Part of Differentiating...:D (1 Viewer)

[Smile12345]

Hello All... Sorry to bother you again...

Could someone please help me with this question?

Q. The tangent and normal to the curve y=x^3 at the point U(1,1) meet the y-axis at P and Q respectively. Find their equations, the co-ordinates of points P and Q, and the area of Triangle QUP.

Thanks heaps, it's much appreciated.

 

[cookiez69]

Equation of tangent:

dy/dx = 3x^2

m = 3(1)^2 = 3

y-1 = 3(x-1)
3x-y-2=0

Equation of the normal:

m = -1/3
y-1 = -1/3(x-1)
3y - 3 = -x+1
x+3y-4=0

---

Coordinates of P: (where the tangent meets the y-axis)

sub x = 0 into the equation of the tangent
3(0)-y-2=0
y=-2

Therefore P(0,-2)

Coordinates of Q: (where the normal meets the y-axis)

sub x = 0 into the equation of the normal
(0)+3y-4=0
3y=4
y=4/3

Therefore Q(0,4/3).

I'll do the area of the triangle one later.

 

[Smile12345]

Equation of tangent:

dy/dx = 3x^2

m = 3(1)^2 = 3

y-1 = 3(x-1)
3x-y-2=0

Equation of the normal:

m = -1/3
y-1 = -1/3(x-1)
3y - 3 = -x+1
x+3y-4=0

---

Coordinates of P: (where the tangent meets the y-axis)

sub x = 0 into the equation of the tangent
3(0)-y-2=0
y=-2

Therefore P(0,-2)

Coordinates of Q: (where the normal meets the y-axis)

sub x = 0 into the equation of the normal
(0)+3y-4=0
3y=4
y=4/3

Therefore Q(0,4/3).

I'll do the area of the triangle one later.

Thanks heaps...

 

[Smile12345]

For the Area... This is how I've approached it... Using 1/2 . b . h

So we need to the distance between Q and P (base)... Use distance formula I got 10/3...

And then 1/2 base x height (1) would= the answer...

Am I right?

 

[Smile12345]

Appreciate your help cookiez69 - +1 to your rep.

 

[cookiez69]

For the Area... This is how I've approached it... Using 1/2 . b . h

So we need to the distance between Q and P (base)... Use distance formula I got 10/3...

And then 1/2 base x height (1) would= the answer...

Am I right?

Whoops forgot to answer back sorry xD

You're on the right track, it helps to graph out the three points so its easier.

Find the distance between Q and U, this will give you your height.
Find the distance between U and P, this will give you your base.

Then use A = 1/2 bh.

(You don't find the distance between Q and P, that's not the base. Like I said, plot the points on a graph)

 

[Smile12345]

Whoops forgot to answer back sorry xD

You're on the right track, it helps to graph out the three points so its easier.

Find the distance between Q and U, this will give you your height.
Find the distance between U and P, this will give you your base.

Then use A = 1/2 bh.

(You don't find the distance between Q and P, that's not the base. Like I said, plot the points on a graph)

No worries at all. ;D

Thanks a million.

 

[Smile12345]

Dist QU: Root 10/9
Dist PU: Root 10

So 1/2 x Root 10/9 x Root 10 = 1 2/3 ... I believe this is right now?