# Make the internet a better place to learn

1

## How do you differentiate y=sin(xy)?

Bill K.
Featured 1 week ago

BTW, the graph of this equation is very beautiful

#### Explanation:

Here's the graph:

1

## How do you find solution for int_0^∞x^9e^(-x^5)dx ?

Featured 1 week ago

The answer is $= \frac{1}{5}$

#### Explanation:

This is an improper integral

Start by calculating the definite integral

Let $u = {x}^{5}$, $\implies$, $\mathrm{du} = 5 {x}^{4} \mathrm{dx}$

Therefore,

$\int {x}^{9} {e}^{- {x}^{5}} \mathrm{dx} = \frac{1}{5} \int u {e}^{- u} \mathrm{du}$

Perform the integration by parts

$p = u$, $\implies$, $p ' = 1$

$q ' = {e}^{-} u$, $\implies$, $q = - {e}^{-} u$

Therefore,

$\int p q ' = p q - \int p ' q$

$\frac{1}{5} \int u {e}^{- u} \mathrm{du} = \frac{1}{5} \left(- u {e}^{-} u + \int {e}^{-} u \mathrm{du}\right) = \frac{1}{5} \left(- u {e}^{-} u - {e}^{-} u\right)$

$= - {e}^{-} \frac{{x}^{5}}{5} \left({x}^{5} + 1\right)$

Therefore,

${\lim}_{t \to + \infty} {\int}_{0}^{t} {x}^{9} {e}^{- {x}^{5}} \mathrm{dx} = {\lim}_{t \to + \infty} {\left[\frac{- {e}^{- {t}^{5}}}{5} \left({t}^{5} + 1\right)\right]}_{0}^{\infty}$

$= {\lim}_{t \to + \infty} \left(- {e}^{- {t}^{5}} / 5 \left({t}^{5} + 1\right)\right) + \frac{1}{5}$

${\lim}_{t \to + \infty} \left(\frac{{t}^{5} + 1}{5 {e}^{{t}^{5}}}\right) + \frac{1}{5}$

$= 0 + \frac{1}{5}$

The integral converges

1

## What's the solution? lim_(nto oo) ((n^2+3)/(2n^2+4))^(n^2)

Cesareo R.
Featured 1 week ago

$0$

#### Explanation:

Making $x = {n}^{2}$

${\lim}_{x \to \infty} {\left(\frac{x + 3}{2 x + 4}\right)}^{x} = {\lim}_{x \to \infty} {\left(\frac{1 + \frac{3}{x}}{2 + \frac{4}{x}}\right)}^{x}$ and for $x$ very large

$\frac{1 + \frac{3}{x}}{2 + \frac{4}{x}} \approx \frac{1}{2}$ then

${\lim}_{x \to \infty} {\left(\frac{x + 3}{2 x + 4}\right)}^{x} = {\lim}_{x \to \infty} {\left(\frac{1}{2}\right)}^{x} = 0$

1

## Lim x->0 (tan x- sin x)/x^3=?

Andrea S.
Featured 5 days ago

${\lim}_{x \to 0} \frac{\tan x - \sin x}{x} ^ 3 = \frac{1}{2}$

#### Explanation:

Transform the function in this way:

$\frac{\tan x - \sin x}{x} ^ 3 = \frac{1}{x} ^ 3 \left(\sin \frac{x}{\cos} x - \sin x\right)$

$\frac{\tan x - \sin x}{x} ^ 3 = \frac{1}{x} ^ 3 \left(\frac{\sin x - \sin x \cos x}{\cos} x\right)$

$\frac{\tan x - \sin x}{x} ^ 3 = \sin \frac{x}{x} ^ 3 \frac{1 - \cos x}{\cos} x$

$\frac{\tan x - \sin x}{x} ^ 3 = \left(\sin \frac{x}{x}\right) \left(\frac{1 - \cos x}{x} ^ 2\right) \left(\frac{1}{\cos} x\right)$

We can use now the well known trigonometric limit:

${\lim}_{x \to 0} \sin \frac{x}{x} = 1$

and using the trigonometric identity:

${\sin}^{2} \alpha = \frac{1 - \cos 2 \alpha}{2}$

we have:

${\lim}_{x \to 0} \frac{1 - \cos x}{x} ^ 2 = {\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(\frac{x}{2}\right)}{x} ^ 2 = \frac{1}{2} {\lim}_{x \to 0} {\left(\sin \frac{\frac{x}{2}}{\frac{x}{2}}\right)}^{2} = \frac{1}{2}$

While the third function is continuous so:

${\lim}_{x \to 0} \frac{1}{\cos} x = \frac{1}{1} = 1$

and we can conclude that:

