Wednesday

July 30, 2014

July 30, 2014

Posted by **mike** on Tuesday, October 5, 2010 at 10:58pm.

- calculus -
**Reiny**, Tuesday, October 5, 2010 at 11:08pmThe point (0,1) lies in the "interior" of the parabola

y = 3x^2.

There is no tangent to the curve that will pass through (0,1).

check your typing.

- calculus -
**mike**, Tuesday, October 5, 2010 at 11:21pmthere are two tangents lines to the curve f(x) = 3x^2 that pass through the point p =0,-1 find the x coordinates of the point where the tangents line intersect the curve, please show working.

- calculus -
**Reiny**, Tuesday, October 5, 2010 at 11:28pmahhh, now it makes sense.

let the point of contact be (a,b)

slope of tangent by the grade 9 way = (b+1)/a

slope of tangent by Calculus is

dy/dx = 6x

so at the point (a,b), slope = 6a

then 6a = (b+1)/a

6a^2 = b+1

but since (a,b) lies on the curve, b = 3a^2

so

6a^2 = 3a^2 + 1

a^2 = 1/3

a = ± 1/√3

so the x coordinates of the two tangents are 1/√3 and -1/√3

- calculus -
**mike**, Tuesday, October 5, 2010 at 11:39pmfind the equation of the tangent line of the curve y=root x divideed by x+1 at the point p = 2, root 2 divided by 3. show working please.

- calculus -
**mike**, Wednesday, October 6, 2010 at 10:54amfind the equation of the tangent line to the curve y = 2√2/x+1 at the point p=1,2/3.please show working. i would appreciate if you do it before 12:05pm today

- calculus -
**mike**, Tuesday, October 19, 2010 at 11:18pmuse implcit differenciation to find an eqaution of both the tangent line to the ellipse:

2x^2 + 4y^2 = 36

that passes through the points: 14,3

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