Draw a tangent that is perpendicular to the line

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Homework Help Overview

The discussion revolves around finding a tangent line to the curve defined by the equation \(y = -\sqrt{2x^3}\) that is perpendicular to the line \(y = \frac{4}{3}x + \frac{1}{3}\). Participants are exploring the relationship between the slopes of the tangent and the given line.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the requirement for the tangent's slope to be -3/4, which is the negative reciprocal of the slope of the given line. There are inquiries about how to find the specific point on the curve where this condition holds true, including taking the derivative of the curve.

Discussion Status

Several participants have suggested taking the derivative of the curve and equating it to -3/4 to find the corresponding x-value. There is acknowledgment of the need for calculus knowledge to proceed, and some participants express confusion about the steps involved in this process.

Contextual Notes

There is an assumption that participants have a background in calculus, as the discussion hinges on the concept of derivatives and tangents. Some participants question the clarity of the instructions and the connection between tangents and derivatives.

TSN79
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I have a curve [itex]-\sqrt {2x^3 }[/itex] on which I'm supposed to draw a tangent that is perpendicular to the line [itex]\[y = \frac{4}{3}x + \frac{1}{3}[/itex].

I know that this tangent must have a "steepness" of -3/4 in order to make it perpendicular, but how do I now find the point on the graph?
 
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Solve for the x from the curve formula that would make its slope equal to that of the tangent.
 
Given that the tangent needs to be perpendicular to the line y = 4/3x + 1/3, he's right that the slope must be -3/4. Negative reciprocal for the normal line.

TSN79 - What you need is the point on the curve where the first derivative is -3/4. Just take the derivative, plug in that value, and solve for x.

Note: I'm presuming you're taking or have taken calculus, as I don't see any other way to do this.
 
Diane_ said:
Note: I'm presuming you're taking or have taken calculus, as I don't see any other way to do this.
He could do it geometrically I suppose, but that may or may not be exact.
 
Diane_ said:
Just take the derivative, plug in that value, and solve for x.
Take the derivative of what? And plug in what value? -3/4? I tried with [itex] - \sqrt {2x^3 }[/itex] but didn't really get anywhere...
 
Take the derivative of [itex]- \sqrt {2x^3 }[/itex] then equate it to -3/4 and solve for x.
 
Ah, now where getting somewhere, thanks you all!
 
Did you really need to be told that? You were told that the line had to be tangent to the curve given by [itex]y=-\sqrt {2x^3 }[/itex]. Didn't you connect "tangent" with "derivative"- after you had been told the find the derivative- the only tangent mentioned was to [itex]y= -\sqrt {2x^3 }[/itex].
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