${\lim}_{x \to 0} \frac{\tan x - \sin x}{x} ^ 3 = {\lim}_{x \to 0} \left(\sin \frac{x}{x}\right) \left(\frac{1 - \cos x}{x} ^ 2\right) \left(\frac{1}{\cos} x\right) = 1 \times \frac{1}{2} \times 1 = \frac{1}{2}$

graph{(tanx-sinx)/x^3 [-1.25, 1.25, -0.025, 1]}

1

## If u and v are two vectors having the same magnitude, how do I show that u + v and u - v are orthogonal?

Featured 5 days ago

See the proof below

#### Explanation:

Let $\vec{u} = < a , b , c >$ in the basis $\beta = \left(\hat{i} , \hat{j} , \hat{k}\right)$

and $\vec{v} = < p , q , r >$

It is given that $| | \vec{u} | | = | | \vec{v} | |$

Then

$\vec{u} + \vec{v} = < a , b , c > + < p , q , r >$

$= < a + p , b + q , c + r >$

$\vec{u} - \vec{v} = < a , b , c > - < p , q , r >$

$= < a - p , b - q , c - r >$

The dot product is

$\left(\vec{u} + \vec{v}\right) . \left(\vec{u} - \vec{v}\right)$

$= < a + p , b + q , c + r > . < a - p , b - q , c - r >$

$= \left(a + p\right) \left(a - p\right) + \left(b + q\right) \left(b - q\right) + \left(c + r\right) \left(c - r\right)$

$= {a}^{2} - {p}^{2} + {b}^{2} - {q}^{2} + {c}^{2} - {r}^{2}$

$= \left({a}^{2} + {b}^{2} + {c}^{2}\right) - \left({p}^{2} + {q}^{2} + {r}^{2}\right)$

$= | | \vec{u} | {|}^{2} - | | \vec{v} | {|}^{2} = 0$

Therefore,

$\left(\vec{u} + \vec{v}\right)$ and $\left(\vec{u} - \vec{v}\right)$ are orthogonal since their dots products $= 0$

1

## If f'(x) = -8x on what intervals is f concave down?

HSBC244
Featured 3 days ago

The function is concave down on $\left(- \infty , \infty\right)$.

#### Explanation:

Concavity is determined by the second derivative.

$f ' ' \left(x\right) = - 8$

This is negative on all $x$, thus it will be concave down on all $x$.

Hopefully this helps!

2

## Lim to infinity of (5x^2-3x+4)/(x-1)^2?

mizoo
Featured 3 days ago

lim_(x→∞) f(x) = 5

#### Explanation:

lim_(x→∞) (5x^2-3x+4)/(x-1)^2 =

lim_(x→∞) (5x^2-3x+4)/(x^2-2x+1) =

lim_(x→∞) (5x^2((5x^2)/(5x^2)-(3x)/(5x^2)+4/(5x^2)))/(x^2(x^2/x^2-(2x)/x^2+1/x^2)) =

lim_(x→∞) (5x^2(1-(3)/(5x)+4/(5x^2)))/(x^2(1-(2)/x+1/x^2)) =

lim_(x→∞) (5x^2(1-0+0))/(x^2(1-0+0)) = (5x^2)/x^2 = 5

1

## (calculus)At what rate is the angle shown changing at that instant?

Jason K.
Featured 3 days ago

$\frac{d \theta}{\mathrm{dt}} = \frac{37}{20}$

#### Explanation:

You were on the correct track with your setup on number 8, but when you took the derivative of the $\frac{y}{z}$ term on the right side with respect to time $t$, you did not use the Quotient Rule to perform the derivative, which rendered the rest of the problem incorrect.

The proper setup is $\sin \theta = \frac{y}{z}$ as you noted. From there:

$\left(\cos \theta\right) \frac{d \theta}{\mathrm{dt}} = \frac{z \cdot \frac{\mathrm{dy}}{\mathrm{dt}} - y \cdot \frac{\mathrm{dz}}{\mathrm{dt}}}{{z}^{2}}$

It is difficult to read the image included, but I believe the setup involves $y = 8$, $z = 10$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 10$, $\frac{\mathrm{dz}}{\mathrm{dt}} = - 6$, and $\cos \theta = \frac{8}{10}$, which means you now have:

$\left(\cos \theta\right) \frac{d \theta}{\mathrm{dt}} = \frac{z \cdot \frac{\mathrm{dy}}{\mathrm{dt}} - y \cdot \frac{\mathrm{dz}}{\mathrm{dt}}}{{z}^{2}}$

$\left(\frac{8}{10}\right) \frac{d \theta}{\mathrm{dt}} = \frac{10 \left(10\right) - 8 \left(- 6\right)}{{10}^{2}}$

$\left(\frac{4}{5}\right) \frac{d \theta}{\mathrm{dt}} = \frac{100 + 48}{100}$

$\left(\frac{4}{5}\right) \frac{d \theta}{\mathrm{dt}} = \frac{148}{100}$

$\frac{d \theta}{\mathrm{dt}} = \frac{37}{25} \cdot \frac{5}{4} = \frac{37}{20}$

A quick sanity check on the sign of the result can help catch errors as they are in progress. As the ships continue moving, ship A will get closer to the lighthouse, while ship B will move farther away.

