Point P(a,b) is on the curve
square root of x + square root of y =1.
Show that the slope of the tangent at P is : - square root of (b/a).
√x + √y = 1 or
x^(1/2) + y^(1/2) = 1
differentiate implicitly
(1/2)x^(-1/2) + (1/2)y^(-1/2)dy/dx = 0
dy/dx = -x^(-1/2)/y^(-1/2)
= - y^(1/2)/x^(1/2)
= - √y/√x
= - √(y/x)
so at the point (a,b)
dy/dx = - √(b/a)
Given
√x + √y = 1
Apply implicit differentiation:
1/(2√x) + 1/(2√y)*(dy/dx) = 0
Transposing and solving for dy/dx:
dy/dx = -√(y/x)
Find dy/dx by implicit differentiation
(1-2xy^3)^5=x+4y
To find the slope of the tangent at point P(a, b) on the curve √x + √y = 1, we can use implicit differentiation.
Step 1: Differentiate both sides of the equation with respect to x.
d/dx (√x + √y) = d/dx 1
Step 2: For the left-hand side, we need to apply the chain rule since there are two functions being added together.
(1/2) * (1/√x) + (1/2) * (1/√y) * dy/dx = 0
Step 3: Simplify and solve for dy/dx.
(1/2√x) + (1/2√y) * dy/dx = 0
Multiply through by 2√x√y:
√y + √x(dy/dx) = 0
Subtract √y from both sides:
√x(dy/dx) = -√y
Divide through by √x:
dy/dx = -√y / √x
Step 4: Substitute the coordinates of point P(a, b) into the expression dy/dx = -√y / √x.
dy/dx = -√b / √a
Therefore, the slope of the tangent at point P(a, b) is -√(b/a).