use implicit differentiantion to find the equation of the tangent line to the curve xy^3+xy=20 at the point (10,1) . The equation of this tangent line can be written in the form y=mx+b where m is=...

use implicit differentiantion to find the equation of the tangent line to the curve xy^3+xy=20 at the point (10,1) . The equation of this tangent line can be written in the form y=mx+b where m is=

and where b is =

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lemjay | High School Teacher | (Level 3) Senior Educator

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`xy^3+xy=20`

First, solve for the slope of the tangent line. To do so, differentiate both sides of the equation with respect to x.
`d/(dx) (xy^3+xy)=d/(dx)20`

Since the derivative of constant is zero, right sideĀ  becomes:

`d/(dx) (xy^3+xy)= 0`

At the left side, take the derivative of each term.

`d/(dx)(xy^3)+d/dx(xy)=0`

Then, apply product rule of derivative.

`y^3*1 + x*3y^2 dy/dx + y*1+x*dy/dx =0`

`y^3+3xy^2dy/dx+y+xdy/dx=0`

Move the terms without dy/dx to the other side of the equation.

`y^3-y^3+3xy^2dy/dx+y-y+xdy/dx=0-y^3-y`

`3xy^2dy/dx+xdy/dx=-y^3-y`

Then, factor out dy/dx.

`dy/dx(3xy^2+x)=-y^3-y`

And, divide both sides by 3xy^2+x to get dy/dx.

`dy/dx=(-y^3-y)/(3xy^2+x)`

Now that dy/dx is known, plug-in (10,1) to get the slope of the tangent line.

`dy/dx=(-1^3-1)/(3*10*1^2+10)=(-1-1)/(30+10)=-2/40`

`dy/dx=-1/20`

So, the slope of the line tangent to the curve at (10,1) is -1/20.

Next, apply the point-slope form to get the equation of the line.

`y-y_1=m(x-x_1)`

`y-1=-1/20(x-10)`

`y-1=-x/20+10/20`

`y-1=-x/20+1/2`

And, add both sides by 1.

`y+1-1=-x/20+1/2+1`

`y=-x/20+1/2+1*2/2`

`y=-x/20+1/2+2/2`

`y=-x/20+3/2`

Hence, the equation of the line is `y=-x/20+3/2` . Its slope is `m=-1/20` and y-intercept is `b=3/2` .

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