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Animated Solution for Mathematics - Differentiation: A point on the parabola $y^{2} = 18 x$ at which the ordinate increases at twice the rate of the abscissa is

Q: Wait, why do we differentiate with respect to $t$ instead of $x$?

Great question! While differentiating with respect to $x$ gives you the slope of the tangent, the problem explicitly mentions 'rates of change' in the context of motion. By introducing time $t$ as a parameter, we can treat $x$ and $y$ as dynamic variables, which is exactly what the physics of the problem demands.

Q: Why can we cancel $\frac{dx}{dt}$? What if it is zero?

If $\frac{dx}{dt}$ were zero, the point would be stationary or moving purely vertically. However, the problem implies a dynamic point moving along the curve. In JEE problems of this type, we assume the point is in motion unless stated otherwise, allowing us to safely divide by the non-zero rate.

Q: Is it possible to have two points?

In this specific case, the condition $\frac{dy}{dt} = 2 \frac{dx}{dt}$ leads to a unique value for $y$. Since $y^2 = 18x$ is a function where $x$ is determined by $y^2$, a single $y$ value yields a single $x$ value. Always check if your algebra leads to a quadratic in $y$—if it did, you might indeed have two points!

Select Answer:

Visualized Solution

Visualizing the Parabola

Equation of the curve: $y^{2} = 18 x$
Let $P (x, y)$ be a point moving along this curve.
The position of $P$ changes with time $t$ .

Understanding Rates of Change

Abscissa ( $x$ -coordinate) rate of change: $\frac{d x}{d t}$
Ordinate ( $y$ -coordinate) rate of change: $\frac{d y}{d t}$

The Given Condition

The ordinate increases at twice the rate of the abscissa.
Mathematical translation: $\frac{d y}{d t} = 2 \frac{d x}{d t}$

Differentiating the Curve Equation

Curve equation: $y^{2} = 18 x$
Differentiate both sides with respect to time $t$ :
$\frac{d}{d t} (y^{2}) = \frac{d}{d t} (18 x)$

Applying the Chain Rule

Left Hand Side (LHS): $\frac{d}{d t} (y^{2}) = 2 y \frac{d y}{d t}$
Right Hand Side (RHS): $\frac{d}{d t} (18 x) = 18 \frac{d x}{d t}$

Equating the Derivatives

Equating LHS and RHS:
$2 y \frac{d y}{d t} = 18 \frac{d x}{d t}$
Simplifying: $y \frac{d y}{d t} = 9 \frac{d x}{d t}$

Substituting the Condition

We know: $\frac{d y}{d t} = 2 \frac{d x}{d t}$
Substitute this into our simplified equation:
$y (2 \frac{d x}{d t}) = 9 \frac{d x}{d t}$

Canceling the Common Terms

Equation: $2 y \frac{d x}{d t} = 9 \frac{d x}{d t}$
Assuming the point is moving, $\frac{d x}{d t} \neq = 0$ .
Cancel $\frac{d x}{d t}$ from both sides:
$2 y = 9$

Solving for the Ordinate

$2 y = 9$
$y = \frac{9}{2}$

Finding the Abscissa

We have $y = \frac{9}{2}$ .
We need to find $x$ .
Substitute $y = \frac{9}{2}$ back into the original parabola equation: $y^{2} = 18 x$
$(\frac{9}{2})^{2} = 18 x$

Squaring the Ordinate

Calculate $(\frac{9}{2})^{2}$ :
$\frac{81}{4} = 18 x$

Solving for $x$

Isolate $x$ :
$x = \frac{81}{4 \times 18}$
Simplify the fraction:
$x = \frac{9}{4 \times 2}$
$x = \frac{9}{8}$

The Final Point

The abscissa is $x = \frac{9}{8}$
The ordinate is $y = \frac{9}{2}$
The required point on the parabola is $(\frac{9}{8}, \frac{9}{2})$

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The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

Imagine a particle moving along the parabola defined by the equation

y^{2} = 18 x

. As the particle slides, its coordinates

(x, y)

change with respect to time

t

We are given that the ordinate (

y

) is increasing at twice the rate of the abscissa (

x

). Mathematically, this is expressed as:

\frac{d y}{d t} = 2 \frac{d x}{d t}

The Master Equation

To relate these rates of change, we differentiate the equation of the parabola

y^{2} = 18 x

with respect to time

t

using the Chain Rule.

