Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (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

OpenStudy (anonymous):

dy/dx=-(x/y)

OpenStudy (anonymous):

So, how?

OpenStudy (anonymous):

so we're finding dy/dx right?

OpenStudy (anonymous):

that'll do it you can even solve for y and then differentiate if you like. get the same answer

OpenStudy (anonymous):

if the gradient of the line joining centre and the point is m, then the gradient of the tangent will be -1/m

OpenStudy (anonymous):

and the differentiation method, anonymous... is right ^^

OpenStudy (anonymous):

intermediate step is \[x^2+y^2=1\] \[2x+2yy'=0\] then you get \[y'=-\frac{x}{y}\]

OpenStudy (anonymous):

Still a bit lost.

OpenStudy (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

OpenStudy (anonymous):

Why equal to 1? I thought 25?

OpenStudy (anonymous):

typo, although it cannot make much of a difference

OpenStudy (anonymous):

my cirlce has radius 1 and yours has radius 5

OpenStudy (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

OpenStudy (anonymous):

so would it be x/ sqrt {25 - x^2}?

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

yes plus or minus

OpenStudy (anonymous):

i already did, it is the derivative

OpenStudy (anonymous):

you have a choice starting with \[x^2+y^2=25\]a) solve for y or b) pretend you have solved for y

OpenStudy (anonymous):

if you solve for y you get \[y=\pm\sqrt{25-x^2}\] and then take the derivative of that you get \[y'=\pm\frac{x}{\sqrt{25-x^2}}\]

OpenStudy (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 \[x^2+y^2=25\] \[2x+2yy'=0\] \[y'=-\frac{x}{y}\]

OpenStudy (anonymous):

So how would you get the values in the first equation once plugging in (-3 and 4)?

OpenStudy (anonymous):

Oh, Ok, -3/4 right?

OpenStudy (anonymous):

But the answer says 3/4?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!