If the angle made by the tangent at the point (x₀, y₀) on the curve x = 12(t + sin t cos t), y = 12(

Q: Wait, why do we use the derivative to find the slope? Can't we just use the coordinates directly?

Great question! The coordinates $(x, y)$ tell us where the point is, but the derivative $\frac{dy}{dx}$ tells us the 'steepness' or the direction of the curve at that specific point. Since the problem gives us the angle of the tangent, we need the slope, which is the geometric interpretation of the derivative.

Q: Why did we divide by 2 when solving $\sqrt{3} \cos(t) - \sin(t) = 1$?

That is a classic trick! We divide by 2 to transform the expression into the form $R \cos(t + \alpha)$, which is much easier to solve. By dividing by 2, we get $\frac{\sqrt{3}}{2} \cos(t) - \frac{1}{2} \sin(t) = \frac{1}{2}$, which is exactly $\cos(t + \frac{\pi}{6}) = \frac{1}{2}$. It turns a messy equation into a standard one!

Q: Is it possible that $t$ could have other values in the range $0 < t < \frac{\pi}{2}$?

In this specific problem, the equation $\cos(t + \frac{\pi}{6}) = \frac{1}{2}$ within the given interval $0 < t < \frac{\pi}{2}$ yields a unique solution. Always check your domain constraints—they are the boundaries that keep our mathematical journey on the right track!

Q: Why did we simplify $\frac{dx}{dt}$ using $2 \cos^2(t)$?

Simplification is the art of physics and math! By using the identity $1 + \cos(2t) = 2 \cos^2(t)$, we were able to cancel out the $\cos(t)$ term in the numerator and denominator later on. Without that simplification, the algebra would have become a nightmare of trigonometric ratios.

Q: Does the value of $x_0$ matter for finding $y_0$?

Actually, no! The problem asks for $y_0$ based on the tangent condition. Since the tangent condition depends only on $t$, we find $t$ first and then plug it into the expression for $y(t)$. $x_0$ is just a passenger on this journey!

Question

If the angle made by the tangent at the point $(x_0, y_0)$ on the curve $x = 12(t + \sin t \cos t)$, $y = 12(1 + \sin t)^2, 0 < t < \frac{\pi}{2}$, with the positive x-axis is $\frac{\pi}{3}$, then $y_0$ is equal to

Options:

(A) $6(3 + 2\sqrt{2})$

(B) $3(7 + 4\sqrt{3})$

(C) 27

(D) 48

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle tracing a path through a two-dimensional plane where the position is governed by a parameter $t$. The curve is defined by the parametric equations:

$$x = 12(t + \sin t \cos t)$$

$$y = 12(1 + \sin t)^2$$

Our objective is to determine the specific point $(x_0, y_0)$ where the tangent to this path makes an angle of $\frac{\pi}{3}$ with the $x$-axis.

### The Velocity of the Curve

To find the slope of the tangent, we calculate the derivative $\frac{dy}{dx}$ using the ratio of the vertical velocity to the horizontal velocity:

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

First, we simplify the horizontal component $x = 12(t + \frac{1}{2} \sin 2t)$. Differentiating with respect to $t$, we obtain:

$$\frac{dx}{dt} = 12(1 + \cos 2t)$$

Using the trigonometric identity $1 + \cos 2t = 2 \cos^2 t$, this expression simplifies to:

$$\frac{dx}{dt} = 24 \cos^2 t$$

Next, we differentiate the vertical component $y = 12(1 + \sin t)^2$ using the chain rule:

$$\frac{dy}{dt} = 12 \cdot 2(1 + \sin t) \cdot \cos t = 24(1 + \sin t) \cos t$$

### The Elegant Cancellation

We now combine these components to find the slope $m$:

$$\frac{dy}{dx} = \frac{24(1 + \sin t) \cos t}{24 \cos^2 t}$$

By canceling the common factors of $24$ and $\cos t$, we arrive at the remarkably clean expression for the slope:

$$\frac{dy}{dx} = \frac{1 + \sin t}{\cos t}$$

### Solving the Trigonometric Puzzle

The problem states that the tangent makes an angle of $\frac{\pi}{3}$ with the $x$-axis. Therefore, the slope $m$ must satisfy:

$$m = 	an\left(\frac{\pi}{3}ight) = \sqrt{3}$$

Setting our derived expression equal to this value, we have:

$$\frac{1 + \sin t}{\cos t} = \sqrt{3} \implies 1 + \sin t = \sqrt{3} \cos t$$

Rearranging the terms yields $\sqrt{3} \cos t - \sin t = 1$. Dividing the entire equation by $2$ allows us to use the cosine addition formula:

$$\frac{\sqrt{3}}{2} \cos t - \frac{1}{2} \sin t = \frac{1}{2}$$

$$\cos\left(t + \frac{\pi}{6}ight) = \frac{1}{2}$$

Within the interval $0 < t < \frac{\pi}{2}$, this implies $t + \frac{\pi}{6} = \frac{\pi}{3}$, which results in **$t = \frac{\pi}{6}$**.

