William Burnett

2021-12-29

Consider the curve created by $2{x}^{2}+3{y}^{2}\u20134xy=36$

(a) Show that $\frac{dy}{dx}=2y-2x3y-2x$

(b) Calculate the slope of the line perpendicular to the curve at each location where x is. $x=6$.

(c) A vertical tangent line to the curve exists at a positive value of x.

usumbiix

Beginner2021-12-30Added 33 answers

So,

$2{x}^{2}+3{y}^{2}-4xy=36$

$\frac{d}{dx}(2{x}^{2}+3{y}^{2}-4xy)=\frac{d}{dx}\left(36\right)$

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

let $u=3{y}^{2}$

$\frac{du}{dx}=\frac{du}{dy}\cdot \frac{dy}{dx}=\frac{d\left(3{y}^{2}y\right)}{\cdot}\frac{dy}{dx}=6y\cdot \frac{dy}{dx}$

$4x+6y\cdot \frac{dy}{dx}-(4x\cdot \frac{dy}{dx}+4y)=0$

$4x+6y\cdot \frac{dy}{dx}-4x\cdot \frac{dy}{dx}-4y=0$

$\frac{dy}{dx}(6y-4x)+4x-4y=0$

$\frac{dy}{dx}(6y-4x)=4y-4x$

$\frac{dy}{dx}=\frac{4y-4x}{6y-4x}=\frac{2(2y-2x)}{2(3y-2x)}$

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

Corgnatiui

Beginner2021-12-31Added 35 answers

Part B : slope of the line tangent to the curve at each point on the curve where $x=6$ .

The slope of the curve will be$\frac{dy}{dx}=\frac{2y-2x}{3y-2x}$

we need to find slope of line tangent to curve at$x=6$

first we need find y value at$x=6$ . For that substitute $x=6$ in equation of curve.

$2{x}^{2}+3{y}^{2}-4xy=36$

$2\left({6}^{2}\right)+3{y}^{2}-4\times 6\times y=36$

$72+3{y}^{2}-24y-36=0$

$3{y}^{2}-24y+36=0$

divide by 3 we get,

${y}^{2}-8y+12=0$

$(y-6)(y-2)=0$

therefore$y=6$ or $y=2$

therefore the point off tangency are

$(x,y)=(6,6)$ and $(x,y)=(6,2)$

equation of slope$=\frac{dy}{dx}=\frac{2y-2x}{3y-2x}=0$

at$(x,y)=(6,6)$

slope${m}_{1}=\frac{2\times 6-2\times 6}{3\times 6-2\times 6}=0$

${m}_{1}=0$

$(x,y)=(6,2)$

slope$m}_{2}=\frac{2\times 2-2\times 6}{3\times 2-2\times 6}=\frac{4-12}{6-12}=\frac{8}{6}=\frac{4}{3$

$m}_{2}=\frac{4}{3$

The slope of the curve will be

we need to find slope of line tangent to curve at

first we need find y value at

divide by 3 we get,

therefore

therefore the point off tangency are

equation of slope

at

slope

slope

karton

Expert2022-01-09Added 613 answers

Part C : value of x at which the curve has a vertical tangent line

The vertical tangent to a curve occurs at a point where slope is undefined or infinite.

This is nothing but the points where derivative of function is not defined.

we know the derivative of function is,

that is 3y − 2x = 0

3y = 2x

substitute this in equation of curve,

therefore vertical tangent to the curve occurs at points x = +7.348 and x = - 7.348

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