Calculus Examples

$\frac{d r}{d θ} = \frac{r θ + r}{r θ + θ}$ , $r (1) = e$

Step 1

Separate the variables.

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Step 1.1

Factor $r$ out of $r θ + r$ .

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Step 1.1.1

Factor $r$ out of $r θ$ .

$\frac{d r}{d θ} = \frac{r (θ) + r}{r θ + θ}$

Step 1.1.2

Raise $r$ to the power of $1$ .

$\frac{d r}{d θ} = \frac{r (θ) + r^{1}}{r θ + θ}$

Step 1.1.3

Factor $r$ out of $r^{1}$ .

$\frac{d r}{d θ} = \frac{r (θ) + r \cdot 1}{r θ + θ}$

Step 1.1.4

Factor $r$ out of $r (θ) + r \cdot 1$ .

$\frac{d r}{d θ} = \frac{r (θ + 1)}{r θ + θ}$

Step 1.2

Factor $θ$ out of $r θ + θ$ .

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Step 1.2.1

Factor $θ$ out of $r θ$ .

$\frac{d r}{d θ} = \frac{r (θ + 1)}{θ r + θ}$

Step 1.2.2

Raise $θ$ to the power of $1$ .

$\frac{d r}{d θ} = \frac{r (θ + 1)}{θ r + θ^{1}}$

Step 1.2.3

Factor $θ$ out of $θ^{1}$ .

$\frac{d r}{d θ} = \frac{r (θ + 1)}{θ r + θ \cdot 1}$

Step 1.2.4

Factor $θ$ out of $θ r + θ \cdot 1$ .

$\frac{d r}{d θ} = \frac{r (θ + 1)}{θ (r + 1)}$

Step 1.3

Regroup factors.

$\frac{d r}{d θ} = \frac{r}{r + 1} \cdot \frac{θ + 1}{θ}$

Step 1.4

Multiply both sides by $\frac{r + 1}{r}$ .

$\frac{r + 1}{r} \frac{d r}{d θ} = \frac{r + 1}{r} (\frac{r}{r + 1} \cdot \frac{θ + 1}{θ})$

Step 1.5

Simplify.

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Step 1.5.1

Multiply $\frac{r}{r + 1}$ by $\frac{θ + 1}{θ}$ .

$\frac{r + 1}{r} \frac{d r}{d θ} = \frac{r + 1}{r} \cdot \frac{r (θ + 1)}{(r + 1) θ}$

Step 1.5.2

Cancel the common factor of $r + 1$ .

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Step 1.5.2.1

Factor $r + 1$ out of $(r + 1) θ$ .

$\frac{r + 1}{r} \frac{d r}{d θ} = \frac{r + 1}{r} \cdot \frac{r (θ + 1)}{(r + 1) (θ)}$

Step 1.5.2.2

Cancel the common factor.

$\frac{r + 1}{r} \frac{d r}{d θ} = \frac{r + 1}{r} \cdot \frac{r (θ + 1)}{(r + 1) θ}$

Step 1.5.2.3

Rewrite the expression.

$\frac{r + 1}{r} \frac{d r}{d θ} = \frac{1}{r} \cdot \frac{r (θ + 1)}{θ}$

Step 1.5.3

Cancel the common factor of $r$ .

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Step 1.5.3.1

Cancel the common factor.

$\frac{r + 1}{r} \frac{d r}{d θ} = \frac{1}{r} \cdot \frac{r (θ + 1)}{θ}$

Step 1.5.3.2

Rewrite the expression.

$\frac{r + 1}{r} \frac{d r}{d θ} = \frac{θ + 1}{θ}$

Step 1.6

Rewrite the equation.

$\frac{r + 1}{r} d r = \frac{θ + 1}{θ} d θ$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{r + 1}{r} d r = \int \frac{θ + 1}{θ} d θ$

Step 2.2

Integrate the left side.

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Step 2.2.1

Split the fraction into multiple fractions.

$\int \frac{r}{r} + \frac{1}{r} d r = \int \frac{θ + 1}{θ} d θ$

Step 2.2.2

Split the single integral into multiple integrals.

$\int \frac{r}{r} d r + \int \frac{1}{r} d r = \int \frac{θ + 1}{θ} d θ$

Step 2.2.3

Cancel the common factor of $r$ .

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Step 2.2.3.1

Cancel the common factor.

$\int \frac{r}{r} d r + \int \frac{1}{r} d r = \int \frac{θ + 1}{θ} d θ$

Step 2.2.3.2

Rewrite the expression.

$\int d r + \int \frac{1}{r} d r = \int \frac{θ + 1}{θ} d θ$

Step 2.2.4

Apply the constant rule.

$r + C_{1} + \int \frac{1}{r} d r = \int \frac{θ + 1}{θ} d θ$

Step 2.2.5

The integral of $\frac{1}{r}$ with respect to $r$ is $\ln (| r |)$ .