### Final Calculation

We substitute $t = \frac{\pi}{6}$ back into the expression for $y$ to find the vertical coordinate $y_0$:

$$y_0 = 12\left(1 + \sin\left(\frac{\pi}{6}ight)ight)^2$$

Given that $\sin\left(\frac{\pi}{6}ight) = \frac{1}{2}$, the calculation proceeds as follows:

$$y_0 = 12\left(1 + \frac{1}{2}ight)^2 = 12\left(\frac{3}{2}ight)^2 = 12 \cdot \frac{9}{4} = 27$$

The vertical coordinate of the point is **$y_0 = 27$**.

Animated Solution for Mathematics - Differentiation: If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x = 12 (t + sin t cos t)$ , $y = 12 (1 + sin t)^{2}, 0 < t < \frac{π}{2}$ , with the positive x-axis is $\frac{π}{3}$ , then $y_{0}$ is equal to

Select Answer:

Visualized Solution

Visualizing the Parametric Curve

The Slope Formula

Differentiating $x (t)$

Differentiating $y (t)$

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

Simplifying the Slope

Applying the Tangent Condition

Solving for $t$

Calculating $y_{0}$

Final Result

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Velocity of the Curve

The Elegant Cancellation

Solving the Trigonometric Puzzle

Final Calculation

Similar Questions

If the tangent to the curve $y = x + sin y$ at a point $(a, b)$ is parallel to the line joining $(0, \frac{3}{2})$ and $(\frac{1}{2}, 2)$ , then:

If the tangent to the curve $y = \frac{x}{x ^{2} - 3}, x \in R, x \neq = \pm 3$ , at a point $(α, β) \neq = (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then :

The point(s) on the curve $y^{3} + 3 x^{2} = 12 y$ where the tangent is vertical, is (are)

The equation of the tangent to the curve $y = x + \frac{4}{x ^{2}}$ , that is parallel to the $x$ -axis, is

The equation of the normal to the curve $y = (1 + x)^{2 y} + cos^{2} (sin^{- 1} x)$ at $x = 0$ is:

If the normal to the curve $y = f (x)$ at the point $(3, 4)$ makes an angle $\frac{3 π}{4}$ with the positive $x$ -axis, then $f^{'} (3) =$

Consider: $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})$ . A normal to $y = f (x)$ at $x = \frac{π}{6}$ also passes through the point

The normal to the curve $x = a (cos θ + θ sin θ)$ , $y = a (sin θ - θ cos θ)$ at any point ' $θ$ ' is such that

Find the equation of the normal to the curve $y = (1 + x)^{y} + sin^{- 1} (sin^{2} x)$ at $x = 0$

Angle between the tangents to the curve $y = x^{2} - 5 x + 6$ at the points $(2, 0)$ and $(3, 0)$ is

Select Answer:

Visualized Solution

Visualizing the Parametric Curve

The Slope Formula

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

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

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

Simplifying the Slope

Applying the Tangent Condition

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y0​

Final Result

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Velocity of the Curve

The Elegant Cancellation

Solving the Trigonometric Puzzle

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the curve y=x+siny at a point (a,b) is parallel to the line joining (0,23​) and (21​,2), then:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the curve y=x2−3x​,x∈R,x=±3​, at a point (α,β)=(0,0) on it is parallel to the line 2x+6y−11=0, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The point(s) on the curve y3+3x2=12y where the tangent is vertical, is (are)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the tangent to the curve y=x+x24​, that is parallel to the x-axis, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the normal to the curve y=(1+x)2y+cos2(sin−1x) at x=0 is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normal to the curve y=f(x) at the point (3,4) makes an angle 43π​ with the positive x-axis, then f′(3)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider: f(x)=tan−1(1−sinx1+sinx​​),x∈(0,2π​). A normal to y=f(x) at x=6π​ also passes through the point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The normal to the curve x=a(cosθ+θsinθ), y=a(sinθ−θcosθ) at any point 'θ' is such that

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equation of the normal to the curve y=(1+x)y+sin−1(sin2x) at x=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle between the tangents to the curve y=x2−5x+6 at the points (2,0) and (3,0) is