Recent questions in Calculus 2

Calculus 2Answered question

nuseldW4r 2022-12-03

Why is Taylor series expansion for $1/(1-x)$ valid only for $x\in (-1,1)$ ?

Calculus 2Answered question

alistianyStu 2022-12-02

Prove that the equation ${4}^{x}=8x+1$ has only one solution

Calculus 2Answered question

Julius Ho 2022-12-02

Consider the differential equation $\frac{dy}{dt}=ay-b$.

Find the equilibrium solution ${y}_{e}$

Find the equilibrium solution ${y}_{e}$

Calculus 2Answered question

e3r2a1cakCh7 2022-12-01

Let f be a differentiable function such that f(3) = 2 and f'(3) = 5. If the tangent line to the graph of f at x = 3 is used to find an approximation to a zero of f, that approximation is?

Calculus 2Open question

hsupaing moe2022-11-29

A Norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window. Find the dimensions of a Norman window of maximum area when the total perimeter is 16 feet.

Calculus 2Answered question

Brenda Leach 2022-11-29

Solve the given initial-value problem. The DE is homogeneous.

$x{y}^{2}\frac{dy}{dx}=\frac{{y}^{3}}{{x}^{3}}$

$x{y}^{2}\frac{dy}{dx}=\frac{{y}^{3}}{{x}^{3}}$

Calculus 2Answered question

Naomi Rowland 2022-11-27

Let $w=\frac{yz}{x}$ where $x={t}^{2},\text{}y=r+t$ and $z=r-t$. Find $\frac{\mathrm{\partial}w}{\mathrm{\partial}t}$ and $\frac{\mathrm{\partial}w}{\mathrm{\partial}r}$ by using Chain Rule.

Calculus 2Answered question

Harmony Oneal 2022-11-25

Suppose $\sum _{n=1}^{\mathrm{\infty}}{x}_{n}<\mathrm{\infty}$. Then prove that $\sum _{n=1}^{\mathrm{\infty}}{x}_{n}{y}_{n}$ converges.

Calculus 2Answered question

Jamir Summers 2022-11-25

How to prove that $\underset{n\to \mathrm{\infty}}{lim}\sqrt[n]{n}=1$ ?

Calculus 2Answered question

valahanyHcm 2022-11-25

I'm struggling somewhat to understand how to use implicit differentiation to solve the following equation:

$\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2})-x\mathrm{cot}y=-2y$

I figublack that the calculation requires the chain rule to differentiate the composite function, but I'm not sure how to 'remove' the y with respect to x from inside the composite function. My calculations are:

$\frac{dy}{dx}[\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2})-x\mathrm{cot}y]=\frac{dy}{dx}[-2y]$

$\frac{dy}{dx}[\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2})]=\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2}\cdot {y}^{\prime}(x))\cdot \mathrm{sin}({x}^{3}{y}^{2}\cdot {y}^{\prime}(x))\cdot 6{x}^{2}y\cdot {y}^{\prime}(x)$

This seems a bit long and convoluted. I'm also not sure how this will allow me to solve for ${y}^{\prime}(x)$. Carrying on...

$\frac{dy}{dx}[x\mathrm{cot}y]=-{\mathrm{csc}}^{2}y\cdot {y}^{\prime}(x)$

$\frac{dy}{dx}[-2y]=-2$

Is my calculation correct so far? This seems to be a very complex derivative. Any comments or feedback would be appreciated.

$\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2})-x\mathrm{cot}y=-2y$

I figublack that the calculation requires the chain rule to differentiate the composite function, but I'm not sure how to 'remove' the y with respect to x from inside the composite function. My calculations are:

$\frac{dy}{dx}[\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2})-x\mathrm{cot}y]=\frac{dy}{dx}[-2y]$

$\frac{dy}{dx}[\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2})]=\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2}\cdot {y}^{\prime}(x))\cdot \mathrm{sin}({x}^{3}{y}^{2}\cdot {y}^{\prime}(x))\cdot 6{x}^{2}y\cdot {y}^{\prime}(x)$

This seems a bit long and convoluted. I'm also not sure how this will allow me to solve for ${y}^{\prime}(x)$. Carrying on...

$\frac{dy}{dx}[x\mathrm{cot}y]=-{\mathrm{csc}}^{2}y\cdot {y}^{\prime}(x)$

$\frac{dy}{dx}[-2y]=-2$

Is my calculation correct so far? This seems to be a very complex derivative. Any comments or feedback would be appreciated.

Calculus 2Answered question

hEorpaigh3tR 2022-11-25

What is the value of $\frac{100}{0}$?