$r + C_{1} + \ln (| r |) + C_{2} = \int \frac{θ + 1}{θ} d θ$

Step 2.2.6

Simplify.

$r + \ln (| r |) + C_{3} = \int \frac{θ + 1}{θ} d θ$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the fraction into multiple fractions.

$r + \ln (| r |) + C_{3} = \int \frac{θ}{θ} + \frac{1}{θ} d θ$

Step 2.3.2

Split the single integral into multiple integrals.

$r + \ln (| r |) + C_{3} = \int \frac{θ}{θ} d θ + \int \frac{1}{θ} d θ$

Step 2.3.3

Cancel the common factor of $θ$ .

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Step 2.3.3.1

Cancel the common factor.

$r + \ln (| r |) + C_{3} = \int \frac{θ}{θ} d θ + \int \frac{1}{θ} d θ$

Step 2.3.3.2

Rewrite the expression.

$r + \ln (| r |) + C_{3} = \int d θ + \int \frac{1}{θ} d θ$

Step 2.3.4

Apply the constant rule.

$r + \ln (| r |) + C_{3} = θ + C_{4} + \int \frac{1}{θ} d θ$

Step 2.3.5

The integral of $\frac{1}{θ}$ with respect to $θ$ is $\ln (| θ |)$ .

$r + \ln (| r |) + C_{3} = θ + C_{4} + \ln (| θ |) + C_{5}$

Step 2.3.6

Simplify.

$r + \ln (| r |) + C_{3} = θ + \ln (| θ |) + C_{6}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$r + \ln (| r |) = θ + \ln (| θ |) + C$

Step 3

Use the initial condition to find the value of $C$ by substituting $1$ for $θ$ and $e$ for $r$ in $r + \ln (| r |) = θ + \ln (| θ |) + C$ .

$e + \ln (| e |) = 1 + \ln (| 1 |) + C$

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $1 + \ln (| 1 |) + C = e + \ln (| e |)$ .

$1 + \ln (| 1 |) + C = e + \ln (| e |)$

Step 4.2

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| 1 |) - \ln (| e |) = - 1 - C + e$

Step 4.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| 1 |}{| e |}) = - 1 - C + e$

Step 4.4

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\ln (\frac{1}{| e |}) = - 1 - C + e$

Step 4.5

$e$ is approximately $2.71828182$ which is positive so remove the absolute value

$\ln (\frac{1}{e}) = - 1 - C + e$

Step 4.6

Rewrite $\frac{1}{e}$ as $1 e^{- 1}$ .

$\ln (1 e^{- 1}) = - 1 - C + e$

Step 4.7

Rewrite $\ln (1 e^{- 1})$ as $\ln (1) + \ln (e^{- 1})$ .

$\ln (1) + \ln (e^{- 1}) = - 1 - C + e$

Step 4.8

Use logarithm rules to move $- 1$ out of the exponent.

$\ln (1) - \ln (e) = - 1 - C + e$

Step 4.9

The natural logarithm of $e$ is $1$ .

$\ln (1) - 1 \cdot 1 = - 1 - C + e$

Step 4.10

Multiply $- 1$ by $1$ .

$\ln (1) - 1 = - 1 - C + e$

Step 4.11

The natural logarithm of $1$ is $0$ .

$0 - 1 = - 1 - C + e$

Step 4.12

Subtract $1$ from $0$ .

$- 1 = - 1 - C + e$

Step 4.13

Since $C$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- 1 - C + e = - 1$

Step 4.14

Move all terms not containing $C$ to the right side of the equation.

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Step 4.14.1

Add $1$ to both sides of the equation.

$- C + e = - 1 + 1$

Step 4.14.2

Subtract $e$ from both sides of the equation.

$- C = - 1 + 1 - e$

Step 4.14.3

Add $- 1$ and $1$ .

$- C = 0 - e$

Step 4.14.4

Subtract $e$ from $0$ .

$- C = - e$

Step 4.15

Divide each term in $- C = - e$ by $- 1$ and simplify.

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Step 4.15.1

Divide each term in $- C = - e$ by $- 1$ .

$\frac{- C}{- 1} = \frac{- e}{- 1}$

Step 4.15.2

Simplify the left side.

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Step 4.15.2.1

Dividing two negative values results in a positive value.

$\frac{C}{1} = \frac{- e}{- 1}$

Step 4.15.2.2

Divide $C$ by $1$ .

$C = \frac{- e}{- 1}$

Step 4.15.3

Simplify the right side.

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Step 4.15.3.1

Dividing two negative values results in a positive value.

$C = \frac{e}{1}$

Step 4.15.3.2

Divide $e$ by $1$ .

$C = e$

Step 5

Substitute $e$ for $C$ in $r + \ln (| r |) = θ + \ln (| θ |) + C$ and simplify.

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Step 5.1

Substitute $e$ for $C$ .

$r + \ln (| r |) = θ + \ln (| θ |) + e$