JEE Main 2022 (28 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $x (t) = 22 cos t sin 2 t$ and $y (t) = 22 sin t sin 2 t$ , $t \in (0, \frac{π}{2})$ . Then $\frac{1 + ( \frac{d y}{d x} ) ^{2}}{\frac{d ^{2} y}{d x ^{2}}}$ at $t = \frac{π}{4}$ is equal to

Select Answer:

Progressive Hints

Hint 1

Hint 1: Stop! Do not attempt to eliminate the parameter t by squaring and adding. Instead, observe the structure of x(t) and y(t). They are begging for the product rule, and the resulting derivatives will reveal a beautiful trigonometric identity. Keep your eyes peeled for \cos(A+B) and \sin(A+B).

Hint 2

Hint 2: You have the first derivative \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. Once you simplify this, you will find a surprisingly elegant function of t. Now, for the second derivative, remember the golden rule of parametric differentiation: \frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}. Never forget that \frac{dt}{dx} term!

Hint 3

Hint 3: Start by calculating \frac{dy}{dx} = \tan 3t. Now, differentiate this with respect to t to get 3 \sec^2 3t. Finally, multiply this by the reciprocal of \frac{dx}{dt}, which you calculated earlier. Evaluate at t = \frac{\pi}{4} only after you have the full expression for the second derivative.

Visualized Solution

Visualizing the Parametric Curve

Given parametric equations:
$x (t) = 22 cos t sin 2 t$
$y (t) = 22 sin t sin 2 t$
Goal: Evaluate $\frac{1 + ( \frac{d y}{d x} ) ^{2}}{\frac{d ^{2} y}{d x ^{2}}}$ at $t = \frac{π}{4}$

The Strategy

To evaluate the expression, we need two key components:
First derivative: $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Second derivative: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \cdot \frac{d t}{d x}$

Differentiating $x (t)$

Differentiate $x (t)$ with respect to $t$ using the Product Rule:
$\frac{d x}{d t} = 22 [\frac{d}{d t} (cos t) \cdot sin 2 t + cos t \cdot \frac{d}{d t} (sin 2 t)]$
$\frac{d x}{d t} = 22 [- sin t sin 2 t + cos t \cdot \frac{1}{2 s i n 2 t} \cdot (2 cos 2 t)]$

Simplifying $\frac{d x}{d t}$

Simplify the expression by taking the LCM:
$\frac{d x}{d t} = 22 [\frac{- s i n t s i n 2 t + c o s t c o s 2 t}{s i n 2 t}]$
Using the identity $cos A cos B - sin A sin B = cos (A + B)$ :
$\frac{d x}{d t} = \frac{2 2 c o s ( 2 t + t )}{s i n 2 t} = \frac{2 2 c o s 3 t}{s i n 2 t}$

Differentiating $y (t)$

Differentiate $y (t)$ with respect to $t$ using the Product Rule:
$\frac{d y}{d t} = 22 [\frac{d}{d t} (sin t) \cdot sin 2 t + sin t \cdot \frac{d}{d t} (sin 2 t)]$
$\frac{d y}{d t} = 22 [cos t sin 2 t + sin t \cdot \frac{c o s 2 t}{s i n 2 t}]$

Simplifying $\frac{d y}{d t}$

Simplify the expression by taking the LCM:
$\frac{d y}{d t} = 22 [\frac{c o s t s i n 2 t + s i n t c o s 2 t}{s i n 2 t}]$
Using the identity $sin A cos B + cos A sin B = sin (A + B)$ :
$\frac{d y}{d t} = \frac{2 2 s i n ( 2 t + t )}{s i n 2 t} = \frac{2 2 s i n 3 t}{s i n 2 t}$

Finding $\frac{d y}{d x}$

Apply the parametric differentiation rule:
$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Substitute the simplified expressions:
$\frac{d y}{d x} = \frac{\frac{2 2 s i n 3 t}{s i n 2 t}}{\frac{2 2 c o s 3 t}{s i n 2 t}} = \frac{s i n 3 t}{c o s 3 t} = tan 3 t$

Evaluating $\frac{d y}{d x}$ at $t = \frac{π}{4}$

At $t = \frac{π}{4}$ :
$\frac{d y}{d x} = tan (3 \cdot \frac{π}{4}) = tan \frac{3 π}{4}$
Since $tan \frac{3 π}{4} = - 1$ :
$\frac{d y}{d x} = - 1$

