Question

In Exercises 17-26, find the lines that are (a) tangent and (b) normal to the curve at the given point. $$x \sin 2 y=y \cos 2 x, \quad(\pi / 4, \pi / 2)$$

Short answer

The equations of the tangent and normal lines to the curve \(x \sin 2 y=y \cos 2 x\) at the point \((\pi / 4, \pi / 2)\) are obtained after finding the derivative (slope of the tangent), using the point-slope form of the linear equation to obtain the equation of the tangent, and finding the negative reciprocal to get the slope of the normal, again using the point-slope form.

Step-by-step solution

Find the Derivative of the Function

Differentiate implicitly given function \(x \sin 2 y=y \cos 2 x\) with respect to \(x\). Apply the product rule and chain rule correctly to get the derivative.

Solve for dy/dx (slope of tangent)

After differentiating, solve the resulting equation for \(\frac{dy}{dx}\). Remember that you are interested in finding slope of the tangent, which is given by \(\frac{dy}{dx}\) at the particular point.

Substitute the Point into the Derivative

Substitute the point \((\pi / 4, \pi / 2)\) in place of \(x\) and \(y\) in your equation gotten from Step 2 to find the value of the slope of the tangent at the given point.

Find the Equation of the Tangent Line

Use the point-slope form of linear equation to get the equation of the tangent line. You know that the form is \(y - y1 = m(x - x1)\) where \(m\) is the slope of the line and \((x1, y1)\) is the given point on the line.

Find the slope of the normal line

Slope of the normal line is the negative reciprocal of the slope of the tangent line. So, compute the negative reciprocal using the slope obtained in Step 3.

Find the Equation of the Normal Line

Just like you found the equation of the tangent line in Step 4, now find the equation of the normal line using the slope obtained in Step 5 and the given point.

Key concepts

Implicit Differentiation

Implicit differentiation is a powerful technique in calculus that is used when dealing with equations not explicitly solved for one variable in terms of another. In problems involving curves, like the given equation \(x \sin 2y = y \cos 2x\), not all relationships between variables can be expressed in the form \(y = f(x)\).

In implicit differentiation, we differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\) (noted as \(y(x)\)). This means when we differentiate \(y\), we apply the chain rule, multiplying by \(\frac{dy}{dx}\) after taking the derivative with respect to \(x\). The ultimate goal is to solve for \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\), which represents the slope of the tangent line we are seeking.

Finding Slope of a Tangent

The slope of a tangent line to a curve at a given point reflects how steep the line is at that intersection. When we know the function explicitly, \(y = f(x)\), we find this slope by simply differentiating \(f(x)\) to get \(f'(x)\). However, with an implicit function, after using implicit differentiation, we must evaluate \(\frac{dy}{dx}\) at the specific point of interest.

For the given problem, once implicit differentiation is conducted and we solve for \(\frac{dy}{dx}\), we substitute the values of \(x\) and \(y\) from the given point \((\pi/4, \pi/2)\) to find the precise slope of the tangent line at that point. This value is critical as it dictates the angle at which the tangent line touches the curve.

Point-Slope Form Equation

The point-slope form of a linear equation is a quick way to write the equation of a line when you know a point on the line and its slope. The general form is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point the line passes through and \(m\) is the slope of the line.

Once we've found the slope of the tangent line, we use the coordinates of our known point and the calculated slope to plug into this formula. In our case, the point given is \((\pi/4, \pi/2)\), and by substituting the slope from previous steps, we get the particular equation of the tangent line. This equation is incredibly valuable for visualizing and understanding the specific linear behavior of the curve at the given point.

Slope of Normal Line

The normal line to a curve at a certain point is perpendicular to the tangent line at that point. In the coordinate plane, if one line has a slope \(m\), the line perpendicular to it will have a slope that is the negative reciprocal, or \(-\frac{1}{m}\).

In the exercise, finding the slope of the normal line means taking the negative reciprocal of the tangent line's slope we already calculated. With the slope of the normal line and the given point on the curve, we can apply the same point-slope formula to find the equation of the normal line. This line signifies the direction in which the curve is flattest at the point of tangency, offering a complete geometric perspective of the function's behavior at \((\pi/4, \pi/2)\).