IB Mathematics: Analysis and Approaches Quiz: Parametric Functions

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Parametric FunctionsQuestion 1 of 20

A curve is defined by the parametric equations $x = t^2$ and $y = t^3 - 3t$ . Find the value of $\frac{d^2y}{dx^2}$ at the point where $t=2$ .

\frac{9}{32}

\frac{15}{32}

\frac{15}{8}

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Parametric Functions

Practice Parametric Functions in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Parametric Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A curve is defined by the parametric equations $x = t^2$ and $y = t^3 - 3t$ . Find the value of $\frac{d^2y}{dx^2}$ at the point where $t=2$ .

$\frac{9}{32}$
$\frac{15}{32}$ (correct answer)
$\frac{15}{8}$
6

Explanation: First, find the first derivatives with respect to

t

\frac{dx}{dt} = 2t

\frac{dy}{dt} = 3t^2 - 3

Next, find

\frac{dy}{dx}

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 3}{2t} = \frac{3}{2}t - \frac{3}{2t}

Now, differentiate

\frac{dy}{dx}

with respect to

t

\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} + \frac{3}{2t^2}

Finally, use the formula for the second derivative:

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{3}{2} + \frac{3}{2t^2}}{2t} = \frac{3(t^2+1)}{2t^2} \cdot \frac{1}{2t} = \frac{3(t^2+1)}{4t^3}

Substitute

t=2

into the expression for

\frac{d^2y}{dx^2}

\frac{d^2y}{dx^2}\bigg|_{t=2} = \frac{3(2^2+1)}{4(2^3)} = \frac{3(5)}{4(8)} = \frac{15}{32}

A) $\frac{9}{32}$ : This result comes from a sign error when differentiating $-\frac{3}{2t}$ , getting $-\frac{3}{2t^2}$ instead of $+\frac{3}{2t^2}$ . This gives $\frac{d}{dt}(\frac{dy}{dx}) = \frac{3}{2} - \frac{3}{2t^2}$ , which is $\frac{9}{8}$ at $t=2$ . Then, dividing by $\frac{dx}{dt}=4$ gives $\frac{9}{32}$ .
C) $\frac{15}{8}$ : This is the value of $\frac{d}{dt}(\frac{dy}{dx})$ at $t=2$ , but without the final step of dividing by $\frac{dx}{dt}$ .
D) 6: This result comes from calculating the incorrect formula $\frac{d^2y/dt^2}{d^2x/dt^2}$ . Here, $\frac{d^2y}{dt^2} = 6t$ and $\frac{d^2x}{dt^2} = 2$ . At $t=2$ , this would be $\frac{12}{2} = 6$ .

Question 2

At what values of $t$ does the curve defined by $x=t^2-2t$ and $y=t^2+2t$ intersect the line $y=x$ ?

$t=0$ only (correct answer)
$t=-1$ only
$t=0$ and $t=1$
The curve never intersects the line.

Explanation: To find the points of intersection, we set

y=x

and substitute the parametric expressions:

y(t) = x(t)

t^2 + 2t = t^2 - 2t

Subtract

t^2

from both sides:

2t = -2t

Add

2t

to both sides:

4t = 0

t = 0

So, the curve intersects the line

y=x

only at

t=0

. The point of intersection is

x(0)=0^2-2(0)=0

and

y(0)=0^2+2(0)=0

, which is the origin

(0,0)

B) $t=-1$ only: This could be a miscalculation such as $t^2+2t=-(t^2-2t)$ , which gives $2t^2=0$ , still $t=0$ . A different error might lead to this.
C) $t=0$ and $t=1$ : This might result from an error like $t^2+2t-1 = t^2-2t$ , leading to $4t=1$ .
D) The curve never intersects the line.: This would be the answer if the equation $4t=0$ had no solution, which is incorrect.

Question 3

A curve is given by $x=t+\frac{1}{t}$ and $y=t-\frac{1}{t}$ for $t>0$ . Which of the following is the Cartesian equation of the curve?

