Thread: Gradient and parametric equations help!

Could anyone help me with the following problems?:

a)Find the gradient of the curve:

8x^2/4x^2-3y^3 = 3y at (3,2)

b) The curve C has parametric equations x=4cos2t and y=3sint for t being between -pi/2 and pi/2.
A is the point (2,3/2) and lies on C.
The normal to C at A is 6y-16x+23. The normal at A cuts C again at point B. Find the y coordinate of point B.

Thank you so much if you can help me

Hello, Confuzzled!

Can't read the first one . . .

b) The curve C has parametric equations: .x = 4·cos2t and y = 3·sin t
for -π/2 < t < π/2.

A is the point (2,3/2) and lies on C.

The normal to C at A is: .6y - 16x + 23 .= .0.

The normal at A cuts C again at point B.
Find the y-coordinate of point B.

Eliminate the parameter:

. . x .= .4·cos2t .= .4(1 - 2·sin²t) .[1]

. . y .= .3·sin t . → . sin t .= .y/3 [2]


Substitute [2] into [1]: .x .= .4(1 - 2[y²/9])

. . and we get: .x .= .4 - 8y²/9 . [3] . . . a parabola

The equation of the normal is: .x .= .3y/8 + 23/16 . [4]


Equate [3] and [4]: .4 - 8y²/9 .= .3y/8 + 23/16

. . which simplifies to: .128y² + 54y - 369 .= .0

. . which factors: .(2y - 3)(64y + 123) .= .0 . **

. . and has roots: .y .= .3/2, -123/64


So the normal at A cuts the parabola at: .y = -123/64


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Okay, okay! . . . I didn't really factor it.

I used the Quadratic Formula and was pleasantly surprised
. . to find that the roots were rational.

From the roots I constructed the factors,
. . then created the illusion that I'm a genius.

Originally Posted by Confuzzled?

Could anyone help me with the following problems?:

a)Find the gradient of the curve:

8x^2/4x^2-3y^3 = 3y at (3,2)

...

Hi,

I hope I read your problem correctly:

Originally Posted by Confuzzled?

Could anyone help me with the following problems?:

a)Find the gradient of the curve:

8x^2/4x^2-3y^3 = 3y at (3,2)

...

Ooops,

sorry, I forgot to attach the diagram

Hello, Confuzzled?

Pay no attention to the man behind the curtain . . .

a) Find the gradient of the curve: .8x²/(4x² - 3y³) = 3y .at (3,2)

We have: 8x² .= .12x²y - 9y^4

Differentiate implicitly: .16x .= .12x²y' + 24xy - 36y³y'

Then: . 36y³y' - 12x²y' .= .24xy - 16x

Factor: .12(3y³ - x²)y' .= .8x(3y - 2)

. . . . . . . . . . . . .2x(3y - 2)
Finally: . y' . = . --------------
. . . . . . . . . . . . 3(3y³ - x²)


. . . . . . . . . . . . .2(3)(3·2 - 2) . . . . .8
At (3,2): .y' .= . ----------------- .= .- ---
. . . . . . . . . . . . .3(3·2³ - 3²) . . . . .15