The Second Derivative Trap

To find $\frac{d ^{2} y}{d x ^{2}}$ , we must use the chain rule carefully:
$\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d t} (tan 3 t) \cdot \frac{d t}{d x}$
Recall that $\frac{d t}{d x} = \frac{1}{\frac{d x}{d t}}$

Calculating $\frac{d ^{2} y}{d x ^{2}}$

Differentiate $tan 3 t$ : $\frac{d}{d t} (tan 3 t) = 3 sec^{2} 3 t$
Substitute $\frac{d t}{d x} = \frac{s i n 2 t}{2 2 c o s 3 t}$ :
$\frac{d ^{2} y}{d x ^{2}} = 3 sec^{2} 3 t \cdot \frac{s i n 2 t}{2 2 c o s 3 t}$
$\frac{d ^{2} y}{d x ^{2}} = \frac{3 s e c ^{3} 3 t s i n 2 t}{2 2}$

Evaluating $\frac{d ^{2} y}{d x ^{2}}$ at $t = \frac{π}{4}$

At $t = \frac{π}{4}$ :
$sin 2 t = sin \frac{π}{2} = 1$
$sec 3 t = sec \frac{3 π}{4} = - 2$
Substitute into the formula:
$\frac{d ^{2} y}{d x ^{2}} = \frac{3 ( - 2 ) ^{3} \cdot 1}{2 2} = \frac{3 ( - 2 2 )}{2 2} = - 3$

Final Calculation

Substitute the values into the target expression:
Expression $= \frac{1 + ( \frac{d y}{d x} ) ^{2}}{\frac{d ^{2} y}{d x ^{2}}}$
$= \frac{1 + ( - 1 ) ^{2}}{- 3}$
$= \frac{1 + 1}{- 3} = - \frac{2}{3}$
Final Answer: $- \frac{2}{3}$

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The Sigma Insight: Techniques of Differentiation

Solution Diagram

Analyzing the Setup

Welcome, future engineers. Today, we are peeling back the layers of a parametric curve defined by the equations:

x (t) = 22 cos t sin 2 t

y (t) = 22 sin t sin 2 t

Our goal is to evaluate the expression

\frac{1 + ( \frac{d y}{d x} ) ^{2}}{\frac{d ^{2} y}{d x ^{2}}}

at the specific parameter value

t = \frac{π}{4}

The First Derivative

To find the slope of the tangent, we utilize the relation

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

. When we differentiate

x (t)

and

y (t)

using the product rule, the terms involving

sin 2 t

simplify elegantly.

By combining terms over a common denominator, the numerators transform into trigonometric identities. Specifically, we obtain:

\frac{d x}{d t} = \frac{2 2 cos 3 t}{sin 2 t}

\frac{d y}{d t} = \frac{2 2 sin 3 t}{sin 2 t}

Dividing these expressions causes the square root terms to vanish entirely, leaving us with the simplified slope:

\frac{d y}{d x} = tan 3 t

The Second Derivative Trap

We must now calculate the second derivative with respect to

x

. By the chain rule, we apply the following transformation:

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (tan 3 t) \cdot \frac{d t}{d x}

We know that

\frac{d}{d t} (tan 3 t) = 3 sec^{2} 3 t

. Since

\frac{d t}{d x}

is the reciprocal of

\frac{d x}{d t}

, we substitute the previously derived expression to obtain:

\frac{d ^{2} y}{d x ^{2}} = 3 sec^{2} 3 t \cdot \frac{sin 2 t}{2 2 cos 3 t} = \frac{3 sec ^{3} 3 t sin 2 t}{2 2}

Final Calculation

We are now at the finish line. At

t = \frac{π}{4}

, we observe that

sin 2 t = 1

and

sec 3 t = sec (\frac{3 π}{4}) = - 2

Plugging these values into our expressions, we find:

\frac{d y}{d x} = tan (\frac{3 π}{4}) = - 1

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

Substituting these into our target expression, we arrive at the final result:

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

The final value is $- \frac{2}{3}$ . Remember, the key to JEE Advanced is the ability to see the structure beneath the symbols.

