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Animated Solution for Mathematics - Differentiation: The maximum distance from origin of a point on the curve $x = a sin t - b sin (\frac{a t}{b})$ , $y = a cos t - b cos (\frac{a t}{b})$ , both $a, b > 0$ is

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Progressive Hints

Hint 1

The Spark: Do not let the parametric equations intimidate you. When you see a point (x, y) and need the distance from the origin, your first instinct should be the distance formula D = \sqrt{x^2 + y^2}. The secret here is not in the individual x or y values, but in how they interact when squared and summed. Visualize the point tracing a path; we are looking for the furthest point on that path.

Hint 2

The Arsenal: You will need two powerful tools from your trigonometry toolkit. First, the fundamental Pythagorean identity: \sin^2 \theta + \cos^2 \theta = 1. Second, the cosine difference identity: \cos(A - B) = \cos A \cos B + \sin A \sin B. Keep these ready; they are the keys to unlocking the simplification.

Hint 3

The Execution: Start by writing D^2 = x^2 + y^2. Substitute the expressions for x and y. Expand both (a \sin t - b \sin(\frac{at}{b}))^2 and (a \cos t - b \cos(\frac{at}{b}))^2. When you group the terms, you will find that the a^2 and b^2 terms simplify beautifully, leaving you with a single cosine term to optimize.

Visualized Solution

Coordinate System Setup

Given parametric equations:
$x = a sin t - b sin (\frac{a t}{b})$
$y = a cos t - b cos (\frac{a t}{b})$
Constants: $a, b > 0$

Distance from Origin

Distance $D$ from origin $(0, 0)$ to $P (x, y)$ is given by:
$D = x^{2} + y^{2}$
To avoid square roots, we maximize $D^{2} = x^{2} + y^{2}$

Substituting Coordinates

$D^{2} = [a sin t - b sin (\frac{a t}{b})]^{2} + [a cos t - b cos (\frac{a t}{b})]^{2}$

Expanding $x^{2}$

Using $(A - B)^{2} = A^{2} + B^{2} - 2 A B$ :
$x^{2} = a^{2} sin^{2} t + b^{2} sin^{2} (\frac{a t}{b}) - 2 ab sin t sin (\frac{a t}{b})$

Expanding $y^{2}$

Similarly for $y$ :
$y^{2} = a^{2} cos^{2} t + b^{2} cos^{2} (\frac{a t}{b}) - 2 ab cos t cos (\frac{a t}{b})$

Summing $x^{2}$ and $y^{2}$

Grouping the $a^{2}$ , $b^{2}$ , and $- 2 ab$ terms:
$D^{2} = a^{2} (sin^{2} t + cos^{2} t) + b^{2} [sin^{2} (\frac{a t}{b}) + cos^{2} (\frac{a t}{b})]$
$- 2 ab [cos t cos (\frac{a t}{b}) + sin t sin (\frac{a t}{b})]$

Applying Pythagorean Identity

Since $sin^{2} θ + cos^{2} θ = 1$ :
$a^{2} (1) + b^{2} (1) = a^{2} + b^{2}$
The equation simplifies to:
$D^{2} = a^{2} + b^{2} - 2 ab [cos t cos (\frac{a t}{b}) + sin t sin (\frac{a t}{b})]$

Applying Cosine Difference Formula

Using the identity: $cos (A - B) = cos A cos B + sin A sin B$
Let $A = t$ and $B = \frac{a t}{b}$
$D^{2} = a^{2} + b^{2} - 2 ab cos (t - \frac{a t}{b})$

Maximizing the Distance

To maximize $D^{2}$ , we need to maximize the entire expression.
Since there is a negative sign before $2 ab$ , we must minimize the cosine term.
Minimum value of $cos θ$ is $- 1$ .

Calculating Maximum Distance

Substitute $cos (t - \frac{a t}{b}) = - 1$ :
$D_{ma x}^{2} = a^{2} + b^{2} - 2 ab (- 1)$
$D_{ma x}^{2} = a^{2} + b^{2} + 2 ab = (a + b)^{2}$

Final Result

Taking the square root:
$D_{ma x} = a + b$
The maximum distance from the origin is $a + b$ .

