Find the slope of the tangent line to the curve below at the point ( 5 , 5 ) .
√3x+2y+√4xy=15
slope=
Your given point does not even lie on the curve, so any calculations would
be futile.
Check your typing
√3x+2y + √4xy =15
√ this goes on for the whole equation for equation for both
√3x+2y √4xy
√(3x)+2y+2√(xy) = 15
3/(2√(3x)) + 2y' + 1/(√(xy)) (y + xy') = 0
2y' + x/√(xy)y' = -3/(2√(3x)) - y/√(xy)
y' = -(3/(2√(3x)) + y/√(xy)) / (2 + x/√(xy))
= -(y√(3x) + 2y√(xy))/(4xy + 2x√(xy))
= -(√(3y) + 2) / √x(4√y + 2x)
so whats the slope
so, at (5,5),
y' = -(√15 + 2) / √5(4√5 + 10)
better check my algebra above
isnt it a whole number the slope?
ahhh, I suspect you meant:
√(3x+2y) + √(4xy) = 15 , strange how a few innocent brackets make the
whole thing come alive.
(3x+2y)^(1/2) + (4xy)^1/2 = 15
(1/2)(3x+2y)^(-1/2) (3 + 2dy/dx) + (1/2)(4xy)^(-1/2)(4xdy/dx + 4y) = 0
(3 + 2dy/dx)/√(3x+2y) + (4xdy/dx + 4y)/√(4xy) = 0
sub in your point (5,5)
(3+2dy/dx)/5 + (20dy/dx + 20)/10 = 0
multiply each term by 10
6 + 4dy/dx + 20dy/dx + 20 = 0
24dy/dx = -26
dy/dx = -13/12
thanks for the help, ill note this formula
I graphed √(3x+2y) + √(4xy) = 15
and the tangent line of y-5 = (-13/12)(x-5) on Desmos and it
verfied. copy and paste the link below
www.desmos.com/calculator
To find the slope of the tangent line at a given point, you need to take the derivative of the equation with respect to x and then evaluate the derivative at the given point.
Let's start by rearranging the equation to make it easier to find the derivative.
√3x + 2y + √4xy = 15
To find the derivative, we will treat y as a function of x, so we can use the chain rule.
Take the derivative of both sides of the equation with respect to x:
d/dx (√3x + 2y + √4xy) = d/dx(15)
To find the derivative of each term, we can use the power rule and the chain rule.
The derivative of √3x with respect to x is (1/2)*(3x)^(-1/2) * 3 = (3/2)√(3/x)
The derivative of 2y with respect to x is 0 because y is not a function of x.
The derivative of √4xy with respect to x is (1/2)*(4xy)^(-1/2) * (4y + 4x(dy/dx)) = (2/y)√(x/y)*(y + x(dy/dx))
The derivative of 15 with respect to x is 0 because 15 is a constant.
Therefore, the derivative of the equation is:
(3/2)√(3/x) + (2/y)√(x/y)*(y + x(dy/dx)) = 0
Now we need to substitute the x and y values of the given point (5, 5) into the derivative equation and solve for dy/dx.
(3/2)√(3/5) + (2/5)√(5/5)*(5 + 5(dy/dx)) = 0
Simplify the equation and solve for dy/dx:
(3/2)√(3/5) + 2(dy/dx) = 0
2(dy/dx) = -(3/2)√(3/5)
dy/dx = -3√(3/10)
Therefore, the slope of the tangent line to the curve at the point (5, 5) is -3√(3/10).