Find the equation of the tangent line to the graph of √(xy)=x-2y at (4,1)
(1/2)(xy)^(-1/2) (x dy/dx + y) = 1 - 2dy/dx
at the given point
(1/2)(4)^(-1/2) (4dy/dx + 1) = 1 - 2dy/dx
(1/4)(4dy/dx + 1 = 1 - 2dy/dx
times 4
4dy/dx + 4 = 1 - 2dy/dx
6dy/dx = -3
dy/dx = -1/2
y-1 = -(1/2)(x-4)
2y-2= -x+4
x + 2y = 6
or
y = (-1/2)x + 6
To find the equation of the tangent line to the graph of the equation √(xy) = x - 2y at the point (4,1), we need to determine the slope of the tangent line and then use the point-slope form of a line equation.
First, let's start by differentiating the given equation implicitly with respect to x. This will give us the slope of the tangent line at any point on the curve:
Differentiating both sides of the equation √(xy) = x - 2y with respect to x, we get:
d/dx(√(xy)) = d/dx(x - 2y)
To differentiate the left side, we can use the chain rule:
(d/dx(√(xy))) = (1/2√(xy))(d/dx(xy))
Applying the product rule on the right side, we have:
d/dx(x - 2y) = d/dx(x) - d/dx(2y)
= 1 - 0
= 1
Now, let's simplify the left side:
(1/2√(xy))(d/dx(xy)) = 1
(1/2√(xy))(x(dy/dx) + y) = 1
Next, let's substitute the given point (4,1) into the equation:
(1/2√(4*1))(4(dy/dx) + 1) = 1
Simplifying further:
(1/2√(4))(4(dy/dx) + 1) = 1
(1/4)(4(dy/dx) + 1) = 1
(dy/dx) + 1/4 = 1
(dy/dx) = 1 - 1/4
(dy/dx) = 3/4
Now that we have the slope of the tangent line, which is 3/4, we can use the point-slope form of a line equation to find the equation of the tangent line. The point-slope form is given as:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) are the coordinates of the given point (4,1).
Substituting the values, we have:
y - 1 = (3/4)(x - 4)
Simplifying the equation further:
y - 1 = (3/4)x - 3
y = (3/4)x - 2
Therefore, the equation of the tangent line to the graph of √(xy) = x - 2y at the point (4,1) is y = (3/4)x - 2.