$x^2+y^2=4$
$xy=1$
$y^2-x^2=4$
$x^2-y^2=4$ (correct answer)

Explanation: When you encounter parametric equations like these, your goal is to eliminate the parameter

t

to find the relationship between

x

and

y

. This process reveals the curve's Cartesian equation. Given

x = t + \frac{1}{t}

and

y = t - \frac{1}{t}

, let's find

x^2

and

y^2

x^2 = \left(t + \frac{1}{t}\right)^2 = t^2 + 2 + \frac{1}{t^2}

y^2 = \left(t - \frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2}

Now subtract:

x^2 - y^2 = (t^2 + 2 + \frac{1}{t^2}) - (t^2 - 2 + \frac{1}{t^2}) = 4

This gives us

x^2 - y^2 = 4

, which is answer D. Looking at the wrong answers: A)

x^2 + y^2 = 4

would result from incorrectly adding the squared expressions instead of subtracting them. B)

xy = 1

comes from a common mistake of trying to multiply the parametric equations directly, but

(t + \frac{1}{t})(t - \frac{1}{t}) = t^2 - \frac{1}{t^2}

, which isn't constant. C)

y^2 - x^2 = 4

reverses the correct subtraction order. Study tip: When eliminating parameters from equations involving

t + \frac{1}{t}

and

t - \frac{1}{t}

, always square both expressions and look for the difference of squares pattern. The cross terms will cancel beautifully, leaving you with a clean constant.

Question 4

The curve defined by $x = t^2 - 1$ and $y = t^3 - t$ intersects itself at a point P. Find the coordinates of P.

(-1, 0)
(0, 0) (correct answer)
(0, 2)
(3, 0)

Explanation: A self-intersection occurs at a point

(x, y)

that is generated by two different parameter values, say

t_1

and

t_2

, where

t_1 \neq t_2

. We need to solve the system of equations:

x(t_1) = x(t_2) \implies t_1^2 - 1 = t_2^2 - 1

y(t_1) = y(t_2) \implies t_1^3 - t_1 = t_2^3 - t_2

From the first equation:

t_1^2 = t_2^2 \implies t_1 = \pm t_2

Since

t_1 \neq t_2

, we must have

t_1 = -t_2

. Now substitute

t_2 = -t_1

into the second equation:

t_1^3 - t_1 = (-t_1)^3 - (-t_1)

t_1^3 - t_1 = -t_1^3 + t_1

2t_1^3 - 2t_1 = 0

2t_1(t_1^2 - 1) = 0

2t_1(t_1 - 1)(t_1 + 1) = 0

This gives possible values

t_1 = 0, 1, -1

. If

t_1 = 0

, then

t_2 = -0 = 0

, which means

t_1 = t_2

. This is not a self-intersection. If

t_1 = 1

, then

t_2 = -1

. These are distinct values. If

t_1 = -1

, then

t_2 = 1

. This is the same pair of values. So, the self-intersection occurs when

t = 1

and

t = -1

. Let's find the coordinates of the point P by substituting either value into the parametric equations: For

t = 1

x(1) = 1^2 - 1 = 0

y(1) = 1^3 - 1 = 0

For

t = -1

x(-1) = (-1)^2 - 1 = 0

y(-1) = (-1)^3 - (-1) = -1 + 1 = 0

Both parameter values give the point

(0, 0)

. Therefore, the coordinates of P are

(0, 0)

A) (-1, 0): This is the point on the curve when $t=0$ .
C) (0, 2): This point is not on the curve. It might result from an algebraic error.
D) (3, 0): This is the point on the related curve $y=t^3-4t$ when $t=2$ .

Question 5

A curve is given by $x = \cos(2t)$ and $y = \sin(t)$ for $0 \le t \le \pi$ . Find the coordinates of the point where the tangent to the curve is vertical.

(-1, 1)
(0, 1)
(1, 0) and (-1, 1)
(1, 0) (correct answer)

Explanation: When you encounter parametric curves and need to find where the tangent is vertical, you're looking for points where

\frac{dx}{dt} = 0

while

\frac{dy}{dt} \neq 0

. This is because the slope of the tangent line is

\frac{dy/dt}{dx/dt}

, which becomes undefined (vertical) when the denominator is zero but the numerator isn't. Given

x = \cos(2t)

and

y = \sin(t)

, let's find the derivatives:

$\frac{dx}{dt} = -2\sin(2t)$
$\frac{dy}{dt} = \cos(t)$

For a vertical tangent, set

\frac{dx}{dt} = 0

-2\sin(2t) = 0

\sin(2t) = 0

On the interval

[0, \pi]

, this occurs when

2t = 0, \pi, 2\pi

, so

t = 0, \frac{\pi}{2}, \pi

. Check that

\frac{dy}{dt} \neq 0

at these points:

At $t = 0$ : $\cos(0) = 1 \neq 0$ ✓
At $t = \frac{\pi}{2}$ : $\cos(\frac{\pi}{2}) = 0$ ✗
At $t = \pi$ : $\cos(\pi) = -1 \neq 0$ ✓

So vertical tangents occur at

t = 0

and

t = \pi

. The corresponding coordinates are:

$t = 0$ : $(1, 0)$
$t = \pi$ : $(-1, 0)$

Answer D gives

(1, 0)

, which corresponds to

t = 0

. Answer A gives

(-1, 1)

, but at

t = \pi

we get

(-1, 0)

, not

(-1, 1)

. Answer B gives

(0, 1)

, which occurs at

t = \frac{\pi}{2}

where the tangent is horizontal, not vertical. Answer C incorrectly includes

(-1, 1)

. Remember: vertical tangents require

\frac{dx}{dt} = 0

AND

\frac{dy}{dt} \neq 0

. Always verify both conditions and calculate the actual coordinates carefully.

