True or False? Eliminating the parameter loses information about the curve.

True. The Cartesian equation records the shape but not the direction of travel, nor the order in which the points are reached. Those are only visible from the parameter.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 9: Parametric Equations, Vector-Valued Functions & Polar CoordinatesParametric Equations

Parametric Equations (College Board AP® Calculus BC): Flashcards

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FrontDefining Parametric Equations
Define parametric equations.

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Cards in this collection (33)

Define parametric equations.
A way of writing a curve using a third variable called a parameter, usually $t$ .
The curve is given by a pair of equations $x = f (t)$ and $y = g (t)$ .
How do you find the direction of flow along a parametric curve?
Work out the coordinates at two values of $t$ and see which way the point moves as $t$ increases.
The curve is traced in that direction.
A parametric curve has $x = f (t)$ and $y = g (t)$ . How do you find its $y$ -intercepts?
Solve $f (t) = 0$ , since the $y$ -axis is where $x = 0$ .
Substituting each solution into $g (t)$ gives the intercept, provided that value of $t$ is in range.
To eliminate the parameter from $x = t - 5$ and $y = t^{2} + e^{t}$ , make $t$ the subject of the first equation and substitute it into the second. The result is
$y = {(______)}^{2} + e^{x + 5}$
The completed equation is $y = {(x + 5)}^{2} + e^{x + 5}$ .
Rearranging $x = t - 5$ gives $t = x + 5$ , which then replaces $t$ everywhere in the equation for $y$ .
How can a trigonometric identity eliminate the parameter?
Rearrange each equation for $\sin t$ and $\cos t$ , then substitute into $\sin^{2} t + \cos^{2} t = 1$ .
From $x = 2 + \sin t$ and $y = 4 \cos t$ this gives ${(x - 2)}^{2} + {(\frac{y}{4})}^{2} = 1$ .
True or False?
Eliminating the parameter loses information about the curve.
True.
The Cartesian equation records the shape but not the direction of travel, nor the order in which the points are reached.
Those are only visible from the parameter.
Can the parameter always be eliminated?
No.
Curves such as $x = t + e^{t}$ and $y = \sin t + \ln t$ cannot be rearranged to remove $t$ .
Even where it works, the result may not be solvable for $y$ .
The curve $x = 3 - e^{t}$ , $y = 2 + t$ has $0 \leq t \leq 2$ . Write it as $y = f (x)$ with an $x$ -range.
Rearranging gives $t = \ln (3 - x)$ , so $y = 2 + \ln (3 - x)$ .
The endpoints $t = 0$ and $t = 2$ turn the range into $3 - e^{2} \leq x \leq 2$ .
A curve is given by $x = f (t)$ and $y = g (t)$ . The slope of the tangent to the curve is
$\frac{d y}{d x} = \frac{______}{\frac{d x}{d t}}$
The completed formula is $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ .
It comes from the chain rule, $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$ , with the last factor written as the reciprocal of $\frac{d x}{d t}$ .
What condition must hold before you can use the parametric first derivative?
$\frac{d x}{d t}$ must not be zero at the point concerned, so that the quotient does not divide by zero.
Find $\frac{d y}{d x}$ at $t = 4$ for the curve $x = t + e^{t}$ and $y = t^{2} + 3$ .
Differentiate each equation separately to get $\frac{d x}{d t} = 1 + e^{t}$ and $\frac{d y}{d t} = 2 t$ .
Then $\frac{d y}{d x} = \frac{2 t}{1 + e^{t}}$ , which at $t = 4$ is $\frac{8}{1 + e^{4}}$ .
When does a parametric curve have a horizontal tangent?
When $\frac{d y}{d t} = 0$ and $\frac{d x}{d t} \neq 0$ at that value of $t$ .
The quotient is then zero.
When does a parametric curve have a vertical tangent?
When $\frac{d x}{d t} = 0$ and $\frac{d y}{d t} \neq 0$ at that value of $t$ .
The quotient becomes infinite, and the tangent is a line $x = k$ .
True or False?
If $\frac{d x}{d t}$ and $\frac{d y}{d t}$ are both zero, the curve has a vertical tangent there.
False.
With both derivatives zero the quotient is $\frac{0}{0}$ , which says nothing at all.
The limit of $\frac{d y}{d x}$ has to be investigated separately.
The curve $x = t^{3} - 12 t$ , $y = t^{3} + 3 t$ has vertical tangents. Find their equations.
Setting $3 t^{2} - 12 = 0$ gives $t = \pm 2$ , and $\frac{d y}{d t} = 15$ there, which is not zero.
Substituting into $x$ gives the tangents $x = - 16$ and $x = 16$ .
The parametric second derivative is found by differentiating the first derivative with respect to $t$ , then dividing. The formula is
$\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{______}$
The completed formula is $\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}$ .
Both the first and the second parametric derivative divide by $\frac{d x}{d t}$ , which is a useful pattern to remember.
True or False?
