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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Captain on March 09, 2008, 01:43:55 pm

Title: Specialist exam 1 2007 Q 3
Post by: Captain on March 09, 2008, 01:43:55 pm: Got bored so looking at some of the questions from last year.

Got stuck on

Quote
Find the equation of the tangent to the curve x^3−2(x^2)y+2(y^2)=2 at the point P(2, 3).

Working out would be appreciated, thanks :)
Title: Re: Specialist exam 1 2007 Q 3
Post by: unknown id on March 09, 2008, 02:12:02 pm: u hav to use implicit differentiation to find the derivative
Title: Re: Specialist exam 1 2007 Q 3
Post by: Mao on March 09, 2008, 02:18:30 pm: $x^3-2x^2y+2y^2=2$

$\frac{d}{dx} \left(x^3-2x^2y+2y^2\right)=\frac{d}{dx} 2$

When we implicitly differentiate u (an expression in terms of y), we'll have:
$\frac{d}{dx} u=\frac{du}{dx}=\frac{du}{dy} \times \frac{dy}{dx}=u' \times \frac{dy}{dx}$

so

$3x^2 - 2\frac{d}{dx}x^2 \cdot y + 2\frac{d}{dx}y^2 = 0$

using the product rule on the middle term:
$3x^2-2 \left( 2xy + x^2 \frac{dy}{dx} \right) + 4y \frac{dy}{dx}=0$

$3x^2-4xy-2x^2\frac{dy}{dx} + 4y \frac{dy}{dx}=0$

$3x^2-4xy+\left(4y-2x^2\right) \frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{3x^2-4xy}{4y-2x^2}$

with that, you can sub in P(2,3) and find you tangent :D
Title: Re: Specialist exam 1 2007 Q 3
Post by: midas_touch on March 10, 2008, 08:17:37 pm: Quote from: Mao on March 09, 2008, 02:18:30 pm
$x^3-2x^2y+2y^2=2$

$\frac{d}{dx} \left(x^3-2x^2y+2y^2\right)=\frac{d}{dx} 2$

When we implicitly differentiate u (an expression in terms of y), we'll have:
$\frac{d}{dx} u=\frac{du}{dx}=\frac{du}{dy} \times \frac{dy}{dx}=u' \times \frac{dy}{dx}$

so

$3x^2 - 2\frac{d}{dx}x^2 \cdot y + 2\frac{d}{dx}y^2 = 0$

using the product rule on the middle term:
$3x^2-2 \left( 2xy + x^2 \frac{dy}{dx} \right) + 4y \frac{dy}{dx}=0$

$3x^2-4xy-2x^2\frac{dy}{dx} + 4y \frac{dy}{dx}=0$

$3x^2-4xy+\left(4y-2x^2\right) \frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{3x^2-4xy}{4y-2x^2}$

with that, you can sub in P(2,3) and find you tangent :D

That only gives you the gradient of the tangent. Once you have this, you can find the equation by simply subbing in the x and y values into y = mx + c, where m is the gradient you just found.
Title: Re: Specialist exam 1 2007 Q 3
Post by: AppleXY on March 14, 2008, 11:13:35 pm: or you could use the point-slope method.

y-y1 = dy/dx|(2,3)*[x-x1]