Question 6

For which interval of $t$ is the curve defined by $x = t^2$ and $y = t^3 - t$ concave up?

$t < 0$
$t > 0$ (correct answer)
$-\frac{1}{\sqrt{3}} < t < \frac{1}{\sqrt{3}}$
For all real $t$

Explanation: Concavity is determined by the sign of the second derivative,

\frac{d^2y}{dx^2}

. The curve is concave up when

\frac{d^2y}{dx^2} > 0

. First, find

\frac{dy}{dx}

\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 - 1

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 1}{2t}

Next, find

\frac{d^2y}{dx^2}

. We need to differentiate

\frac{dy}{dx}

with respect to

t

using the quotient rule:

\frac{d}{dt}\left(\frac{3t^2 - 1}{2t}\right) = \frac{(6t)(2t) - (3t^2 - 1)(2)}{(2t)^2} = \frac{12t^2 - 6t^2 + 2}{4t^2} = \frac{6t^2 + 2}{4t^2} = \frac{3t^2 + 1}{2t^2}

Now, divide by

\frac{dx}{dt}

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{(3t^2 + 1)/(2t^2)}{2t} = \frac{3t^2 + 1}{4t^3}

For the curve to be concave up, we need

\frac{d^2y}{dx^2} > 0

\frac{3t^2 + 1}{4t^3} > 0

The numerator,

3t^2 + 1

, is always positive for any real

t

. Therefore, the sign of the fraction is determined by the sign of the denominator,

4t^3

4t^3 > 0 \implies t^3 > 0 \implies t > 0

So, the curve is concave up for

t > 0

A) $t < 0$ : This is the interval where the curve is concave down, as $4t^3 < 0$ .
C) $-\frac{1}{\sqrt{3}} < t < \frac{1}{\sqrt{3}}$ : This interval comes from solving $\frac{dy}{dt} = 3t^2 - 1 = 0$ , which relates to horizontal tangents, not concavity.
D) For all real $t$ : This would be the case if the sign of $\frac{d^2y}{dx^2}$ was constant, which would happen if the student only looked at the numerator $3t^2+1$ or the term $\frac{d}{dt}(\frac{dy}{dx}) = \frac{3t^2+1}{2t^2}$ , which is always positive for $t \neq 0$ .

Question 7

A curve is defined by the parametric equations $x = 3 + 2\cos t$ and $y = -1 + 2\sin t$ . Find the Cartesian equation of the curve.

$(x-3)^2 + (y+1)^2 = 4$ (correct answer)
$(x+3)^2 + (y-1)^2 = 4$
$(x-3)^2 + (y+1)^2 = 2$
$\frac{(x-3)^2}{4} - \frac{(y+1)^2}{4} = 1$

Explanation: The goal is to eliminate the parameter

t

. We can use the trigonometric identity

\cos^2 t + \sin^2 t = 1

. First, rearrange the parametric equations to isolate

\cos t

and

\sin t

x = 3 + 2\cos t \implies x - 3 = 2\cos t \implies \cos t = \frac{x-3}{2}

y = -1 + 2\sin t \implies y + 1 = 2\sin t \implies \sin t = \frac{y+1}{2}

Now substitute these expressions into the identity

\cos^2 t + \sin^2 t = 1

\left(\frac{x-3}{2}\right)^2 + \left(\frac{y+1}{2}\right)^2 = 1

\frac{(x-3)^2}{4} + \frac{(y+1)^2}{4} = 1

Multiply by 4 to get the standard form of the equation of a circle:

(x-3)^2 + (y+1)^2 = 4

This is the equation of a circle with center

(3, -1)

and radius 2.

B) $(x+3)^2 + (y-1)^2 = 4$ : This results from sign errors when rearranging the parametric equations, leading to a circle with the wrong center $(-3, 1)$ .
C) $(x-3)^2 + (y+1)^2 = 2$ : This results from not squaring the radius. The equation is $r^2$ , so if $r=2$ , then $r^2=4$ .
D) $\frac{(x-3)^2}{4} - \frac{(y+1)^2}{4} = 1$ : This equation represents a hyperbola and would arise from incorrectly using an identity like $\sec^2 t - \tan^2 t = 1$ instead of $\cos^2 t + \sin^2 t = 1$ .