The parametric second derivative is $\frac{d^{2} y / d t^{2}}{d^{2} x / d t^{2}}$ .
False.
You must differentiate the first derivative $\frac{d y}{d x}$ with respect to $t$ , then divide by $\frac{d x}{d t}$ .
Differentiating the top and bottom of the first-derivative quotient separately gives something else entirely.
Where does the parametric second derivative formula come from?
From the chain rule applied to $\frac{d}{d x} (\frac{d y}{d x})$ , rewritten as $\frac{d}{d t} (\frac{d y}{d x}) \times \frac{d t}{d x}$ .
The last factor is the reciprocal of $\frac{d x}{d t}$ .
Which differentiation rule is usually needed for the numerator?
The quotient rule, because $\frac{d y}{d x}$ is itself a quotient of two functions of $t$ .
A parametric curve has $\frac{d y}{d x} = 0$ at $t = t_{0}$ . How do you classify that point?
Substitute $t_{0}$ into $\frac{d^{2} y}{d x^{2}}$ .
A positive value gives a local minimum and a negative value a local maximum, while zero means further investigation is needed.
For $x = t^{2} + t$ and $y = t^{3} - 3 t$ , find $\frac{d^{2} y}{d x^{2}}$ .
The first derivative is $\frac{3 t^{2} - 3}{2 t + 1}$ .
Differentiating it by the quotient rule gives $\frac{6 t^{2} + 6 t + 6}{{(2 t + 1)}^{2}}$ , and dividing by $2 t + 1$ gives $\frac{6 (t^{2} + t + 1)}{{(2 t + 1)}^{3}}$ .
The length of the parametric curve $x = f (t)$ , $y = g (t)$ from $t = t_{1}$ to $t = t_{2}$ is
$L = \int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(______)}^{2}} d t$
The completed formula is $L = \int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$ .
Both parametric derivatives are squared and added, and the square root of the total is integrated.
What are the limits in a parametric arc length integral?
Values of the parameter $t$ , not $x$ -coordinates.
The integration is with respect to $t$ , so the limits have to match.
True or False?
A parametric curve must be converted to $y = f (x)$ before its arc length can be found.
False.
The parametric formula works directly on $\frac{d x}{d t}$ and $\frac{d y}{d t}$ .
Eliminating the parameter first is unnecessary and often impossible.
Show that the length of $x = 1 - \cos t$ , $y = t - \sin t$ from $0$ to $\frac{π}{2}$ is $\sqrt{2} \int_{0}^{\frac{π}{2}} \sqrt{1 - \cos t} d t$ .
The derivatives are $\sin t$ and $1 - \cos t$ , so the integrand is $\sqrt{\sin^{2} t + 1 - 2 \cos t + \cos^{2} t}$ .
The Pythagorean identity turns this into $\sqrt{2 - 2 \cos t}$ , and factoring out the $2$ gives the result.
Find the length of $x = 3 t + 1$ , $y = 2 t^{\frac{3}{2}}$ from $t = 0$ to $t = 8$ .
The derivatives are $3$ and $3 t^{\frac{1}{2}}$ , giving $\int_{0}^{8} \sqrt{9 + 9 t} d t = 3 \int_{0}^{8} \sqrt{1 + t} d t$ .
This evaluates to $52$ units.
A particle moves in the $x y$ -plane with position $(x (t), y (t))$ . What do $x (t)$ and $y (t)$ represent?
The displacement in the $x$ -direction and the displacement in the $y$ -direction, each measured from a fixed origin.
Together they give the particle's coordinates at time $t$ .
How is a particle's speed in the plane related to its velocity components?
Speed is the magnitude of the velocity, so it is the hypotenuse of a right triangle with legs $\frac{d x}{d t}$ and $\frac{d y}{d t}$ .
That gives $\sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}}$ .
The total distance a particle travels in the plane from $t_{1}$ to $t_{2}$ is the arc length of its path, given by
$\int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}}______$
The completed formula is $\int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$ .
The integrand is the particle's speed, so integrating speed across the time interval gives the total distance traveled, which is always positive.
True or False?
The definite integral of a particle's velocity in the $x$ -direction gives the distance it travels.
False.
It gives the change in the $x$ -coordinate, which can be negative.
Total distance is the integral of speed and is never negative.
What does it mean for a particle in the plane to be at rest?
Its speed is zero, which requires $\frac{d x}{d t}$ and $\frac{d y}{d t}$ both to be zero at that instant.
One component being zero on its own is not enough.
A particle has $\frac{d x}{d t} = 6 t^{2} - 2 t$ and is at $x = 50$ when $t = 3$ . Find $x$ in terms of $t$ .
Integrating gives $x = 2 t^{3} - t^{2} + C$ .
Substituting $t = 3$ and $x = 50$ gives $50 = 45 + C$ , so $C = 5$ and $x = 2 t^{3} - t^{2} + 5$ .
A particle has $\frac{d x}{d t} = 6 t^{2} - 2 t$ and $\frac{d y}{d t} = \sqrt{4 e^{t} + 1}$ . Show it is not initially at rest.
At $t = 0$ the two components are $0$ and $\sqrt{5}$ , so the speed is $\sqrt{0 + 5} = 2.236$ .
That is not zero, so the particle is moving.

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