Question

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$\sin y+5 x=y^{2} ;(0,0)$$ (Graph cant copy)

Short answer

Answer: The equation of the tangent line to the curve at the point (0, 0) is $$y = -5x$$.

Step-by-step solution

a. Verify that the given point lies on the curve

First, we will check if the given point (0,0) satisfies the equation. Plug the coordinates of the point into the equation: $$\sin(0) + 5(0) = 0^2$$ This simplifies to: $$0 = 0$$ Since the point (0,0) satisfies the equation, it lies on the curve.

b. Determine an equation of the line tangent to the curve at the given point

Now, we will find the equation of the tangent line to the curve at the point (0,0). Since the equation is given implicitly, we need to find the derivative of y with respect to x using implicit differentiation. Differentiate both sides of the equation $$\sin y + 5x = y^2$$ with respect to x, to get: $$\frac{d}{dx}(\sin y + 5x) = \frac{d}{dx}(y^2)$$ By using the chain rule with implicit differentiation, we have $$\cos y \frac{dy}{dx} + 5 = 2y\frac{dy}{dx}$$ Now, we want to solve for $$\frac{dy}{dx}$$ at the point (0, 0). Substitute y = 0 and x = 0, to get: $$\cos(0) \frac{dy}{dx} + 5 = 2(0)\frac{dy}{dx}$$ This simplifies to: $$1\frac{dy}{dx} + 5 = 0$$ Now, solve for $$\frac{dy}{dx}$$: $$\frac{dy}{dx} = -5$$ The slope of the tangent line to the curve at the point (0, 0) is -5. To find the equation of the tangent line, we will use the point-slope form of a line, which is given by: $$y - y_{1} = m(x - x_{1})$$ Substitute the point (0, 0) and the slope m = -5, to get: $$y - 0 = -5(x - 0)$$ This simplifies to: $$y = -5x$$ The equation of the tangent line to the curve at the point (0, 0) is $$y = -5x$$.

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used to find derivatives of functions that aren't explicitly solved for one variable in terms of another. Instead, these functions are intertwined, or "implicit." This method is particularly helpful when dealing with complex equations that describe curves.
To carry out implicit differentiation, we differentiate both sides of the equation with respect to one variable, typically x, while treating other variables as implicit functions of x. This often involves applying the chain rule and product rule of differentiation.

• Differentiate terms as usual, applying the chain rule where needed.
• Treat other variables as functions of x, not constants.
• Collect terms containing the derivative and solve for it.

In the example from the original exercise, implicit differentiation allows us to find how y changes concerning x, even though y isn't isolated on one side of the equation.

Point-Slope Form

Point-slope form is a favored method for writing the equation of a line when you know a point on the line and its slope. The formula is written as:
\[y - y_1 = m(x - x_1)\]where:

• \(m\) is the slope of the line,
• (\(x_1, y_1\)) is a known point on the line.

This form is straightforward to use because it substitutes directly from known values, avoiding those complex algebraic manipulations that other forms sometimes require.
In the exercise, once we determined the slope \(m = -5\) at the point (0, 0), inserting these values into the point-slope form immediately gives us the tangent line's equation: \(y = -5x\). This illustrates how point-slope form simplifies writing equations of tangent lines.

Slope of Tangent Line

The slope of a tangent line to a curve at a given point is critical for understanding the behavior of the curve at that location. It essentially tells us the immediate "direction" or inclination of the curve.
The tangent slope is found by computing the derivative of the function at that point.

• If considering the function’s derivative at point (0,0), the process involves substitution of these coordinates into the derivative equation obtained through implicit differentiation.
• Thus, you can calculate the slope, which is denoted as \(\frac{dy}{dx}\) or simply \(m\) in point-slope form equations.

In our example, we calculated the slope of the tangent line to be -5 at the point (0,0). This value helps define the exact orientation and steepness of the tangent line, leading us to understand how our curve behaves at that precise point.

Derivative

Derivatives, in the broader sense of calculus, measure how a function changes as its input changes. They represent the function's rate of change and are foundational to understanding tangents, rates, and many other concepts in calculus.
In the context of our exercise:

• We use derivatives to find tangent slopes, determining how the curve behaves locally around a point.
• The derivative, calculated using implicit differentiation, tells us exactly how y changes in relation to x in the original implicit function.

Derivatives transform abstract functional relationships into tangible, precise data about their behaviors, such as speed, acceleration, or in this context, direction at a given point, making them indispensable in both theoretical and applied mathematics.