Question 8

A parametric curve is defined by $x = 2\sin^2\theta$ and $y = 2\cos\theta$ for $0 \le \theta \le \pi$ . Find its Cartesian equation and domain.

$y^2 = 2(2-x)$ for $0 \le x \le 2$ (correct answer)
$x^2 + y^2 = 4$ for $-2 \le x \le 2$
$y = 2\sqrt{1-x/2}$ for $0 \le x \le 2$
$x = 2-y$ for $-2 \le y \le 2$

Explanation: We use the identity

\sin^2\theta + \cos^2\theta = 1

. From the parametric equations, we have:

\sin^2\theta = \frac{x}{2}

\cos\theta = \frac{y}{2} \implies \cos^2\theta = \frac{y^2}{4}

Substitute these into the identity:

\frac{x}{2} + \frac{y^2}{4} = 1

To clear the denominators, multiply by 4:

2x + y^2 = 4 \implies y^2 = 4 - 2x \implies y^2 = 2(2-x)

Now we need to determine the domain for

x

. The parameter

\theta

is in

[0, \pi]

. For

x = 2\sin^2\theta

, since

0 \le \sin\theta \le 1

for

\theta \in [0, \pi]

, we have

0 \le \sin^2\theta \le 1

. Thus,

0 \le 2\sin^2\theta \le 2

. So the domain for

x

0 \le x \le 2

. For

y = 2\cos\theta

, since

-1 \le \cos\theta \le 1

for

\theta \in [0, \pi]

, the range for

y

-2 \le y \le 2

. The equation

y^2 = 2(2-x)

is consistent with this range for the given domain of

x

B) $x^2 + y^2 = 4$ : This would be the case if $x=2\cos\theta, y=2\sin\theta$ .
C) $y = 2\sqrt{1-x/2}$ : This is only part of the curve. Since $\cos\theta$ can be negative on $[0, \pi]$ , we cannot just take the positive square root.
D) $x = 2-y$ : This is a linear equation and does not represent the parabolic relationship derived.

Question 9

A curve is defined by the parametric equations $x = t^2$ and $y = t^3 - 3t$ . Find the value of $\frac{d^2y}{dx^2}$ at the point where $t=2$ .

$\frac{9}{32}$
$\frac{15}{32}$ (correct answer)
$\frac{15}{8}$
6

Explanation: First, find the first derivatives with respect to

t

\frac{dx}{dt} = 2t

\frac{dy}{dt} = 3t^2 - 3

Next, find

\frac{dy}{dx}

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 3}{2t} = \frac{3}{2}t - \frac{3}{2t}

Now, differentiate

\frac{dy}{dx}

with respect to

t

\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} + \frac{3}{2t^2}

Finally, use the formula for the second derivative:

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{3}{2} + \frac{3}{2t^2}}{2t} = \frac{3(t^2+1)}{2t^2} \cdot \frac{1}{2t} = \frac{3(t^2+1)}{4t^3}

Substitute

t=2

into the expression for

\frac{d^2y}{dx^2}

\frac{d^2y}{dx^2}\bigg|_{t=2} = \frac{3(2^2+1)}{4(2^3)} = \frac{3(5)}{4(8)} = \frac{15}{32}

A) $\frac{9}{32}$ : This result comes from a sign error when differentiating $-\frac{3}{2t}$ , getting $-\frac{3}{2t^2}$ instead of $+\frac{3}{2t^2}$ . This gives $\frac{d}{dt}(\frac{dy}{dx}) = \frac{3}{2} - \frac{3}{2t^2}$ , which is $\frac{9}{8}$ at $t=2$ . Then, dividing by $\frac{dx}{dt}=4$ gives $\frac{9}{32}$ .
C) $\frac{15}{8}$ : This is the value of $\frac{d}{dt}(\frac{dy}{dx})$ at $t=2$ , but without the final step of dividing by $\frac{dx}{dt}$ .
D) 6: This result comes from calculating the incorrect formula $\frac{d^2y/dt^2}{d^2x/dt^2}$ . Here, $\frac{d^2y}{dt^2} = 6t$ and $\frac{d^2x}{dt^2} = 2$ . At $t=2$ , this would be $\frac{12}{2} = 6$ .

Question 10

Find the equation of the tangent to the curve defined by $x = \sqrt{t}$ and $y = t^2 - \frac{1}{t}$ at the point where $t=1$ .

