<p>given curve x^2 - xy + y^2 = 9</p>
<p>a) find coordinates of points where on the curve the tangents are vertical</p>
<p>i got (y/2 , 2x)</p>
<p>b) at point (0,3) find the rate of change in the slope of the curve with respect to x.</p>
<p>i got 1/2</p>
<p>are these answers right, can someone please check? if i am wrong please show your work</p>
<p>anyone????</p>
<p>vertical or horizontal?</p>
<p>question asks find coordinates of points where on the curve the tangents are "vertical"</p>
<p>if the tangents are vertical....the gradients will be undefined.....so how do we make use of dy/dx?</p>
<p>i thought it was horizontal too but its not, its asking for vertical, i redid part a just now and i got (2y,x/2) is that right? and i redid part b also since its asking "with respect to x", and i got 2. Can someone please check these answers.</p>
<p>i found dx/dy and set that equal to 0 to find vertical tangents
i ended up with 0=x-2y so the coordinants are (2y, x/2)
And for part b i simply plugged in 0 for x and 3 for y into dx/dy and i got 2.</p>
<p>dy/dx can only be zero if the tangents are horizontal.......wait....are differentiating with respect to x or y?</p>
<p>since it is asking for where the tangents are vertical i found dx/dy.</p>
<p>then i guess you are correct</p>
<p>i hope so can someone else check this</p>
<p>This is a harder problem than you guys are making it out to be.</p>
<p>You need implicit differentiation for this problem. Differentiate both sides of the equation with respect to x, and you get:</p>
<p>2x-(y+xy')+2yy'=0</p>
<p>Solve this for y' and you get:</p>
<p>y' = (y-2x)/(x-2y)</p>
<p>For vertical tangents, you need the derivative to be undefined, so set the denominator equal to zero and you get x=2y. However, you need the coordinates of points on the curve, so substitute x/2 for y in the original equation and get that x=2sqrt(3) or x=-2sqrt(3) (the reason we substitute for y instead of x becomes clear if you try it the other way and get 4 possible answers for x). Using those x-coordinates, you can substitute them back into the original equation to solve for y, and thus you get two points: (2sqrt(3), sqrt(3)) and (-2sqrt(3), -sqrt(3)). That answers part (a).</p>
<p>(b) is talking about the second derivative at that point. You already found the first derivative for part (a), but it's not in the form you want. It's a quotient and that will make life very difficult. Put it back in the plain, simple form: 2x-y-xy'+2yy'=0 and implicitly differentiate that to get:</p>
<p>2-y'-xy''-y'+2yy''+2y'y' = 0</p>
<p>That looks pretty awful, but just use the formula derived in part (a) for y' to get y'(0,3) = 1/2. Now plug that (and x and y) into the ugly equation with y'' in it and solve for y''. The answer is -1/4.</p>
<p>If you have a really smart graphing program, try graphing this curve. It's actually an ellipse that's been rotated. Since it's an implicit equation you can't graph it on a calculator.</p>
<p>U did something wrong in the beginning it should be y' = (y-2x)/(2y-x) the denominator you had (x-2y) is wrong</p>
<p>never mind thinkdifferent u are right , u didnt do anything wrong, thank you</p>