so i suppose to find slope.
i have trouble this problem. my math book is not really clear.
Find the slope of the tangent line to the curve √(3x+2y)+√(4xy)=11.6 at the point (6,2).
√(3x+2y) is the square root(3x+2y)
√(4xy)is the square root 4xy.
1/2sqrt(3x+2y)* (3 +2yy')+ 1/2sqrt(4xy)*(4y+2xy')=0
now at the point 6,2, put that into the equation for x,y
and solve for y'
1/2 √(3(6)+2(2)) * 3(2(2))+ 1/2 √(4(6)(2)) * 4(2)+2(6) right?
To find the slope of the tangent line to a curve at a specific point, we need to find the derivative of the curve at that point. In this case, we are given the equation of the curve as √(3x+2y) + √(4xy) = 11.6.
Step 1: Solve for y in terms of x:
Start by isolating the square root term involving y:
√(3x+2y) = 11.6 - √(4xy)
Square both sides to eliminate the square root:
3x + 2y = (11.6 - √(4xy))^2
Simplify the right side:
3x + 2y = 134.56 - 23.2√(4xy) + 4xy
Rearrange the equation to get y in terms of x:
2y = 134.56 - 3x - 23.2√(4xy) + 4xy
Divide the entire equation by 2 to solve for y:
y = (134.56 - 3x) / 2 - 11.6√(4xy) + 2xy
This equation gives us the equation of the curve in terms of x.
Step 2: Find the derivative of y with respect to x:
Differentiate y with respect to x to find the derivative(dy/dx):
(dy/dx) = (-3/2) + (11.6/2) * (1 / (2√(4xy))) + 2x
Simplify the derivative:
(dy/dx) = (-3/2) + (5.8/√(4xy)) + 2x
Step 3: Substitute the coordinates of the given point (6, 2) into the derivative to find the slope:
Substitute x = 6 and y = 2 into the derivative:
(dy/dx) = (-3/2) + (5.8/√(4*6*2)) + 2*6
(dy/dx) = (-3/2) + (5.8/4) + 12
(dy/dx) = (-3/2) + 1.45 + 12
(dy/dx) = 9.95
Therefore, the slope of the tangent line to the curve √(3x+2y) + √(4xy) = 11.6 at the point (6, 2) is 9.95.