anonymous
  • anonymous
(STILL STUCK!) The following equation represents a circle. Find the gradient of the tangent at the given point (i) by finding the coordinates of the centre, and (ii) by differentiating the implicit equation. (a) x^2 +y^2 =25
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
dy/dx=-(x/y)
anonymous
  • anonymous
So, how?
anonymous
  • anonymous
so we're finding dy/dx right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
that'll do it you can even solve for y and then differentiate if you like. get the same answer
anonymous
  • anonymous
if the gradient of the line joining centre and the point is m, then the gradient of the tangent will be -1/m
anonymous
  • anonymous
and the differentiation method, anonymous... is right ^^
anonymous
  • anonymous
intermediate step is \[x^2+y^2=1\] \[2x+2yy'=0\] then you get \[y'=-\frac{x}{y}\]
anonymous
  • anonymous
Still a bit lost.
anonymous
  • anonymous
or you can start \[x^2+y^2=1\] \[y=\pm\sqrt{1-x^2}\] \[y'=\pm\frac{x}{\sqrt{1-x^2}}\] same thing
anonymous
  • anonymous
Why equal to 1? I thought 25?
anonymous
  • anonymous
typo, although it cannot make much of a difference
anonymous
  • anonymous
my cirlce has radius 1 and yours has radius 5
anonymous
  • anonymous
using partial derivatives, implicit differentiation is less laborous... all you have to do is set get this to a form F(x,y)=0. In this case x^2+y^2-25=0 and we know that \[dy/dx=-F_x/F_y\] where \[F_x\] is the partial derivative of F(x,y) with respect to x and \[F_y\] is the partial derivative with respect to y
anonymous
  • anonymous
so would it be x/ sqrt {25 - x^2}?
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
yes plus or minus
anonymous
  • anonymous
i already did, it is the derivative
anonymous
  • anonymous
you have a choice starting with \[x^2+y^2=25\]a) solve for y or b) pretend you have solved for y
anonymous
  • anonymous
if you solve for y you get \[y=\pm\sqrt{25-x^2}\] and then take the derivative of that you get \[y'=\pm\frac{x}{\sqrt{25-x^2}}\]
anonymous
  • anonymous
or you pretend \[y=f(x)\] and take the derivative of \[x^2+f^2(x)=25\] to get \[2x+2f(x)f'(x)=0\] solving for \[f'(x)\] you get \[f'(x)=-\frac{x}{f(x)}\] more easily written as \[x^2+y^2=25\] \[2x+2yy'=0\] \[y'=-\frac{x}{y}\]
anonymous
  • anonymous
So how would you get the values in the first equation once plugging in (-3 and 4)?
anonymous
  • anonymous
Oh, Ok, -3/4 right?
anonymous
  • anonymous
But the answer says 3/4?

Looking for something else?

Not the answer you are looking for? Search for more explanations.