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Find the equation of the tangent line to the curve y = x^2 at the point (1, 1). The slope of the tangent line is given to be 2.
The equation I need to find is the line that goes through or touches the curve at the point (1, 1) located in quadrant 1.
I am going to use the point-slope formula.
y - y_1 = m(x - x_1)
Let x_1 = 1
Let y_1 = 1
Let m = slope of the tangent line.
y - 1 = 2(x - 1)
y - 1 = 2x - 2
y = 2x - 2 +1
y = 2x - 1
The equation y = 2x - 1 is the tangent line at the point (1, 1).
Is any of this right?
Follow-up questions for the reader.
1. How do I find the slope of the secant line?
2. What about the slope of the normal line?
3. What is the BASIC DIFFERENCE between the slope of the tangent line, secant line and normal line?
4. In the question above, the slope of the tangent line is given. What if the slope is not given? How do I proceed?
Thank you.
Last edited by nycmathguy (2021-06-03 01:03:46)
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hi
Your working is all correct. You can differentiate the function (dy/dx = 2x) to get the gradient function and then substitute x = 1.
The normal to a curve at a point A is the line through A at right angles to the tangent.
So in your example, dy/dx = 2x (= 2 at x = 1). I think you already know that a line perpendicular to the tangent has gradient n
where nm = -1. So the gradient of the normal is -1/2. And it goes through (1,1) so you can use the same method as above to get the equation.
If you know two points on a curve, A and B, the secant line is the line AB.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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hi
Your working is all correct. You can differentiate the function (dy/dx = 2x) to get the gradient function and then substitute x = 1.
The normal to a curve at a point A is the line through A at right angles to the tangent.
So in your example, dy/dx = 2x (= 2 at x = 1). I think you already know that a line perpendicular to the tangent has gradient n
where nm = -1. So the gradient of the normal is -1/2. And it goes through (1,1) so you can use the same method as above to get the equation.If you know two points on a curve, A and B, the secant line is the line AB.
Bob
You said:
"So in your example, dy/dx = 2x (= 2 at x = 1). I think you already know that a line perpendicular to the tangent has gradient nwhere nm = -1. So the gradient of the normal is -1/2. And it goes through (1,1) so you can use the same method as above to get the equation."
Can you please provide a sample for this statement?
"If you know two points on a curve, A and B, the secant line is the line AB."
Can you please provide an example for this statement?
Can you please provide a graph using different color lines showing the tangent line, secant line and normal line all on the same xy-plane?
Thank you.
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Hope this satisfies all:
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hope this satisfies all:
https://i.imgur.com/umbEqXL.gif
Bob
Beautiful graph. Nice colors. Tell me, how did you upload this graph? I may need to graph some functions in the coming chapters.
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hi nycmathguy
Funnily enough there's another member in NYC who is studying exactly the same topics as you who also asked exactly this question.
He wanted to be told the steps just one step at a time. I'm guessing you would like the same.
The first step is to sign up to www.imgur.com.
Post back when you have done that and I'll tell you step 2.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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hi nycmathguy
Funnily enough there's another member in NYC who is studying exactly the same topics as you who also asked exactly this question.
He wanted to be told the steps just one step at a time. I'm guessing you would like the same.
The first step is to sign up to www.imgur.com.
Post back when you have done that and I'll tell you step 2.
Bob
There are over 20 million people in NYC. Why are you startled? Are you saying I am the other guy? What's his username? NYC is divided into 5 boroughs. Where does he live? Am I the only person studying calculus in the Big Apple?
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I'll assume you don't want help with posting pictures of graphs then.
B
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I'll assume you don't want help with posting pictures of graphs then.
B
All I need is the steps. This is my first time joining a site that does not have an attachment for pictures.
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hi nycmathguy
Funnily enough there's another member in NYC who is studying exactly the same topics as you who also asked exactly this question.
He wanted to be told the steps just one step at a time. I'm guessing you would like the same.
The first step is to sign up to www.imgur.com.
Post back when you have done that and I'll tell you step 2.
Bob
I signed up at www.imgur.com.
I had to download the app.
What is next?
Last edited by nycmathguy (2021-06-04 04:31:58)
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Good. I've given the next step on your other post. So let's call this thread finished.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Good. I've given the next step on your other post. So let's call this thread finished.
Bob
What other post?
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