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

The point

P (x, y)

is defined by the parametric equations:

x = a sin t - b sin (\frac{a t}{b})

y = a cos t - b cos (\frac{a t}{b})

Our mission is to find the maximum distance

D

from the origin

(0, 0)

The Geometry of Distance

The distance

D

of any point

(x, y)

from the origin is given by

D = x^{2} + y^{2}

To simplify the calculation, we maximize

D^{2} = x^{2} + y^{2}

instead. Since the square function is monotonic for positive values, the value of

t

that maximizes

D^{2}

will also maximize

D

The Algebraic Expansion

We compute

x^{2} + y^{2}

by expanding the binomials:

x^{2} = a^{2} sin^{2} t + b^{2} sin^{2} (\frac{a t}{b}) - 2 ab sin t sin (\frac{a t}{b})

y^{2} = a^{2} cos^{2} t + b^{2} cos^{2} (\frac{a t}{b}) - 2 ab cos t cos (\frac{a t}{b})

The Trigonometric Revelation

Summing these expressions and grouping by coefficients, we obtain:

D^{2} = a^{2} (sin^{2} t + cos^{2} t) + b^{2} (sin^{2} (\frac{a t}{b}) + cos^{2} (\frac{a t}{b})) - 2 ab (cos t cos (\frac{a t}{b}) + sin t sin (\frac{a t}{b}))

Using the Pythagorean identity

sin^{2} θ + cos^{2} θ = 1

, the expression simplifies significantly:

D^{2} = a^{2} + b^{2} - 2 ab (cos t cos (\frac{a t}{b}) + sin t sin (\frac{a t}{b}))

Applying the cosine difference formula

cos (A - B) = cos A cos B + sin A sin B

, we reach the elegant form:

D^{2} = a^{2} + b^{2} - 2 ab cos (t - \frac{a t}{b})

Final Calculation

To maximize

D^{2}

, we must minimize the term

- 2 ab cos (t - \frac{a t}{b})

. Since the minimum value of the cosine function is

- 1

, we substitute this into the equation:

D_{ma x}^{2} = a^{2} + b^{2} - 2 ab (- 1) = a^{2} + b^{2} + 2 ab

This expression is a perfect square:

D_{ma x}^{2} = (a + b)^{2}

Taking the square root, we find the maximum distance is: $D_{ma x} = a + b$

Common Traps & FAQs

Why do we maximize D^2 instead of D?

Great question! The distance function D = \sqrt{x^2 + y^2} is strictly increasing for D \ge 0. This means that the value of t that maximizes D^2 will also maximize D. By working with D^2, we eliminate the square root, making our algebra much cleaner and less prone to errors.

How do I know when to use the cosine difference formula?

Look for the pattern! In your expanded expression, you will see terms like \cos t \cos(\frac{at}{b}) + \sin t \sin(\frac{at}{b}). Whenever you see a sum of products of sines and cosines, your brain should immediately scream 'Compound Angle Formula!' It is the standard way to collapse complex trigonometric sums into a single, manageable function.

Wait, why do we pick -1 for the cosine value? Shouldn't we pick 1 to maximize?

This is the classic trap! Look at the expression we derived: D^2 = a^2 + b^2 - 2ab \cos(\dots). Because there is a negative sign in front of the 2ab \cos(\dots) term, we need the cosine part to be as small as possible to make the whole expression as large as possible. The minimum value of cosine is -1, which turns that negative term into a positive addition, giving us the maximum distance.

Animated Solution for Mathematics - Differentiation: The maximum distance from origin of a point on the curve $x = a sin t - b sin (\frac{a t}{b})$ , $y = a cos t - b cos (\frac{a t}{b})$ , both $a, b > 0$ is

Select Answer:

Visualized Solution

Coordinate System Setup

Distance from Origin

Substituting Coordinates

Expanding $x^{2}$

Expanding $y^{2}$

Summing $x^{2}$ and $y^{2}$

Applying Pythagorean Identity

Applying Cosine Difference Formula

Maximizing the Distance

Calculating Maximum Distance

Final Result

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Geometry of Distance

The Algebraic Expansion

The Trigonometric Revelation

Final Calculation

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The maximum distance from origin of a point on the curve x=asint−bsin(bat​), y=acost−bcos(bat​), both a,b>0 is

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Visualized Solution

Coordinate System Setup

Distance from Origin

Substituting Coordinates

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

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Applying Cosine Difference Formula

Maximizing the Distance

Calculating Maximum Distance

Final Result

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Geometry of Distance

The Algebraic Expansion

The Trigonometric Revelation

Final Calculation

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Animated Solution for Mathematics - Differentiation: The maximum distance from origin of a point on the curve $x = a sin t - b sin (\frac{a t}{b})$ , $y = a cos t - b cos (\frac{a t}{b})$ , both $a, b > 0$ is

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