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2. At what point on the curve y = sqrt{1 + 2x} is the tangent line perpendicular to the line 6x + 2y = 1. 3. Consider the conchiod of Nicomedes, given by x^2y^2 = (y + 1)^2(4 - y^2). (a) Use www.desmos.com to sketch this curve. (b) Is the curve a function? Explain your answer. (c) Find an equation of the tangent line to the curve at the point (2sqrt{3}, 1). Check your answer using www.desmos.com.

          2. At what point on the curve y = sqrt{1 + 2x} is the tangent line perpendicular to the line 6x + 2y = 1.
3. Consider the conchiod of Nicomedes, given by
x^2y^2 = (y + 1)^2(4 - y^2).
(a) Use www.desmos.com to sketch this curve.
(b) Is the curve a function? Explain your answer.
(c) Find an equation of the tangent line to the curve at the point (2sqrt{3}, 1). Check your answer using www.desmos.com.

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2. At what point on the curve y = sqrt1 + 2x is the tangent line perpendicular to the line 6x + 2y = 1.
3. Consider the conchiod of Nicomedes, given by
x^2y^2 = (y + 1)^2(4 - y^2).
(a) Use www.desmos.com to sketch this curve.
(b) Is the curve a function? Explain your answer.
(c) Find an equation of the tangent line to the curve at the point (2sqrt3, 1). Check your answer using www.desmos.com.

Added by Born G.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Angela Guo

02/18/2021

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Transcript

00:01 For the first question, we're asked to find the point that's on the curve y is equal to square root of 1 plus 2x that is perpendicular, which we both denote by the symbol, to the line 6x plus 2y is equal to 1.

00:23 So what you need to realize is that in order for two lines to be perpendicular, they need to have the slope has to be negative reciprocals of each other.

00:34 So meaning the first thing we need to do is find the slope of this line.

00:39 So to find the slope, you can just change this to slope intercept form.

00:43 So subtract 6x on both sides and then divide both sides by 2.

00:48 So we end up with y is equal to negative 3x.

00:53 Plus a half.

00:54 So therefore, the slope of this line is negative three.

00:59 Meaning the tangent line to this curve has to have a slope of positive 1 over 3 because it's the negative reciprocal.

01:10 So meaning, want slope, let's call that m1 and we'll call the other one m0, want the slope here to be positive 1 third.

01:24 So how do we find the slope? well, this is what taking the derivative is.

01:29 So if we take the derivative of this curve, then that will give us the slope at any point.

01:35 So let's go ahead and take the derivative.

01:39 So we have y prime is equal to.

01:43 So this is a square root.

01:44 Think of this as 1 plus 2x to the power of 1 half...

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