Equation of the Tangent Line to a Curve at a Given Point

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the curve defined by the function y = x^(1/4) at the point (1,1). The derivative of the function is calculated as y' = (1/4)x^(-3/4), which provides the slope of the tangent line at the specified point. The correct equation for the tangent line is derived as y = (1/4)x + (3/4) by substituting the point into the line equation y - y1 = m(x - x1). Participants clarify the distinction between the general equation of a line and the specific tangent line equation.

PREREQUISITES
  • Understanding of derivatives and their application in calculus
  • Familiarity with the concept of tangent lines
  • Knowledge of the slope-intercept form of a line (y = mx + b)
  • Basic exponent rules and simplification techniques
NEXT STEPS
  • Study the process of finding derivatives using the power rule
  • Learn how to apply the point-slope form of a line
  • Explore the concept of higher-order derivatives and their geometric interpretations
  • Practice problems involving tangent lines to various functions
USEFUL FOR

Students studying calculus, particularly those learning about derivatives and tangent lines, as well as educators seeking to clarify these concepts for their students.

Drakkith
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Homework Statement


Find an equation of the tangent line to the curve at the given point.

Homework Equations


y=x¼ Point = (1,1)

The Attempt at a Solution


[/B]
Derivative of y is y' = ¼x-¾
Plugging in the derivative to the equation for a line: y-1=¼(x-1)-¾.
My book's answer is Y=¼x+¾, but I don't know how they got there. Expanding ¼(x-1)-¾ gives me a nasty fraction with a radical that doesn't seem to come out to equal ¼x+¾. I'm not sure what to do next here.
 
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Drakkith said:

Homework Statement


Find an equation of the tangent line to the curve at the given point.

Homework Equations


y=x¼ Point = (1,1)

The Attempt at a Solution


[/B]
Derivative of y is y' = ¼x-¾
Plugging in the derivative to the equation for a line: y-1=¼(x-1)-¾.
My book's answer is Y=¼x+¾, but I don't know how they got there. Expanding ¼(x-1)-¾ gives me a nasty fraction with a radical that doesn't seem to come out to equal ¼x+¾. I'm not sure what to do next here.

That's not the equation of a line. Where did the exponent -3/4 come from? Review the form for the equation of a line again.
 
Drakkith said:

Homework Statement


Find an equation of the tangent line to the curve at the given point.

Homework Equations


y=x¼ Point = (1,1)

The Attempt at a Solution


[/B]
Derivative of y is y' = ¼x-¾
Plugging in the derivative to the equation for a line: y-1=¼(x-1)-¾.
My book's answer is Y=¼x+¾, but I don't know how they got there. Expanding ¼(x-1)-¾ gives me a nasty fraction with a radical that doesn't seem to come out to equal ¼x+¾. I'm not sure what to do next here.

You use the y' equation to give you the slope of the line. What is the basic equation for a straight line in terms of its slope and y-intercept?

Use the slope and the fact that the line goes through the point (1,1) to figure out the y-intercept to complete the equation...
 
Dick said:
That's not the equation of a line. Where did the exponent -3/4 come from? Review the form for the equation of a line again.

Since ¼x-¾ is the slope at any point, doesn't the -¾ go there?

berkeman said:
You use the y' equation to give you the slope of the line. What is the basic equation for a straight line in terms of its slope and y-intercept?

I believe it's: y=mx+b

Use the slope and the fact that the line goes through the point (1,1) to figure out the y-intercept to complete the equation...

Plugging in values for x and y: 1 = ¼(1)-¾ + b.
Simplifying: 1 = ¼ + b, so b = ¾.

So are they simplifying ¼x-¾ to ¼x at the point since x = 1?
 
Drakkith said:
So are they simplifying ¼x-¾ to ¼x at the point since x = 1?
There is a difference between "The Equation of a line" and "The Equation of a Tangent Line".
Check the links I provided.
 
Drakkith said:
So are they simplifying ¼x-¾ to ¼x at the point since x = 1?

Not exactly... you find y' = 1/4 at the point (1,1), and then separately substitute that into the equation of a line y = mx + b, and solve for b. Since you know the line passes through (1,1) at its point of tangency, you can use that point to solve for b.

There are likely many ways to solve this type of problem, but that's the way I prefer to do it.
 
Okay, so I have a curve given by the first equation. They want the tangent line at a certain point on the curve. The derivative of the original equation gives me another function. At any x-value, this new function gives me the slope of the curve at that x-value, correct? So to find the slope of the curve at (1,1) I'd plug 1 in for X, which gives me 1/4 as the slope. Then, using (1,1), I get y-1 = 1/4(x-1), which simplifies to y= 1/4x + 3/4, which is the equation for the tangent line at that point on the curve.

How's that look?
 
Drakkith said:
Okay, so I have a curve given by the first equation. They want the tangent line at a certain point on the curve. The derivative of the original equation gives me another function. At any x-value, this new function gives me the slope of the curve at that x-value, correct? So to find the slope of the curve at (1,1) I'd plug 1 in for X, which gives me 1/4 as the slope. Then, using (1,1), I get y-1 = 1/4(x-1), which simplifies to y= 1/4x + 3/4, which is the equation for the tangent line at that point on the curve.

How's that look?

Looks good! :smile:
 
  • #10
BTW, I'd probably just plug the point (1,1) into the y = mx + b equation, so 1 = (1/4)*1 + b to solve for b. You know the line goes through (1,1) and has the slope 1/4.
 
  • #11
Dick said:
That's not the equation of a line. Where did the exponent -3/4 come from? Review the form for the equation of a line again.

ToBePhysicist said:
Greetings,
Equation of a line.
Finding the Equation of a Tangent line.
I can't believe I linked to wiki how...:DD

Thanks for the help guys.

berkeman said:
Looks good! :smile:

Thanks, Berk. My main problems were that A.) I didn't really understand the overall problem and how everything worked together, and B.) for some reason I was thinking that x-3/4 = 1/3x1/4... Had to look up my exponent laws again to figure out that its one over the fourth root of x cubed, not three times the fourth root of x.
 
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  • #12
Drakkith said:
Thanks for the help guys.
Thanks, Berk. My main problems were that A.) I didn't really understand the overall problem and how everything worked together, and B.) for some reason I was thinking that x-3/4 = 1/3x1/4... Had to look up my exponent laws again to figure out that its one over the fourth root of x cubed, not three times the fourth root of x.
Any time, sir.
I hope people understand my way...
Instead of noticing that I am new here and not everything I say is accurate...People will click the links to notice some other (Experienced human or resources to get all the needed info) and getting the answers themselves...That or I am too lazy to type the answer! :wink:
 

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