what is the equation of the tangent line that passes through
X^(1/4) + y^(1/4) =4 at the point (16,16)
please give the equation
(1/4)x^(-3/4) + (1/4)y^(-3/4) dy/dx = 0
dy/dx = -x^(-3/4)/y^(-3/4)
= y^(3/4)/x^(3/4)
sub in the point(16,16) to get the slope.
since you have a point on the line and the slope, it becomes an easy question.
btw, 16^(3/4) = [ 16^(1/4) ]^3 = 2^3 = 8
so slope = 8/8 = 1
To find the equation of the tangent line that passes through the point (16, 16) on the curve defined by the equation x^(1/4) + y^(1/4) = 4, we need to do the following:
Step 1: Differentiate the equation with respect to x to find the derivative.
To differentiate x^(1/4) + y^(1/4) = 4, we differentiate each term separately. The derivative of x^(1/4) with respect to x (dx/dx) is 1/4 * x^(-3/4), and the derivative of y^(1/4) with respect to x (dy/dx) is dy/dx * (1/4) * y^(-3/4).
Step 2: Substitute the values for x and y from the point (16, 16) into the derivative.
Substituting x = 16 and y = 16 into the derivative obtained in step 1, we get:
1/4 * 16^(-3/4) + (dy/dx) * (1/4) * 16^(-3/4) = 0
Simplifying this equation gives:
1/4 * (1/2) + (dy/dx) * (1/4) * (1/2) = 0
1/8 + (dy/dx) * (1/8) = 0
Step 3: Solve for (dy/dx) to find the slope of the tangent line.
Isolating (dy/dx) gives:
(dy/dx) * (1/8) = -1/8
(dy/dx) = -1
Step 4: Use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope of the line. Plugging in the values (x1, y1) = (16, 16) and m = -1, we get:
y - 16 = -1(x - 16)
y - 16 = -x + 16
y = -x + 32
Therefore, the equation of the tangent line that passes through the point (16, 16) on the curve defined by x^(1/4) + y^(1/4) = 4 is y = -x + 32.