If the tangent to the curve y = x + sin y at a point (a, b) is parallel to the line joining (0, 3/2)

Q: Why can't we just solve for $y$ in $y = x + \sin y$?

That is a great question! The equation $y = x + \sin y$ is a transcendental equation. This means you cannot isolate $y$ in terms of elementary functions of $x$. That is exactly why we use implicit differentiation—it allows us to find the derivative $\frac{dy}{dx}$ without needing an explicit formula for $y$.

Q: Why is the slope of the tangent equal to the slope of the line?

Because the problem explicitly states that the tangent is parallel to the given line! In coordinate geometry, parallel lines always have identical slopes. By finding the slope of the line connecting the two given points, we effectively find the slope of the tangent at the point $(a, b)$.

Q: How do we know which value of $\sin b$ to choose?

Excellent observation. Since our condition led to $\cos b = 0$, $\sin b$ can indeed be $1$ or $-1$. In such cases, we look at the provided options. If $b - a = 1$ is an option, we check if it corresponds to one of our cases. Since $\sin b = 1$ leads to $b = a + 1$, which rearranges to $b - a = 1$, we have found our match!

Question

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

Options:

(A) $b = a$

(B) $b = \frac{\pi}{2} + a$

(C) $a + b = 1$

(D) $b - a = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at the curve defined by the equation $y = x + \sin y$. We are interested in a specific point on this curve, which we call $(a, b)$.

At this exact location, a tangent line touches the curve. The problem states that this tangent is parallel to a line passing through $(0, \frac{3}{2})$ and $(\frac{1}{2}, 2)$.

In geometry, parallel lines share the same orientation, meaning they must have the same slope. We calculate the slope of the given line using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$:

$$m_{line} = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$$

Since our tangent is parallel to this line, the slope of the tangent at $(a, b)$ must also be $1$. This serves as our anchor point for the remainder of the derivation.

### The Calculus of the Curve: Implicit Differentiation

We now turn our attention to the curve $y = x + \sin y$. To find the slope of the tangent at any point, we calculate the derivative $\frac{dy}{dx}$ using implicit differentiation.

Differentiating both sides of the equation with respect to $x$:

$$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\sin y)$$

Applying the chain rule to the $\sin y$ term, we obtain:

$$\frac{dy}{dx} = 1 + \cos y \cdot \frac{dy}{dx}$$

To isolate $\frac{dy}{dx}$, we group the terms involving the derivative on one side:

$$\frac{dy}{dx} - \cos y \cdot \frac{dy}{dx} = 1$$

Factoring out $\frac{dy}{dx}$, we arrive at the general expression for the slope:

$$\frac{dy}{dx}(1 - \cos y) = 1 \implies \frac{dy}{dx} = \frac{1}{1 - \cos y}$$

### The Synthesis: Bringing It All Together

We know the slope of the tangent at $(a, b)$ must be $1$. Substituting $y = b$ into our derivative expression and setting it equal to $1$, we get:

$$\frac{1}{1 - \cos b} = 1$$

Cross-multiplying yields $1 = 1 - \cos b$, which simplifies to $\cos b = 0$. Using the identity $\sin^2 b + \cos^2 b = 1$, we find that $\sin^2 b = 1$, implying $\sin b = 1$ or $\sin b = -1$.

Finally, we substitute these values into the original curve equation $b = a + \sin b$:

1. If $\sin b = 1$, then $b = a + 1$, which implies **$b - a = 1$**.
2. If $\sin b = -1$, then $b = a - 1$, which implies **$a - b = 1$**.

Depending on the specific constraints of the coordinate system, the relationship **$b - a = 1$** represents the valid geometric condition for the tangent at the point $(a, b)$.

Animated Solution for Mathematics - Differentiation: 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:

Select Answer:

Visualized Solution

Visualizing the Curve

Point on the Curve

Slope of the Given Line

Calculating the Line's Slope

Implicit Differentiation

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

Equating the Slopes

Solving for $cos b$

Finding $sin b$

The Final Relationship

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus of the Curve

The Synthesis

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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; } 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:

Select Answer:

Visualized Solution

Visualizing the Curve

Point on the Curve

Slope of the Given Line

Calculating the Line's Slope

Implicit Differentiation

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

Equating the Slopes

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

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

The Final Relationship

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus of the Curve

The Synthesis

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