Exercise Question from Maths book

[King_Silver]

Homework Statement

"Find all the points on the curve with equation 6y = 2x^3 + 3x^2 + 6x + 5 where the tangent is parallel to the line with equation 3x − y + 2 = 0

Homework Equations

No idea

The Attempt at a Solution

There are no exercises in the book even similar to this example which is fairly frustrating. Its under the differentiation chapter so it obviously involves differentiation. I can do that myself no bother but what is the method to find all points on the curve when a tangent is parallel to the line? Is there some sort of rule?

NOTE: This is not a graded assignment or even homework for that matter, it is revision for an exam I have in 5 days.

 

[micromass]

Can you find the tangent line to your curve in an arbitrary point ##(a,b)##?

 

[King_Silver]

Errrrrr... I might be able to?
I need to find all points on the curve where the tangent is parallel to the line. I've tried finding a video on YouTube that works with a similar problem but I wasn't able to find anything for it.

 

[micromass]

How do you find the tangent line to the curve? Do you see the relation with derivatives?

 

[King_Silver]

How do you find the tangent line to the curve? Do you see the relation with derivatives?

You differentiate the 6y - 2x^3 +3x^2 +6x +5
but I am not given the points on the curve I am asked to find them :/

 

[SteamKing]

Homework Statement

"Find all the points on the curve with equation 6y = 2x^3 + 3x^2 + 6x + 5 where the tangent is parallel to the line with equation 3x − y + 2 = 0

Homework Equations

No idea

The Attempt at a Solution

There are no exercises in the book even similar to this example which is fairly frustrating. Its under the differentiation chapter so it obviously involves differentiation. I can do that myself no bother but what is the method to find all points on the curve when a tangent is parallel to the line? Is there some sort of rule?

NOTE: This is not a graded assignment or even homework for that matter, it is revision for an exam I have in 5 days.

Well, can you at least find the slope of the line 3x - y + 2 = 0?

 

[King_Silver]

y'(x) = 3 is the slope of the line 3x -y +2 = 0 right? :)

 

[SteamKing]

y'(x) = 3 is the slope of the line 3x -y +2 = 0 right? :)

Yes, this is correct.

Now, can you turn 6y = 2x^3 + 3x^2 + 6x + 5 into y = something ?

 

[King_Silver]

Yes, this is correct.

Now, can you turn 6y = 2x^3 + 3x^2 + 6x + 5 into y = something ?

Well if 6y = 2x^3 + 3x^2 +6x +5 then y = (2x^3 + 3x^2 +6x +5)/6 no?

 

[SteamKing]

Well if 6y = 2x^3 + 3x^2 +6x +5 then y = (2x^3 + 3x^2 +6x +5)/6 no?

That's correct.

Now do you understand that the derivative dy/dx for a function can be used to calculate the slope of a tangent to the function when given a particular value of x?

 

[King_Silver]

Yea I understand that I just don't get how to find the points on the curve where the tangent is parallel to the line.

 

[SteamKing]

Yea I understand that I just don't get how to find the points on the curve where the tangent is parallel to the line.

Well, have you calculated dy/dx for y = (2x^3 + 3x^2 +6x +5)/6 yet?

BTW: I appreciate the likes, but it's not necessary to like every reply.

 

[micromass]

How can you see whether two lines are parellel?

 

[King_Silver]

How can you see whether two lines are parellel?

If their slopes are identical? i.e. the exact same value yes?

 

[micromass]

Yes. So you need to find the points whose tangent line has slope is 3.

 

[King_Silver]

Well, have you calculated dy/dx for y = (2x^3 + 3x^2 +6x +5)/6 yet?

BTW: I appreciate the likes, but it's not necessary to like every reply.

dy/dx is x^2 + x + 1.

 

[SteamKing]

dy/dx is x^2 + x + 1.

Now that you can find the slope along the function, for what values of x would dy/dx = 3?

 

[King_Silver]

Now that you can find the slope along the function, for what values of x would dy/dx = 3?

dy/dx would = 3 for...
x = 1
(1)^2 + 1 + 1 = 3. What does that prove in terms of the question? :)

 

[SteamKing]

dy/dx would = 3 for...
x = 1
(1)^2 + 1 + 1 = 3. What does that prove in terms of the question? :)

Re-read your Post #14 above.

Also, is x = 1 the only value which satisfies the equation dy/dx = 3 when dy/dx = x² + x + 1 ?

 

[King_Silver]

Re-read your Post #14 above.

Also, is x = 1 the only value which satisfies the equation dy/dx = 3 when dy/dx = x² + x + 1 ?

x = 1 and x = -2 (because (-2)^2 + (-2) + 1 => 4 -2 + 1 = 3)
so you have x = 1, x = -2.

Then you just do when x = 1, y = whatever you solve.
x = -2, y = whatever you solve
?

 

[SteamKing]

x = 1 and x = -2 (because (-2)^2 + (-2) + 1 => 4 -2 + 1 = 3)
so you have x = 1, x = -2.

Then you just do when x = 1, y = whatever you solve.
x = -2, y = whatever you solve
?

When x = 1 or x = -2, dy/dx = 3 for the function 6y = 2x³ + 3x² + 6x + 5

The slope of the line 3x − y + 2 = 0 was found to be 3.

Now, the problem statement asked:

Homework Statement

"Find all the points on the curve with equation 6y = 2x^3 + 3x^2 + 6x + 5 where the tangent is parallel to the line with equation 3x − y + 2 = 0

Do you see how to do this now?

 

[King_Silver]

When x = 1 or x = -2, dy/dx = 3 for the function 6y = 2x³ + 3x² + 6x + 5

The slope of the line 3x − y + 2 = 0 was found to be 3.

Now, the problem statement asked:Do you see how to do this now?

(1,2 and 2/3s) and (-2, -1 and 5/6s) ?

 

[SteamKing]

(1,2 and 2/3s) and (-2, -1 and 5/6s) ?

It's probably better to use simple rational expressions than improper fractions here:

point 1 : (1, 8/3)
point 2 : (-2, -11/6)

 

[King_Silver]

It's probably better to use simple rational expressions than improper fractions here:

point 1 : (1, 8/3)
point 2 : (-2, -11/6)

Thanks for that :) I understand how to do it now cheers!