$y = \frac{1}{6}x - \frac{1}{6}$
$y = \frac{3}{2}x - \frac{3}{2}$
$y = 6x - 6$ (correct answer)
$x + 6y = 1$

Explanation: First, find the coordinates of the point of tangency when

t=1

x(1) = \sqrt{1} = 1

y(1) = 1^2 - \frac{1}{1} = 0

The point is

(1, 0)

. Next, find the derivatives with respect to

t

\frac{dx}{dt} = \frac{1}{2\sqrt{t}}

\frac{dy}{dt} = 2t + \frac{1}{t^2}

Evaluate the derivatives at

t=1

\frac{dx}{dt}\bigg|_{t=1} = \frac{1}{2\sqrt{1}} = \frac{1}{2}

\frac{dy}{dt}\bigg|_{t=1} = 2(1) + \frac{1}{1^2} = 3

Calculate the gradient of the tangent,

m = \frac{dy}{dx}

m = \frac{dy/dt}{dx/dt} = \frac{3}{1/2} = 6

Use the point-slope form of a linear equation,

y - y_1 = m(x - x_1)

y - 0 = 6(x - 1)

y = 6x - 6

A) $y = \frac{1}{6}x - \frac{1}{6}$ : This equation uses the reciprocal of the correct gradient, $m = \frac{dx/dt}{dy/dt} = \frac{1}{6}$ .
B) $y = \frac{3}{2}x - \frac{3}{2}$ : This uses an incorrect gradient calculation, possibly from multiplying the derivatives instead of dividing: $m = (dy/dt) \times (dx/dt) = 3 \times \frac{1}{2} = \frac{3}{2}$ .
D) $x + 6y = 1$ : This is the equation of the normal to the curve at the point. The normal has a gradient of $-\frac{1}{m} = -\frac{1}{6}$ .

Question 11

The motion of a particle is described by $x(t) = 3t^2$ and $y(t) = 2t^3$ . For what value of $t>0$ is the tangent to the path parallel to the line $y=2x$ ?

1
$\sqrt{2}$
2 (correct answer)
4

Explanation: The line

y=2x

has a gradient of 2. We need to find the value of

t

for which the gradient of the tangent to the parametric curve,

\frac{dy}{dx}

, is equal to 2. First, find the derivatives with respect to

t

\frac{dx}{dt} = 6t

\frac{dy}{dt} = 6t^2

Next, find the gradient

\frac{dy}{dx}

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^2}{6t} = t

We want the gradient to be 2, so we set

\frac{dy}{dx} = 2

t = 2

Since the question specifies

t>0

, this is the correct solution.

A) 1: This would mean the gradient is 1.
B) $\sqrt{2}$ : This value might arise from a calculation error.
D) 4: This could arise if one equated $y/x = 2$ , i.e., $2t^3 / (3t^2) = 2 \Rightarrow 2t/3 = 2 \Rightarrow t=3$ . Or perhaps from another calculation error.

Question 12

A parametric curve is defined by $x = 2\sin^2\theta$ and $y = 2\cos\theta$ for $0 \le \theta \le \pi$ . Find its Cartesian equation and domain.

$y^2 = 2(2-x)$ for $0 \le x \le 2$ (correct answer)
$x^2 + y^2 = 4$ for $-2 \le x \le 2$
$y = 2\sqrt{1-x/2}$ for $0 \le x \le 2$
$x = 2-y$ for $-2 \le y \le 2$

Explanation: We use the identity

\sin^2\theta + \cos^2\theta = 1

. From the parametric equations, we have:

\sin^2\theta = \frac{x}{2}

\cos\theta = \frac{y}{2} \implies \cos^2\theta = \frac{y^2}{4}

Substitute these into the identity:

\frac{x}{2} + \frac{y^2}{4} = 1

To clear the denominators, multiply by 4:

2x + y^2 = 4 \implies y^2 = 4 - 2x \implies y^2 = 2(2-x)

Now we need to determine the domain for

x

. The parameter

\theta

is in

[0, \pi]

. For

x = 2\sin^2\theta

, since

0 \le \sin\theta \le 1

for

\theta \in [0, \pi]

, we have

0 \le \sin^2\theta \le 1

. Thus,

0 \le 2\sin^2\theta \le 2

. So the domain for

x

0 \le x \le 2

. For

y = 2\cos\theta

, since

-1 \le \cos\theta \le 1

for

\theta \in [0, \pi]

, the range for

y

-2 \le y \le 2

. The equation

y^2 = 2(2-x)

is consistent with this range for the given domain of

x

B) $x^2 + y^2 = 4$ : This would be the case if $x=2\cos\theta, y=2\sin\theta$ .
C) $y = 2\sqrt{1-x/2}$ : This is only part of the curve. Since $\cos\theta$ can be negative on $[0, \pi]$ , we cannot just take the positive square root.
D) $x = 2-y$ : This is a linear equation and does not represent the parabolic relationship derived.

