help me set it up? I can take it from there.

A balloon, initially of radius 1 cm, is being inflated in such a way that its radius at time t seconds is (1+2t) cm. What is the rate of increase of its volume when t = 2?

If we assume the balloon is spherical, then we can begin with the volume of a sphere:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

Differentiating w.r.t time $t$, we obtain:

\(\displaystyle \d{V}{t}=4\pi r^2\d{r}{t}\)

We are told:

\(\displaystyle r=2t+1\implies \d{r}{t}=2\)

So, can you now express the time rate of change of volume as a function of $t$?

 

[harpazo]

If we assume the balloon is spherical, then we can begin with the volume of a sphere:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

Differentiating w.r.t time $t$, we obtain:

\(\displaystyle \d{V}{t}=4\pi r^2\d{r}{t}\)

We are told:

\(\displaystyle r=2t+1\implies \d{r}{t}=2\)

So, can you now express the time rate of change of volume as a function of $t$?

Are you saying to solve r = 2t + 1 for t?

 

[MarkFL]

Are you saying to solve r = 2t + 1 for t?

No, just plug in for $r$ and \(\displaystyle \d{r}{t}\) in the expression for the time rate of change of volume. :D

 

[harpazo]

r = 2t + 1

r = 2(2) = 1

r = 5

dv/dt = 4πr^2 (dr/dt)

dv/dtb= 4π(5)^2(2)

dv/dt = 4π50

dv/dt = 200π cm/second

Correct?

 

[MarkFL]

Plugging in, we obtain

\(\displaystyle \d{V}{t}=4\pi(2t+1)^2(2)=8\pi(2t+1)^2\)

Hence, in \(\displaystyle \frac{\text{cm}^3}{\text{s}}\), we have:

\(\displaystyle \left.\d{V}{t}\right|_{t=2}=8\pi(2(2)+1)^2=200\pi\quad\checkmark\)

 

[harpazo]

Plugging in, we obtain

\(\displaystyle \d{V}{t}=4\pi(2t+1)^2(2)=8\pi(2t+1)^2\)

Hence, in \(\displaystyle \frac{\text{cm}^3}{\text{s}}\), we have:

\(\displaystyle \left.\d{V}{t}\right|_{t=2}=8\pi(2(2)+1)^2=200\pi\quad\checkmark\)

I got it right. Cool. I love related rates. I wish I had a total understanding of the set up. Back to calculus 3. I am in the triple integral chapter. By the way, I test myself at the end of each chapter. The passing grade is 70 percent per chapter. If my grade is less than 70, I must repeat the chapter. This is my method. What do you say?

 

[MarkFL]

...By the way, I test myself at the end of each chapter. The passing grade is 70 percent per chapter. If my grade is less than 70, I must repeat the chapter. This is my method. What do you say?

I say whatever works best for you is fine by me. :)

To save time, I would likely only review the sections pertaining to the problems I missed.

 

[harpazo]

I say whatever works best for you is fine by me. :)

To save time, I would likely only review the sections pertaining to the problems I missed.

Most people do not remember what they had for breakfast this morning much less what they learned in calculus long ago. This is my problem. I am learning calculus 3 now. I am learning triple integrals. But, to be honest, I cannot recall most of calculus 2 not to mention related rates. I have not played with related rates since 2015.

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How can I remember what I studied years ago? I am now learning triple integrals. Next month, double integrals will become a distant chapter. There are people in this website that can answer precalculus through calculus 3 like drinking water. How is that possible? The human mind can only remember so much.

 

[I like Serena]

Most people do not remember what they had for breakfast this morning much less what they learned in calculus long ago. This is my problem. I am learning calculus 3 now. I am learning triple integrals. But, to be honest, I cannot recall most of calculus 2 not to mention related rates. I have not played with related rates since 2015.

- - - Updated - - -

How can I remember what I studied years ago? I am now learning triple integrals. Next month, double integrals will become a distant chapter. There are people in this website that can answer precalculus through calculus 3 like drinking water. How is that possible? The human mind can only remember so much.

When someone tells me something new, I have already forgotten pretty much what they're saying before they even finish their sentence.
If someone teaches me something and let's me practice with it, it'll stay for a couple of weeks or months.
If I figure something out for myself, preferably the 'hard' way, it's there for the rest of my life.

 

[harpazo]

When someone tells me something new, I have already forgotten pretty much what they're saying before they even finish their sentence.
If someone teaches me something and let's me practice with it, it'll stay for a couple of weeks or months.
If I figure something out for myself, preferably the 'hard' way, it's there for the rest of my life.

I am the same way. Through self-study and hard work, I learn so much. Through classroom instructions, I grasp very little.