1) 0

2) 1

3) 6

4) Not defined

1) 0

2) 1

3) 6

4) Not defined

Calculus 2Answered question

Cherish Rivera 2022-11-24

What is the integral of log(z) over the unit circle?

I tried three times, but in the end I am concluding that it equals infinity, after parametrizing, making a substitution and integrating directly (since the Residue Theorem is not applicable, because the unit circle encloses an infinite amount of non-isolated singularities.)

Any ideas are welcome.

Thanks,

I tried three times, but in the end I am concluding that it equals infinity, after parametrizing, making a substitution and integrating directly (since the Residue Theorem is not applicable, because the unit circle encloses an infinite amount of non-isolated singularities.)

Any ideas are welcome.

Thanks,

Calculus 2Answered question

Lorena Becker 2022-11-24

My final for my introductory analysis course is tomorrow and my teacher gave us a list of possible theorems to prove. If anyone could please show me a proof for The Intermediate Value Theorem that is short and easy to follow, so even if I still cannot understand it I can at least memorize it. Also, I have looked through numerous texts and the internet, but they all seem to confuse me. I know that itis an insult to all you math experts to memorize proofs, but I am desperate at this point. Thank you

Calculus 2Answered question

Barrett Osborn 2022-11-23

Use the intermediate value theorem to prove that if $f:[1,2]\to R$ is a continuous function, that there is at least one number $c$ in the interval $(1,2)$ such that $f(c)=1/(1-c)+1/(2-c)$

This is a question for my intro calc class that I am having a hard time understanding.

This is a question for my intro calc class that I am having a hard time understanding.

Calculus 2Answered question

Talia Frederick 2022-11-23

I need to find $\frac{dy}{dx}$:

$\frac{d}{dx}{e}^{y}\mathrm{cos}x=1+\mathrm{sin}(xy)$

using implicit differentiation.

So far I've gotten to this point which I don't think I'm doing right because then I get stuck:

${e}^{y}(-\mathrm{sin}x)+(\mathrm{cos}x)({e}^{y})(\frac{dy}{dx})=\mathrm{cos}(xy)(x)(\frac{dy}{dx})+y(1)$

Am I right? Am I wrong? If I'm right I definitely have no clue where to go...and if I'm wrong...then again I have no idea what to do.

$\frac{d}{dx}{e}^{y}\mathrm{cos}x=1+\mathrm{sin}(xy)$

using implicit differentiation.

So far I've gotten to this point which I don't think I'm doing right because then I get stuck:

${e}^{y}(-\mathrm{sin}x)+(\mathrm{cos}x)({e}^{y})(\frac{dy}{dx})=\mathrm{cos}(xy)(x)(\frac{dy}{dx})+y(1)$

Am I right? Am I wrong? If I'm right I definitely have no clue where to go...and if I'm wrong...then again I have no idea what to do.

Calculus 2Answered question

Nicholas Hunter 2022-11-23

Suppose $f:\mathbb{R}\to \mathbb{R}$ is continuous and periodic with period 2a for some $a>0$; that is, $f(x)=f(x+2a)$ for all $x\in R$. Show there is some $c\in [0,a]$ such that $f(c)=f(c+a)$.

The only way I see to do this question is to apply the intermediate value theorem, but I just don't know how to apply it to this question. I know that $f(x)=f(x+2a)$. Therefore, if i can show that $f(c+a)=f(c+2a)$ or $f(c+a)-f(c+2a)=0$, I'd be done. But I just don't know where to go from there. I tried substituting $x=a$, getting $f(2a)-f(3a)$, but I can't show that's less than, equal to, or greater than 0. Any ideas?

The only way I see to do this question is to apply the intermediate value theorem, but I just don't know how to apply it to this question. I know that $f(x)=f(x+2a)$. Therefore, if i can show that $f(c+a)=f(c+2a)$ or $f(c+a)-f(c+2a)=0$, I'd be done. But I just don't know where to go from there. I tried substituting $x=a$, getting $f(2a)-f(3a)$, but I can't show that's less than, equal to, or greater than 0. Any ideas?

When you are dealing with any Calculus 2 homework, it is vital to have a look at the various questions and answers that will help you see whether you are correct in your approach to finding solutions. Even if you are dealing with analytical aspects of Calculus 2, it will be helpful as you are looking at provided equations and learn how the answers relate to original questions and problems specified.

Do not be afraid to take a look at the basic integration and related application if Calculus 2 does not sound clear or start with the Calculus 1 first.