Question 13

Find the coordinates of the point where the curve defined by $x = t^3 - 12t$ and $y = 2t^2 - 4t$ has a horizontal tangent.

(-16, 0)
(-11, -2) (correct answer)
(0, 0)
(16, 16)

Explanation: A horizontal tangent occurs when

\frac{dy}{dt} = 0

and

\frac{dx}{dt} \neq 0

. First, find the derivative of

y

with respect to

t

\frac{dy}{dt} = 4t - 4

Set

\frac{dy}{dt} = 0

to find the value of

t

4t - 4 = 0 \implies t = 1

Next, check that

\frac{dx}{dt} \neq 0

t=1

\frac{dx}{dt} = 3t^2 - 12

t=1

\frac{dx}{dt} = 3(1)^2 - 12 = -9 \neq 0

. So, there is a horizontal tangent at

t=1

. Finally, find the coordinates

(x, y)

t=1

x(1) = (1)^3 - 12(1) = 1 - 12 = -11

y(1) = 2(1)^2 - 4(1) = 2 - 4 = -2

The coordinates of the point are

(-11, -2)

A) (-16, 0): This is a point where the tangent is vertical. This occurs when $\frac{dx}{dt} = 0 \Rightarrow 3t^2 - 12 = 0 \Rightarrow t=2$ . At $t=2$ , $x=-16, y=0$ .
C) (0, 0): This point corresponds to $t=0$ . At $t=0$ , $\frac{dy}{dx} = \frac{-4}{-12} = \frac{1}{3}$ . The tangent is not horizontal.
D) (16, 16): This is another point where the tangent is vertical. This occurs when $t=-2$ . At $t=-2$ , $x=16, y=16$ .

Question 14

Find the equation of the tangent to the curve defined by $x = \sqrt{t}$ and $y = t^2 - \frac{1}{t}$ at the point where $t=1$ .

$y = \frac{1}{6}x - \frac{1}{6}$
$y = \frac{3}{2}x - \frac{3}{2}$
$y = 6x - 6$ (correct answer)
$x + 6y = 1$

Explanation: First, find the coordinates of the point of tangency when

t=1

x(1) = \sqrt{1} = 1

y(1) = 1^2 - \frac{1}{1} = 0

The point is

(1, 0)

. Next, find the derivatives with respect to

t

\frac{dx}{dt} = \frac{1}{2\sqrt{t}}

\frac{dy}{dt} = 2t + \frac{1}{t^2}

Evaluate the derivatives at

t=1

\frac{dx}{dt}\bigg|_{t=1} = \frac{1}{2\sqrt{1}} = \frac{1}{2}

\frac{dy}{dt}\bigg|_{t=1} = 2(1) + \frac{1}{1^2} = 3

Calculate the gradient of the tangent,

m = \frac{dy}{dx}

m = \frac{dy/dt}{dx/dt} = \frac{3}{1/2} = 6

Use the point-slope form of a linear equation,

y - y_1 = m(x - x_1)

y - 0 = 6(x - 1)

y = 6x - 6

A) $y = \frac{1}{6}x - \frac{1}{6}$ : This equation uses the reciprocal of the correct gradient, $m = \frac{dx/dt}{dy/dt} = \frac{1}{6}$ .
B) $y = \frac{3}{2}x - \frac{3}{2}$ : This uses an incorrect gradient calculation, possibly from multiplying the derivatives instead of dividing: $m = (dy/dt) \times (dx/dt) = 3 \times \frac{1}{2} = \frac{3}{2}$ .
D) $x + 6y = 1$ : This is the equation of the normal to the curve at the point. The normal has a gradient of $-\frac{1}{m} = -\frac{1}{6}$ .

Question 15

A particle moves along a curve with its position at time $t$ given by $x(t) = \ln(t)$ and $y(t) = t^2$ . Find the speed of the particle at $t=2$ .

