/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 Find an equation of the line tan... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 69

Find an equation of the line tangent to the following curves at the given point. $$x^{2}=-6 y ;(-6,-6)$$

Short Answer

Expert verified
Answer: The equation of the tangent line is $$y = 2x$$.

Step by step solution

01

Rewrite the given equation as a function of x and y

The given equation is already a function of x and y, so we can proceed to the next step without any changes.
02

Rewrite the equation as a function of y

We need to rewrite the equation in the form of $$y = f(x)$$. By solving for y, we have: $$y = -\frac{1}{6}x^{2}$$
03

Find the derivative dy/dx

To find the tangent line's slope at a given point, we need to calculate the derivative of y with respect to x: $$\frac{dy}{dx} = \frac{d}{dx} (-\frac{1}{6}x^{2})$$ Use the power rule: $$\frac{dy}{dx} = -\frac{1}{3}x$$
04

Evaluate the derivative at the given point

Now, plug in the x-coordinate of the given point (-6) into the derivative to find the tangent line's slope at the point: $$m = -\frac{1}{3}(-6) = 2$$
05

Use the point-slope form to find the equation of the tangent line

The point-slope form of a line's equation is given by: $$y - y_{1} = m (x - x_{1})$$ Plug in the coordinates of the given point and the slope calculated in step 4: $$y - (-6) = 2 (x - (-6))$$ Simplify: $$y + 6 = 2(x + 6)$$ Now, distribute and rearrange the equation to get the final equation of the tangent line: $$y = 2x + 6 - 6$$ $$y = 2x$$ The equation of the tangent line to the curve $$x^2 = -6y$$ at the point $$(-6, -6)$$ is $$y = 2x$$.

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Points at which the graphs of \(r=f(\theta)\) and \(r=g(\theta)\) intersect must be determined carefully. Solving \(f(\theta)=g(\theta)\) identifies some-but perhaps not all-intersection points. The reason is that the curves may pass through the same point for different values of \(\theta .\) Use analytical methods and a graphing utility to find all the intersection points of the following curves. \(r=2 \cos \theta\) and \(r=1+\cos \theta\)

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