Finding the eq. of all tangent lines on a curve

Click For Summary

Homework Help Overview

The problem involves finding the equations of all tangent lines to the curve defined by y = x^2 + 2x that also pass through the point (3, 14). The original poster specifies not to use derivatives in their approach.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the correct form of the curve, with some questioning whether the original equation was miswritten. There is a suggestion to clarify the equation of the curve before proceeding.

Discussion Status

The discussion has clarified the equation of the curve as y = x^2 + 2x. Some participants have begun outlining potential steps to find the tangent lines, including considerations of slope and points of tangency, though no consensus on a method has been reached.

Contextual Notes

There is an explicit constraint against using derivatives, which may affect the approaches discussed. The original poster expresses uncertainty about how to start the problem.

Gus_Chiggins
Messages
3
Reaction score
0

Homework Statement


Problem: Find the equations of all tangent lines to the curve
y = x + 2x so that also go through the point (3, 14).


2. Do not use a derivative



3. I don't even know where to start. I searched my book there isn't really available. Any help would be much appreciated
 
Physics news on Phys.org
Gus_Chiggins said:

Homework Statement


Problem: Find the equations of all tangent lines to the curve
y = x + 2x so that also go through the point (3, 14).


2. Do not use a derivative



3. I don't even know where to start. I searched my book there isn't really available. Any help would be much appreciated

Is that really supposed to be y=x+2x, or have you mis-typed something?
 
What's the correct equation? The "curve" y = x + 2x is a straight line that doesn't go through (3, 14), so no tangent can go through this point either.

Should it be y = x^2 + 2x?
 
sorry everybody,

yes I meant to say

y=x^2 + 2x

sorry
 
OK, now that we've gotten that out of the way...

Let [itex](x_0, y_0)[/itex] be the point of tangency on the graph of the curve. BTW, you have drawn the graph, right?

At the point of tangency, the tangent line has to extend from [itex](x_0, y_0)[/itex] to (3, 14).

Here is an outline of the steps you'll need to carry out for this problem:

1. Find the slope of the line from [itex](x_0, y_0) = (x_0, x_0^2 + 2x_0)[/itex] to (3, 14).
2. By calculating the derivative and evaluating it at [itex]x_0[/itex], find the slope of the tangent line.
3. Equate the value you got in step 1 with the value from step 2, and solve for [itex]x_0[/itex]. (I got two values for [itex]x_0[/itex].)
4. Find the associated y value for each value of [itex]x_0[/itex] from step 3.
5. Using each point [itex](x_0, y_0)[/itex], find the equation of the line from [itex](x_0, y_0)[/itex] to (3, 14). There are two distinct equations.

Is that enough of a hint?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
4K