8
$\frac{\sqrt{65}}{4}$
$\frac{9}{2}$
$\frac{\sqrt{65}}{2}$ (correct answer)

Explanation: The speed of a particle moving along a parametric curve is given by the formula

s = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}

. First, find the derivatives of

x(t)

and

y(t)

with respect to

t

\frac{dx}{dt} = \frac{1}{t}

\frac{dy}{dt} = 2t

Now, evaluate these derivatives at

t=2

\frac{dx}{dt}\bigg|_{t=2} = \frac{1}{2}

\frac{dy}{dt}\bigg|_{t=2} = 2(2) = 4

Substitute these values into the speed formula:

s = \sqrt{\left(\frac{1}{2}\right)^2 + (4)^2} = \sqrt{\frac{1}{4} + 16} = \sqrt{\frac{1}{4} + \frac{64}{4}} = \sqrt{\frac{65}{4}} = \frac{\sqrt{65}}{2}

A) 8: This is the value of the gradient of the tangent, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4}{1/2} = 8$ , not the speed.
B) $\frac{\sqrt{65}}{4}$ : This is the magnitude of the acceleration vector. $\frac{d^2x}{dt^2} = -\frac{1}{t^2}$ and $\frac{d^2y}{dt^2} = 2$ . At $t=2$ , the acceleration vector is $(-\frac{1}{4}, 2)$ . Its magnitude is $\sqrt{(-\frac{1}{4})^2 + 2^2} = \sqrt{\frac{1}{16} + 4} = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4}$ .
C) $\frac{9}{2}$ : This is the sum of the velocity components, $\frac{1}{2} + 4 = \frac{9}{2}$ , not the magnitude of the velocity vector.

Question 16

Find the equation of the normal to the curve defined by $x = 2e^t$ and $y = t^2 + t$ at the point where $t=0$ .

$y = -2x + 4$ (correct answer)
$y = -\frac{1}{2}x + 1$
$y = \frac{1}{2}x - 1$
$y = 2x - 4$

Explanation: First, find the point of interest by substituting

t=0

x(0) = 2e^0 = 2

y(0) = 0^2 + 0 = 0

The point is

(2, 0)

. Next, find the derivatives with respect to

t

\frac{dx}{dt} = 2e^t

\frac{dy}{dt} = 2t + 1

Evaluate the derivatives at

t=0

\frac{dx}{dt}\bigg|_{t=0} = 2e^0 = 2

\frac{dy}{dt}\bigg|_{t=0} = 2(0) + 1 = 1

Calculate the gradient of the tangent,

m_{\text{tan}} = \frac{dy}{dx}

m_{\text{tan}} = \frac{dy/dt}{dx/dt} = \frac{1}{2}

The gradient of the normal is the negative reciprocal of the tangent's gradient:

m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} = -\frac{1}{1/2} = -2

Use the point-slope form

y - y_1 = m(x - x_1)

with the point

(2,0)

and

m_{\text{norm}}=-2

y - 0 = -2(x - 2)

y = -2x + 4

B) $y = -\frac{1}{2}x + 1$ : This equation uses the negative of the tangent gradient, not the negative reciprocal.
C) $y = \frac{1}{2}x - 1$ : This is the equation of the tangent line, not the normal.
D) $y = 2x - 4$ : This equation uses the reciprocal of the tangent gradient, but not the negative reciprocal.

Question 17

Consider the curve given by $x = \sec(t)$ and $y = \tan(t)$ . Find the gradient of the tangent at the point where $t = \pi/4$ .

$\frac{1}{\sqrt{2}}$
1
$\sqrt{2}$ (correct answer)
2

Explanation: The gradient of the tangent is

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

. First, find the derivatives with respect to

t

\frac{dx}{dt} = \sec(t)\tan(t)

\frac{dy}{dt} = \sec^2(t)

Now, form the expression for

\frac{dy}{dx}

\frac{dy}{dx} = \frac{\sec^2(t)}{\sec(t)\tan(t)} = \frac{\sec(t)}{\tan(t)}

This can be simplified:

\frac{dy}{dx} = \frac{1/\cos(t)}{\sin(t)/\cos(t)} = \frac{1}{\sin(t)} = \csc(t)

Evaluate the gradient at

t = \pi/4

m = \csc(\pi/4) = \frac{1}{\sin(\pi/4)} = \frac{1}{1/\sqrt{2}} = \sqrt{2}

A) $\frac{1}{\sqrt{2}}$ : This is the value of $\sin(\pi/4)$ or $\cos(\pi/4)$ , not the final gradient.
B) 1: This is the value of $\tan(\pi/4)$ .
D) 2: This is the value of $\sec^2(\pi/4)$ .

Question 18

The motion of a particle is described by $x(t) = 3t^2$ and $y(t) = 2t^3$ . For what value of $t>0$ is the tangent to the path parallel to the line $y=2x$ ?

1
$\sqrt{2}$
2 (correct answer)
4

Explanation: The line

y=2x

has a gradient of 2. We need to find the value of

t

for which the gradient of the tangent to the parametric curve,

\frac{dy}{dx}

, is equal to 2. First, find the derivatives with respect to

t

\frac{dx}{dt} = 6t

\frac{dy}{dt} = 6t^2

Next, find the gradient

\frac{dy}{dx}

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^2}{6t} = t

We want the gradient to be 2, so we set

\frac{dy}{dx} = 2

t = 2

Since the question specifies

t>0

, this is the correct solution.

