Equation of a Tangent Line to a Polar Curve

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Homework Help Overview

The discussion revolves around finding the equation of the tangent line to a polar curve defined by the parametric equations x=t^4+1 and y=t^3+t, specifically at the point where t=-1. Participants are exploring how to determine the coordinates of the point on the curve at this parameter value, which is necessary for formulating the tangent line's equation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the slope of the tangent line and express uncertainty about finding the corresponding point on the curve. Questions arise regarding how to identify the intersection of the tangent line with the curve and the implications of the parameter t=-1.

Discussion Status

The conversation is ongoing, with participants seeking clarification on the steps needed to find the point on the curve and the tangent line's equation. Some guidance has been offered regarding the relationship between the slope and the point needed to formulate the equation, but there is no explicit consensus on the next steps.

Contextual Notes

Participants note the challenge of determining the intersection of the tangent line and the curve without having the tangent line's equation. There is also a recognition that the parameter t=-1 was used to calculate the slope but may not correspond to a point of intersection.

zmilot
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I need to find the equation of the tangent line to the curve

x=t4+1, y=t3+t; t=-1

I have already found that the slope of the line is -1 by finding (dy/dt)/(dx/dt) I just need to figure out how to solve for y1 and x1

Thanks in advance
 
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Hi zmilot! :smile:

zmilot said:
I need to find the equation of the tangent line to the curve

x=t4+1, y=t3+t; t=-1

I have already found that the slope of the line is -1 by finding (dy/dt)/(dx/dt) I just need to figure out how to solve for y1 and x1

Thanks in advance

So you've got the slope of the tangent line, that's good. Can you find a point on the tangent line?? If you have the slope and any point, then you can easily find the equation with certain formulas.
 
Thats what I was trying to figure out, how do you find a point on the tangent line, all I have is the slope and I don't know where they intersect
 
zmilot said:
Thats what I was trying to figure out, how do you find a point on the tangent line, all I have is the slope and I don't know where they intersect

What's the intersection of the tangent line with the curve?
 
The only way I know to find the intersection is to set the equations equal to each other, but since I don't have the tangent line equation I am not sure how to go about this.
 
When t=-1, what is x? What is y? is this the point for which you calculated the derivative?
 
I don't think they line intersects at t=-1, that was just the number we were given to plug into the first derivative in order to find the slope. I am really lost on this problem so if anyone could make sure my first step is right and then walk me through the rest of the process it would be greatly appreciated.
 
You found the tangent line to the point at which t equals negative one. I wasn't really asking a question so much as inviting you to see it.
 
You're taking the tangent line of the curve at t=-1. What does that mean?
 

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