Equation of Tangent (2 Viewers)

[Smile12345]

Hello...

Could someone please help me with the following question?

Q: Find the equation of the tangent to the curve y=x^2(2x-1)^4 at the point (1,1)...

I began by differentiating... Is this right? u= x^2, u'= 2x, v=(2x-1)^4, v' = 8(2x-1)^3.... Which went to 2x(2x-1)^4 + x^2 x 8(2x-1)^3....

Now.... Not sure what to do

Thanks for your help in advance....

 

[SuperMike96]

Sub x=1 one into the differentiated equation (you were given the point (1,1) ). This will give you the gradient. Now you have all you need to use the point gradient formula (sub the point (1,1) and the gradient into the formula y-y1=m(x-x1)

 

[Najji]

Hello...

Could someone please help me with the following question?

Q: Find the equation of the tangent to the curve y=x^2(2x-1)^4 at the point (1,1)...

I began by differentiating... Is this right? u= x^2, u'= 2x, v=(2x-1)^4, v' = 8(2x-1)^3.... Which went to 2x(2x-1)^4 + x^2 x 8(2x-1)^3....

Now.... Not sure what to do

Thanks for your help in advance....

Does that help? 8x^2

 

[Smile12345]

Sub x=1 one into the differentiated equation (you were given the point (1,1) ). This will give you the gradient. Now you have all you need to use the point gradient formula (sub the point (1,1) and the gradient into the formula y-y1=m(x-x1)

Thanks heaps... Yes, I understand exactly what you mean... (Sorry I might have been thinking a bit too complex!!)

 

[Smile12345]

Does that help? 8x^2

Yes, that is the next step if I was factorising (and simplifying where possible)...

 

[SuperMike96]

Did you get the right answer?

 

[Smile12345]

Did you get the right answer?

Yes I did... 10x-y-9=0. Thanks again.

 

[Smile12345]

What about if it's like ... Find exact values of x for which the gradient of the tangent to the curve y=2x(x+3)^2 is 14... I've differentiated and expanded to 6x^2 + 24x +18....

Now... Could someone please help me?

 

[EpikHigh]

What about if it's like ... Find exact values of x for which the gradient of the tangent to the curve y=2x(x+3)^2 is 14... I've differentiated and expanded to 6x^2 + 24x +18....

Now... Could someone please help me?

Differentiate, since the derivative is the gradient function, sub dy/dx = 14 to find the values for X so

6x^2 + 24x + 18 = 14

6x^2 + 24x - 4 = 0

Use quadratic formula to find values of x for the gradient of the tangent to equal 14.

 

[cookiez69]

What about if it's like ... Find exact values of x for which the gradient of the tangent to the curve y=2x(x+3)^2 is 14... I've differentiated and expanded to 6x^2 + 24x +18....

Now... Could someone please help me?

Given that the gradient of the tangent to the curve at that particular value of x is 14,
dy/dx = 14

Therefore,
6x^2 + 24x + 18 = 14
6x^2 + 24x + 4 = 0
Now solve for x

 

[Smile12345]

Use quadratic formula to find values of x for the gradient of the tangent to equal 14.

Thanks... This is the stage I was up to.

 

[Smile12345]

Given that the gradient of the tangent to the curve at that particular value of x is 14,
dy/dx = 14

Therefore,
6x^2 + 24x + 18 = 14
6x^2 + 24x + 4 = 0
Now solve for x

Thanks heaps... Solving using the quad. formula.

 

[cookiez69]

Yup :]

 

[Smile12345]

Good... Thanks heaps. Have you finished this topic recently???

 

[EpikHigh]

Thanks heaps... Solving using the quad. formula... That right?

That's what I told you?!?!?!!

 

[cookiez69]

At my tutor a fair while ago, at school we're still going through calculus.

 

[Smile12345]

That's what I told you?!?!?!!

Sorry EpikHigh! Please forgive me.... Just out of interest... If you were to be ranked in the top 10 for Mathematics, what kind of marks would you need?

 

[Smile12345]

At my tutor a fair while ago, at school we're still going through calculus.

Yeah ok, I'm just doing the last Exercise for Calculus... And we've started Trig...

 

[EpikHigh]

Sorry! Please forgive me.... Just out of interest... If you were to be ranked in the top 10 for Mathematics, what kind of marks would you need?

Depends on your school ranking, really depends on your cohort can range from 90%+ or 80%+ depending on overall difficulty of your exams etc.

 

[Smile12345]

Hmmm. I'm not getting it... a=6 b=24 c=4

-24+- square root 24^2 -4x 6 x 4 (all over)
12

I'm just not getting -6+-root 30 / 3

Thanks for your help.