Calculating the equations for the tangent/normal lines

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SUMMARY

The discussion focuses on calculating equations for tangent and normal lines in calculus. Participants confirm that the slope of the tangent line at a point can be derived from the derivative, with specific examples including f'(2) = -3 leading to the equation y = -3x + 1. When the tangent slope is zero, the normal line is vertical, represented as x = k, where k is the x-value at the point of tangency. Additionally, when the tangent slope is undefined, the normal slope is zero, resulting in a horizontal line equation.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives and slopes.
  • Familiarity with the equations of lines, including point-slope form.
  • Knowledge of tangent and normal lines in the context of curves.
  • Ability to interpret graphical representations of functions and their derivatives.
NEXT STEPS
  • Study the concept of derivatives in calculus, focusing on how to find slopes at specific points.
  • Learn about the geometric interpretation of tangent and normal lines to curves.
  • Explore examples of vertical and horizontal lines in relation to slopes of tangent lines.
  • Practice solving problems involving tangent and normal lines using various functions.
USEFUL FOR

Students studying calculus, educators teaching derivatives, and anyone looking to deepen their understanding of tangent and normal lines in mathematical analysis.

fulton33
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Homework Statement
Use the table to write the equation for tangent lines at given values of x.
Relevant Equations
y-y0=m(x-x0)
IMG_3607.jpeg

9. When I do this problem I know my slope is -3 because f'(2)=-3. I then went and substituted and got
y+5=-3(x-2) which simplified to y=-3x+1

10. I get lost here because the tangent slope would be 0, which would give me the equation y=-2. The normal means perpendicular and the perpendicular slope to 0 is undefined. Not sure if that is right and what to do after.

11. I did the same steps in 9. The slope is 3 and I get the equation y-4=3(x+1) which simplifies to y=3x+7

12. I am lost here as well. The tangent slope would be DNE, which would mean the normal slope to be 0. When I plug 0 in for m and (2,0) for x and y I get y=0. I think that is wrong.
 
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fulton33 said:
10. I get lost here because the tangent slope would be 0, which would give me the equation y=-2. The normal means perpendicular and the perpendicular slope to 0 is undefined. Not sure if that is right and what to do after.
If the slope of the tangent line is 0, its normal will be a vertical line of the form x = k, where k is the x-value at the point of tangency.
fulton33 said:
12. I am lost here as well. The tangent slope would be DNE, which would mean the normal slope to be 0. When I plug 0 in for m and (2,0) for x and y I get y=0. I think that is wrong.
This is the converse of #10. If the tangent slope is undefined, a line perpendicular to it will have slope 0.
 
Mark44 said:
If the slope of the tangent line is 0, its normal will be a vertical line of the form x = k, where k is the x-value at the point of tangency.
This is the converse of #10. If the tangent slope is undefined, a line perpendicular to it will have slope 0.

10. Does that mean for number 10 the equation would be x=0 because at x=-1 f'(-1)=0?

12. Does that mean that for number 12 y=0 would then be correct?
 
fulton33 said:
Does that mean for number 10 the equation would be x=0 because at x=-1 f'(-1)=0?
No. You're given information about the point (-1, -2).
fulton33 said:
Does that mean that for number 12 y=0 would then be correct?
Yes.
 
Mark44 said:
No. You're given information about the point (-1, -2).
Yes.
That makes sense about 10. That would make it x=-1?
 

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