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