Differentiating both sides yields:

2 y \frac{d y}{d t} = 18 \frac{d x}{d t}

Dividing both sides by 2, we obtain the fundamental relationship:

y \frac{d y}{d t} = 9 \frac{d x}{d t}

Solving for the Coordinates

Now, we substitute the given condition

\frac{d y}{d t} = 2 \frac{d x}{d t}

into our derived relationship:

y (2 \frac{d x}{d t}) = 9 \frac{d x}{d t}

Assuming the particle is in motion such that $\frac{dx}{dt} eq 0$, we can cancel the rate terms from both sides:

2 y = 9 ⟹ y = \frac{9}{2}

Final Calculation

To find the corresponding

x

-coordinate, we substitute

y = \frac{9}{2}

back into the original parabola equation

y^{2} = 18 x

(\frac{9}{2})^{2} = 18 x

\frac{81}{4} = 18 x

Solving for

x

, we get:

x = \frac{81}{4 \times 18} = \frac{81}{72}

Simplifying the fraction by dividing by 9, we find

x = \frac{9}{8}

. Thus, the point on the parabola is $(\frac{9}{8}, \frac{9}{2})$ .

Common Traps & FAQs

Wait, why do we differentiate with respect to

t

instead of

x

Great question! While differentiating with respect to

x

gives you the slope of the tangent, the problem explicitly mentions 'rates of change' in the context of motion. By introducing time

t

as a parameter, we can treat

x

and

y

as dynamic variables, which is exactly what the physics of the problem demands.

Can I just say

\frac{d y}{d x} = 2

Not quite! The problem states the ordinate increases at twice the rate of the abscissa, which is

\frac{d y}{d t} = 2 \frac{d x}{d t}

. If you divide both sides by

\frac{d x}{d t}

, you get

\frac{d y}{d x} = 2

. While this works mathematically here, it is safer to stick to the time-derivative approach to avoid confusion when the rates are not constant.

Why can we cancel

\frac{d x}{d t}

? What if it is zero?

\frac{d x}{d t}

were zero, the point would be stationary or moving purely vertically. However, the problem implies a dynamic point moving along the curve. In JEE problems of this type, we assume the point is in motion unless stated otherwise, allowing us to safely divide by the non-zero rate.

Is it possible to have two points?

In this specific case, the condition

\frac{d y}{d t} = 2 \frac{d x}{d t}

leads to a unique value for

y

. Since

y^{2} = 18 x

is a function where

x

is determined by

y^{2}

, a single

y

value yields a single

x

value. Always check if your algebra leads to a quadratic in

y

—if it did, you might indeed have two points!

Do I need to worry about the sign of the rate?

Absolutely! The word 'increases' is key. If the problem said 'decreases', you would have a negative sign. Always pay attention to the directionality implied by the language of the question.

Animated Solution for Mathematics - Differentiation: A point on the parabola $y^{2} = 18 x$ at which the ordinate increases at twice the rate of the abscissa is

Select Answer:

Visualized Solution

Visualizing the Parabola

Understanding Rates of Change

The Given Condition

Differentiating the Curve Equation

Applying the Chain Rule

Equating the Derivatives

Substituting the Condition

Canceling the Common Terms

Solving for the Ordinate

Finding the Abscissa

Squaring the Ordinate

Solving for $x$

The Final Point

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Master Equation

Solving for the Coordinates

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; } A point on the parabola y2=18x at which the ordinate increases at twice the rate of the abscissa is

Select Answer:

Visualized Solution

Visualizing the Parabola

Understanding Rates of Change

The Given Condition

Differentiating the Curve Equation

Applying the Chain Rule

Equating the Derivatives

Substituting the Condition

Canceling the Common Terms

Solving for the Ordinate

Finding the Abscissa

Squaring the Ordinate

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

The Final Point

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Master Equation

Solving for the Coordinates

Final Calculation

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Animated Solution for Mathematics - Differentiation: A point on the parabola $y^{2} = 18 x$ at which the ordinate increases at twice the rate of the abscissa is

Solving for $x$

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