Common Traps & FAQs

Why can't I just differentiate y with respect to t and divide by the derivative of x with respect to t for the second derivative?

This is the most common trap in JEE Advanced! The formula \frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2} is absolutely incorrect. You must treat \frac{dy}{dx} as a function of t and apply the chain rule: \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}. Respect the chain rule, and you will never fall into this trap.

Is it necessary to simplify the derivatives so much? Can't I just plug in t = \frac{\pi}{4} early?

While you could plug in values early, it is a dangerous game. Simplifying first reveals the underlying structure—in this case, \tan 3t. If you plug in early, you lose the chance to see the beautiful cancellation of terms, and you increase the risk of arithmetic errors. Always simplify first; it is the mark of a master.

Why is the final answer negative? Does that make sense?

Absolutely! A negative value for the second derivative indicates that the curve is concave down at that point. Since our expression involves the radius of curvature, the sign tells us about the orientation of the curve. Trust the math; if your steps are logical, the sign is just the curve telling you its shape.

Animated Solution for Mathematics - Differentiation: Let $x (t) = 22 cos t sin 2 t$ and $y (t) = 22 sin t sin 2 t$ , $t \in (0, \frac{π}{2})$ . Then $\frac{1 + ( \frac{d y}{d x} ) ^{2}}{\frac{d ^{2} y}{d x ^{2}}}$ at $t = \frac{π}{4}$ is equal to

Select Answer:

Visualized Solution

Visualizing the Parametric Curve

The Strategy

Differentiating $x (t)$

Simplifying $\frac{d x}{d t}$

Differentiating $y (t)$

Simplifying $\frac{d y}{d t}$

Finding $\frac{d y}{d x}$

Evaluating $\frac{d y}{d x}$ at $t = \frac{π}{4}$

The Second Derivative Trap

Calculating $\frac{d ^{2} y}{d x ^{2}}$

Evaluating $\frac{d ^{2} y}{d x ^{2}}$ at $t = \frac{π}{4}$

Final Calculation

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The First Derivative

The Second Derivative Trap

Final Calculation

Similar Questions

If $y (α) = 2 (\frac{t a n α + c o t α}{1 + t a n ^{2} α}) + \frac{1}{s i n ^{2} α}$ where $α \in (\frac{3 π}{4}, π)$ , then $\frac{d y}{d α}$ at $α = \frac{5 π}{6}$ is

If $y (α) = 2 (\frac{t a n α + c o t α}{1 + t a n ^{2} α}) + \frac{1}{s i n ^{2} α}$ , $α \in (\frac{3 π}{4}, π)$ then $\frac{d y}{d α}$ at $α = \frac{5 π}{6}$ is

If $(a + 2 b cos x) (a - 2 b cos y) = a^{2} - b^{2}$ , where $a > b > 0$ , then $\frac{d x}{d y}$ at $(\frac{π}{4}, \frac{π}{4})$ is:

If $x = 2^{cosec^{- 1} t}$ and $y = 2^{sec^{- 1} t}$ ( $∣ t ∣ \geq 1$ ), then $\frac{d y}{d x}$ is equal to :

If $2 x^{y} + 3 y^{x} = 20$ , then $\frac{d y}{d x}$ at $(2, 2)$ is equal to:

If $x^{l o g_{e} (l o g_{e} x)} - x^{2} + y^{2} = 4 (y > 0)$ , then $d y / d x$ at $x = e$ is equal to :

For $x > 1$ , if $(2 x)^{2 y} = 4 e^{2 x - 2 y}$ , then $(1 + lo g_{e} 2 x)^{2} \frac{d y}{d x}$ is equal to :

If $2 y = (cot^{- 1} (\frac{3 c o s x + s i n x}{c o s x - 3 s i n x}))^{2}, x \in (0, \frac{π}{2})$ , then $\frac{d y}{d x}$ is equal to :

Let $y = f (x) = sin^{3} (\frac{π}{3} cos (\frac{π}{3 2} (- 4 x^{3} + 5 x^{2} + 1)^{3/2}))$ . Then, at $x = 1$ ,

Find $\frac{d y}{d x}$ at $x = - 1$ , when $(sin y)^{s i n (\frac{π x}{2})} + \frac{3}{2} sec^{- 1} (2 x) + 2^{x} tan (ln (x + 2)) = 0$ .

