Question

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.__x + √xy = 6, (4, 1)

Accepted Answer

First, understand that the problem requires finding the tangent and normal lines to the curve given by the equation $$x + \sqrt{xy} = 6 at $$the point (4, 1). The tangent line is the line that just touches the curve at this point, while the normal line is perpendicular to the tangent line at this point. To find the equation of the tangent line, we need to determine the derivative of the curve with respect to x, which will give us the slope of the tangent line at the point (4, 1). Start by differentiating both sides of the equation $$x + \sqrt{xy} = 6$$ with respect to x. Use implicit differentiation since y is a function of x. Differentiate the left side: The derivative of $$x is 1. $$For $$\sqrt{xy}$$, use the chain rule: $$\frac{d}{dx}(\sqrt{xy}) = \frac{1}{2\sqrt{xy}} \cdot (y + x \frac{dy}{dx}). $$Set the derivative of the right side, which is a constant, to 0. Solve the resulting equation for $$\frac{dy}{dx} to $$find the slope of the tangent line at the point (4, 1). Substitute x = 4 and y = 1 into the derivative to find the specific slope at this point. Once you have the slope of the tangent line, use the point-slope form of a line equation, $$y - y_1 = m(x - x_1)$$, where m is the slope and ($$x_1$$, $$y_1) is $$the point (4, 1), to write the equation of the tangent line. For the normal line, use the negative reciprocal of the tangent slope as the slope of the normal line, and apply the point-slope form again to find its equation.

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.
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x + √xy = 6, (4, 1)

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Key Concepts

Implicit Differentiation

Tangent Line

Normal Line