Calculus Examples

$\frac{d r}{d θ} = - π \sin (π θ)$ , $r (1) = 3$

Step 1

Rewrite the equation.

$d r = - π \sin (π θ) d θ$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d r = \int - π \sin (π θ) d θ$

Step 2.2

Apply the constant rule.

$r + C_{1} = \int - π \sin (π θ) d θ$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $- π$ is constant with respect to $θ$ , move $- π$ out of the integral.

$r + C_{1} = - π \int \sin (π θ) d θ$

Step 2.3.2

Let $u = π θ$ . Then $d u = π d θ$ , so $\frac{1}{π} d u = d θ$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

Let $u = π θ$ . Find $\frac{d u}{d θ}$ .

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Step 2.3.2.1.1

Differentiate $π θ$ .

$\frac{d}{d θ} [π θ]$

Step 2.3.2.1.2

Since $π$ is constant with respect to $θ$ , the derivative of $π θ$ with respect to $θ$ is $π \frac{d}{d θ} [θ]$ .

$π \frac{d}{d θ} [θ]$

Step 2.3.2.1.3

Differentiate using the Power Rule which states that $\frac{d}{d θ} [θ^{n}]$ is $n θ^{n - 1}$ where $n = 1$ .

$π \cdot 1$

Step 2.3.2.1.4

Multiply $π$ by $1$ .

$π$

Step 2.3.2.2

Rewrite the problem using $u$ and $d u$ .

$r + C_{1} = - π \int \sin (u) \frac{1}{π} d u$

Step 2.3.3

Combine $\sin (u)$ and $\frac{1}{π}$ .

$r + C_{1} = - π \int \frac{\sin (u)}{π} d u$

Step 2.3.4

Since $\frac{1}{π}$ is constant with respect to $u$ , move $\frac{1}{π}$ out of the integral.

$r + C_{1} = - π (\frac{1}{π} \int \sin (u) d u)$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{π}$ and $π$ .

$r + C_{1} = - \frac{π}{π} \int \sin (u) d u$

Step 2.3.5.2

Cancel the common factor of $π$ .

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Step 2.3.5.2.1

Cancel the common factor.

$r + C_{1} = - \frac{π}{π} \int \sin (u) d u$

Step 2.3.5.2.2

Rewrite the expression.

$r + C_{1} = - 1 \cdot 1 \int \sin (u) d u$

Step 2.3.5.3

Multiply $- 1$ by $1$ .

$r + C_{1} = - \int \sin (u) d u$

Step 2.3.6

The integral of $\sin (u)$ with respect to $u$ is $- \cos (u)$ .

$r + C_{1} = - (- \cos (u) + C_{2})$

Step 2.3.7

Simplify.

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Step 2.3.7.1

Simplify.

$r + C_{1} = - - \cos (u) + C_{2}$

Step 2.3.7.2

Simplify.

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Step 2.3.7.2.1

Multiply $- 1$ by $- 1$ .

$r + C_{1} = 1 \cos (u) + C_{2}$

Step 2.3.7.2.2

Multiply $\cos (u)$ by $1$ .

$r + C_{1} = \cos (u) + C_{2}$

Step 2.3.8

Replace all occurrences of $u$ with $π θ$ .

$r + C_{1} = \cos (π θ) + C_{2}$

Step 2.4

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

$r = \cos (π θ) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $1$ for $θ$ and $3$ for $r$ in $r = \cos (π θ) + K$ .

$3 = \cos (π \cdot 1) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $\cos (π \cdot 1) + K = 3$ .

$\cos (π \cdot 1) + K = 3$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $π$ by $1$ .

$\cos (π) + K = 3$

Step 4.2.1.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$- \cos (0) + K = 3$

Step 4.2.1.3

The exact value of $\cos (0)$ is $1$ .

$- 1 \cdot 1 + K = 3$

Step 4.2.1.4

Multiply $- 1$ by $1$ .

$- 1 + K = 3$

Step 4.3

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

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Step 4.3.1

Add $1$ to both sides of the equation.

$K = 3 + 1$

Step 4.3.2

Add $3$ and $1$ .

$K = 4$

Step 5

Substitute $4$ for $K$ in $r = \cos (π θ) + K$ and simplify.

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Step 5.1

Substitute $4$ for $K$ .

$r = \cos (π θ) + 4$