Select Answer:

Visualized Solution

Visualizing the Parametric Curve

The Strategy

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x(t)

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

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(t)

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

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

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​ at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=4π​

The Second Derivative Trap

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dx2d2y​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dx2d2y​ at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=4π​

Final Calculation

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The First Derivative

The Second Derivative Trap

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y(α)=2(1+tan2αtanα+cotα​)+sin2α1​​ where α∈(43π​,π), then dαdy​ at α=65π​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y(α)=2(1+tan2αtanα+cotα​)+sin2α1​​, α∈(43π​,π) then dαdy​ at α=65π​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (a+2​bcosx)(a−2​bcosy)=a2−b2, where a>b>0, then dydx​ at (4π​,4π​) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=2cosec−1t​ and y=2sec−1t​ (∣t∣≥1), then dxdy​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 2xy+3yx=20, then dxdy​ at (2,2) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If xloge​(loge​x)−x2+y2=4(y>0), then dy/dx at x=e is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x>1, if (2x)2y=4e2x−2y, then (1+loge​2x)2dxdy​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 2y=(cot−1(cosx−3​sinx3​cosx+sinx​))2,x∈(0,2π​), then dxdy​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=f(x)=sin3(3π​cos(32​π​(−4x3+5x2+1)3/2)). Then, at x=1,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find dxdy​ at x=−1, when (siny)sin(2πx​)+23​​sec−1(2x)+2xtan(ln(x+2))=0.

Differentiating $x (t)$

Simplifying $\frac{d x}{d t}$

Differentiating $y (t)$

Simplifying $\frac{d y}{d t}$

Finding $\frac{d y}{d x}$

Evaluating $\frac{d y}{d x}$ at $t = \frac{π}{4}$

Calculating $\frac{d ^{2} y}{d x ^{2}}$

Evaluating $\frac{d ^{2} y}{d x ^{2}}$ at $t = \frac{π}{4}$

If $y (α) = 2 (\frac{t a n α + c o t α}{1 + t a n ^{2} α}) + \frac{1}{s i n ^{2} α}$ where $α \in (\frac{3 π}{4}, π)$ , then $\frac{d y}{d α}$ at $α = \frac{5 π}{6}$ is

If $y (α) = 2 (\frac{t a n α + c o t α}{1 + t a n ^{2} α}) + \frac{1}{s i n ^{2} α}$ , $α \in (\frac{3 π}{4}, π)$ then $\frac{d y}{d α}$ at $α = \frac{5 π}{6}$ is

If $(a + 2 b cos x) (a - 2 b cos y) = a^{2} - b^{2}$ , where $a > b > 0$ , then $\frac{d x}{d y}$ at $(\frac{π}{4}, \frac{π}{4})$ is:

If $x = 2^{cosec^{- 1} t}$ and $y = 2^{sec^{- 1} t}$ ( $∣ t ∣ \geq 1$ ), then $\frac{d y}{d x}$ is equal to :

If $2 x^{y} + 3 y^{x} = 20$ , then $\frac{d y}{d x}$ at $(2, 2)$ is equal to:

If $x^{l o g_{e} (l o g_{e} x)} - x^{2} + y^{2} = 4 (y > 0)$ , then $d y / d x$ at $x = e$ is equal to :

For $x > 1$ , if $(2 x)^{2 y} = 4 e^{2 x - 2 y}$ , then $(1 + lo g_{e} 2 x)^{2} \frac{d y}{d x}$ is equal to :

If $2 y = (cot^{- 1} (\frac{3 c o s x + s i n x}{c o s x - 3 s i n x}))^{2}, x \in (0, \frac{π}{2})$ , then $\frac{d y}{d x}$ is equal to :

Let $y = f (x) = sin^{3} (\frac{π}{3} cos (\frac{π}{3 2} (- 4 x^{3} + 5 x^{2} + 1)^{3/2}))$ . Then, at $x = 1$ ,

Find $\frac{d y}{d x}$ at $x = - 1$ , when $(sin y)^{s i n (\frac{π x}{2})} + \frac{3}{2} sec^{- 1} (2 x) + 2^{x} tan (ln (x + 2)) = 0$ .