This action will serve to make the angle $\theta$ become larger and larger, much like how the angle made by a ladder to the ground increases as you push a ladder up the side of a house. (By the way, this is a common alternative word problem that tests the exact same math in many Calculus classes.)

Your original attempt ended with a negative answer, which would indicate the angle is decreasing - a result that contradicts the nature of the situation. It might not help you find the issue, but it can at least give you a hint that you're on the wrong track somewhere.

1

## The ideal gas law states pv=nrt, find how fast p is changing when the volume is 10 m^3, p=20 pascals, and volume is decreasing at a rate of 2 m^3 per min, and temperature is increasing at a rate of 4 kelvin per min. How do I solve this?

Jason K.
Featured 2 days ago

$\frac{\mathrm{dP}}{\mathrm{dt}} = \frac{2}{5} n R + 4$

#### Explanation:

In the Ideal Gas Law formula, there are 3 variables and 2 constants. $P$ represents the pressure (in this problem in units of pascals), $V$ represents the volume (in this problem in units of ${m}^{3}$), and $T$ represents the temperature (in this problem in Kelvin). We consider $n$ to be the amount of gas present (usually in moles), while $R$ is the Ideal Gas Constant, or a fixed value.

This is a "classic" related rates problem. We need to take derivatives of every variable ($P , V , T$) with respect to time $t$. To do so, we perform "traditional" derivatives on each of those terms in the equation, but we ensure that we include a $\frac{\mathrm{dP}}{\mathrm{dt}}$, $\frac{\mathrm{dV}}{\mathrm{dt}}$, or $\frac{\mathrm{dT}}{\mathrm{dt}}$ term as necessary.

$P V = n R T$

$P {\underbrace{\left(1 \cdot \frac{\mathrm{dV}}{\mathrm{dt}}\right)}}_{\text{d(V)" + Vunderbrace((1*(dP)/(dt)))_ "d(P)" = nRunderbrace((1*(dT)/(dt)))_ "d(T)}}$

$P \left(\frac{\mathrm{dV}}{\mathrm{dt}}\right) + V \left(\frac{\mathrm{dP}}{\mathrm{dt}}\right) = n R \left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)$

We have all of the values we need provided to us to substitute into this related rates expression:

$\left\{\begin{matrix}P & 20 \text{pascals" & "Pressure" \\ (dV)/(dt) & -2 m^3/min & "Volume Change" \\ V & 10 m^3 & "Volume" \\ (dP)/(dt) & ? & "Pres. Change" \\ (dT)/(dt) & 4 K/min & "Temp Change}\end{matrix}\right.$

$\left(20\right) \left(- 2\right) + \left(10\right) \left(\frac{\mathrm{dP}}{\mathrm{dt}}\right) = n R \left(4\right)$

$- 40 + 10 \left(\frac{\mathrm{dP}}{\mathrm{dt}}\right) = 4 n R$

$10 \left(\frac{\mathrm{dP}}{\mathrm{dt}}\right) = 4 n R + 40$

$\frac{\mathrm{dP}}{\mathrm{dt}} = \frac{4 n R + 40}{10} = \frac{2}{5} n R + 4$

A quick "sanity check" verifies that the sign of our answer is what we'd expect. Since P is inversely related to V and directly related to T, if the volume is decreasing, and the temperature is increasing, we'd definitely expect the pressure P to be increasing (positive). Since $n$ and $R$ are both positive constants, the final expression will be positive.

1

## How do i solve \int xln \frac{x+1}{1-x}dx?

Cem Sentin
Featured 2 days ago

${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - {x}^{2} / 2 - \frac{1}{2} \cdot L n \left({x}^{2} - 1\right) + C$

#### Explanation:

$\int x \cdot \ln \left[\frac{x + 1}{1 - x}\right] \cdot \mathrm{dx}$

After setting $\mathrm{dv} = x \cdot \mathrm{dx}$ an $u = L n \left[\frac{x + 1}{1 - x}\right] = L n \left(x + 1\right) - L n \left(1 - x\right)$,

$v = {x}^{2} / 2$ and $\mathrm{du} = \frac{\mathrm{dx}}{x + 1} - \frac{- \mathrm{dx}}{1 - x} = \frac{\mathrm{dx}}{x + 1} + \frac{\mathrm{dx}}{x - 1} = \frac{2 x \cdot \mathrm{dx}}{{x}^{2} - 1}$

So,

$\int u \cdot \mathrm{dv} = u v - \int v \cdot \mathrm{du}$

$\int x \cdot \ln \left[\frac{x + 1}{1 - x}\right] \cdot \mathrm{dx} = {x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int {x}^{2} / 2 \cdot \frac{2 x \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int \frac{{x}^{3} \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int \frac{\left({x}^{3} - x + x\right) \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int \frac{\left({x}^{3} - x\right) \cdot \mathrm{dx}}{{x}^{2} - 1} - \int \frac{x \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int x \cdot \mathrm{dx} - \frac{1}{2} \cdot \int \frac{2 x \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - {x}^{2} / 2 - \frac{1}{2} \cdot L n \left({x}^{2} - 1\right) + C$