Calculus Examples

$x = 1 - t^{2}$ , $y = t - 3$

Step 1

Set up the parametric equation for $x (t)$ to solve the equation for $t$ .

$x = 1 - t^{2}$

Step 2

Rewrite the equation as $1 - t^{2} = x$ .

$1 - t^{2} = x$

Step 3

Subtract $1$ from both sides of the equation.

$- t^{2} = x - 1$

Step 4

Divide each term in $- t^{2} = x - 1$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- t^{2} = x - 1$ by $- 1$ .

$\frac{- t^{2}}{- 1} = \frac{x}{- 1} + \frac{- 1}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{t^{2}}{1} = \frac{x}{- 1} + \frac{- 1}{- 1}$

Step 4.2.2

Divide $t^{2}$ by $1$ .

$t^{2} = \frac{x}{- 1} + \frac{- 1}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Move the negative one from the denominator of $\frac{x}{- 1}$ .

$t^{2} = - 1 \cdot x + \frac{- 1}{- 1}$

Step 4.3.1.2

Rewrite $- 1 \cdot x$ as $- x$ .

$t^{2} = - x + \frac{- 1}{- 1}$

Step 4.3.1.3

Divide $- 1$ by $- 1$ .

$t^{2} = - x + 1$

Step 5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$t = \pm \sqrt{- x + 1}$

Step 6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.1

First, use the positive value of the $\pm$ to find the first solution.

$t = \sqrt{- x + 1}$

Step 6.2

Next, use the negative value of the $\pm$ to find the second solution.

$t = - \sqrt{- x + 1}$

Step 6.3

The complete solution is the result of both the positive and negative portions of the solution.

$t = \sqrt{- x + 1}$

$t = - \sqrt{- x + 1}$

$t = \sqrt{- x + 1}$

$t = - \sqrt{- x + 1}$

Step 7

Replace $t$ in the equation for $y$ to get the equation in terms of $x$ .

$y = (\sqrt{- x + 1}, - \sqrt{- x + 1}) - 3$