A) 1: This would mean the gradient is 1.
B) $\sqrt{2}$ : This value might arise from a calculation error.
D) 4: This could arise if one equated $y/x = 2$ , i.e., $2t^3 / (3t^2) = 2 \Rightarrow 2t/3 = 2 \Rightarrow t=3$ . Or perhaps from another calculation error.

Question 19

Find the coordinates of the point(s) on the curve $x = t^2 - 1, y = t^3 - 3t$ where the tangent is parallel to the x-axis.

(0, -2) and (0, 2) (correct answer)
(-1, 0)
(2, 2) and (2, -2)
(0, 2)

Explanation: A tangent parallel to the x-axis is a horizontal tangent. This occurs when

\frac{dy}{dt} = 0

and

\frac{dx}{dt} \neq 0

. First, find the derivatives:

\frac{dy}{dt} = 3t^2 - 3

\frac{dx}{dt} = 2t

Set

\frac{dy}{dt} = 0

3t^2 - 3 = 0 \implies 3(t^2 - 1) = 0 \implies t = \pm 1

Now, check

\frac{dx}{dt}

at these values:

At $t=1$ : $\frac{dx}{dt} = 2(1) = 2 \neq 0$ . So there is a horizontal tangent.
At $t=-1$ : $\frac{dx}{dt} = 2(-1) = -2 \neq 0$ . So there is another horizontal tangent. Now find the coordinates for each value of $t$ :
For $t=1$ : $x = 1^2 - 1 = 0$ , $y = 1^3 - 3(1) = -2$ . Point: $(0, -2)$ .
For $t=-1$ : $x = (-1)^2 - 1 = 0$ , $y = (-1)^3 - 3(-1) = -1 + 3 = 2$ . Point: $(0, 2)$ . So the points are $(0, -2)$ and $(0, 2)$ .
B) (-1, 0): This is the point where the tangent is vertical ( $\frac{dx}{dt}=0$ at $t=0$ ).
C) (2, 2) and (2, -2): These points correspond to $t=\pm\sqrt{3}$ , which are the x-intercepts of the curve ( $y=t(t^2-3)=0$ ).
D) (0, 2): This is only one of the two points with a horizontal tangent.

Question 20

A curve is given by $x = \cos(2t)$ and $y = \sin(t)$ for $0 \le t \le \pi$ . Find the coordinates of the point where the tangent to the curve is vertical.

(-1, 1)
(0, 1)
(1, 0) and (-1, 1)
(1, 0) (correct answer)

Explanation: When you encounter parametric curves and need to find where the tangent is vertical, you're looking for points where

\frac{dx}{dt} = 0

while

\frac{dy}{dt} \neq 0

. This is because the slope of the tangent line is

\frac{dy/dt}{dx/dt}

, which becomes undefined (vertical) when the denominator is zero but the numerator isn't. Given

x = \cos(2t)

and

y = \sin(t)

, let's find the derivatives:

$\frac{dx}{dt} = -2\sin(2t)$
$\frac{dy}{dt} = \cos(t)$

For a vertical tangent, set

\frac{dx}{dt} = 0

-2\sin(2t) = 0

\sin(2t) = 0

On the interval

[0, \pi]

, this occurs when

2t = 0, \pi, 2\pi

, so

t = 0, \frac{\pi}{2}, \pi

. Check that

\frac{dy}{dt} \neq 0

at these points:

At $t = 0$ : $\cos(0) = 1 \neq 0$ ✓
At $t = \frac{\pi}{2}$ : $\cos(\frac{\pi}{2}) = 0$ ✗
At $t = \pi$ : $\cos(\pi) = -1 \neq 0$ ✓

So vertical tangents occur at

t = 0

and

t = \pi

. The corresponding coordinates are:

$t = 0$ : $(1, 0)$
$t = \pi$ : $(-1, 0)$

Answer D gives

(1, 0)

, which corresponds to

t = 0

. Answer A gives

(-1, 1)

, but at

t = \pi

we get

(-1, 0)

, not

(-1, 1)

. Answer B gives

(0, 1)

, which occurs at

t = \frac{\pi}{2}

where the tangent is horizontal, not vertical. Answer C incorrectly includes

(-1, 1)

. Remember: vertical tangents require

\frac{dx}{dt} = 0

AND

\frac{dy}{dt} \neq 0

. Always verify both conditions and calculate the actual coordinates carefully.