Find the slope of the line tangent to the curve `(x+2y)^(1/2)+(3xy)^(1/2)=17.7` at the point (8,7) .(1x+2y)^1/2+(3xy)^1/2=17.7 at the points (8,7)

Asked on by playerjb92

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lemjay's profile pic

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

Posted on

Note that the slope of a line which is tangent to a point in a curve is equal to y'.

So to solve for the above problem, take the derivative of both sides of the given equation. Use implicit differentiation since x and y variables are on the same side of the equation.

`(x+2y)^(1/2) + (3xy)^(1/2)=17.7`

`d/(dx)((x+2y)^(1/2) + (3xy)^(1/2))=d/(dx)17.7`

`d/(dx)(x+2y)^(1/2) + d/(dx)(3xy)^(1/2)=d/(dx)17.7`

Then, let's take the derivative of each term separately.

For the first term, use power formula of derivative which is `(u^n)'=n*u^(n-1)*u'` .

>> `d/(dx) (x+2y)^(1/2)=1/2(x+2y)^(-1/2)*(x+2y)'`


  Express the exponent as positive.


For the second term, use the power formula of derivatives, too.

>> `d/(dx)(3xy)^(1/2)=1/2(3xy)^(-1/2)*(3xy)'`

 To determine (3xy)', use the product rule which is`(u*v)'=vu'+uv'`.



 Also, express the exponent as positive.


And at the right side, note that the derivative of a constant is zero. So,

>> `d/(dx)17.7=0`

Thus, taking the derivative of both sides of the equation result to:

`d/(dx)(x+2y)^(1/2) + d/(dx)(3xy)^(1/2)=d/(dx)17.7`

From the given point (8,7), substitute x=8 and y=7.

`(1+2y')/(2(8+2*7)^(1/2))+ (3*7+3*8y')/(2(3*8*7)^(1/2))=0`



To cancel the denominators, multiply both sides by the LCD.




Move the terms without y' to the right side.


Factor out y' at the left side. 


Then, isolate y'.



Hence, the slope of the tangent line is -0.805.

jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

The slope of a tangent to any curve is given by the first derivative.

`(x+2y)^(1/2)+(3xy)^(1/2) = 17.7`

Derivate both sides with respect to x.

`1/(2*(x+2y)^(1/2))*(1+2y')+1/(2*(3xy)^(1/2))*3(x*y'+y) = 0`

`1/(2*(x+2y)^(1/2))+(2y')/(2*(x+2y)^(1/2))+(3y)/(2*(3xy)^(1/2))+(3xy')/(2*(3xy)^(1/2)) = 0`

The slope of curve is given by y' at point (8,7).

`1/(2*(x+2y)^(1/2))+(2y')/(2*(x+2y)^(1/2))+(3y)/(2*(3xy)^(1/2))+(3xy')/(2*(3xy)^(1/2)) = 0`

`1/(2*(8+14)^(1/2))+(2y')/(2*(8+14)^(1/2))+(21)/(2*(3*56)^(1/2))+(24y')/(2*(3*56)^(1/2)) = 0`

`1/(2*(22)^(1/2))+(y')/((22)^(1/2))+(21)/(2*(168)^(1/2))+(12y')/((168)^(1/2)) = 0`


`0.1066+0.213y'+0.81+0.926y' = 0`

`1.139y' = -0.9166`

       `y' = -0.8047`


So the slope of the tangent is -0.8047.





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