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

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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
Plugging in the derivative to the equation for a line: y-1=¼(x-1).
My book's answer is Y=¼x+¾, but I don't know how they got there. Expanding ¼(x-1) 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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  • #2
Dick
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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
Plugging in the derivative to the equation for a line: y-1=¼(x-1).
My book's answer is Y=¼x+¾, but I don't know how they got there. Expanding ¼(x-1) 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.
 
  • #3
berkeman
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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
Plugging in the derivative to the equation for a line: y-1=¼(x-1).
My book's answer is Y=¼x+¾, but I don't know how they got there. Expanding ¼(x-1) 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...
 
  • #5
Drakkith
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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 is the slope at any point, doesn't the -¾ go there?

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) + b.
Simplifying: 1 = ¼ + b, so b = ¾.

So are they simplifying ¼x to ¼x at the point since x = 1?
 
  • #6
So are they simplifying ¼x 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.
 
  • #7
berkeman
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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.
 
  • #8
Drakkith
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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?
 
  • #9
berkeman
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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
berkeman
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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
Drakkith
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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.
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.

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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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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