Find the slope of the tangent line at (8,8) for the function (16-x)y^2=x^3
Round your answer to two decimal places.
using implicit differentiation,
(-1)(y^2) + (16-x)(2yy') = 3x^2
y' = (3x^2+y^2)/(2y(16-x))
So, just evaluate y' at (8,8)
Find equation of all lines having slop -4 that are tangent to the curve y=16/(x-2)
To find the slope of the tangent line at a point on a curve, we need to find the derivative of the curve and then evaluate it at that point.
Let's start by finding the derivative of the given function:
First, rewrite the equation in terms of y to isolate y:
(16-x)y^2 = x^3
y^2 = x^3 / (16-x)
Now differentiate both sides of the equation with respect to x using the Quotient Rule:
d/dx (y^2) = d/dx (x^3 / (16-x))
To differentiate the left side, we can use the Chain Rule:
2y * dy/dx = (3x^2(16-x) - x^3*(-1))/(16-x)^2
Therefore, dy/dx is:
dy/dx = [(3x^2(16-x) - x^3*(-1))/(16-x)^2] / (2y)
Now we have the derivative, but we need to find it at the point (8, 8). Substituting x=8 and y=8 into the derivative equation, we get:
dy/dx = [(3(8)^2(16-8) - (8)^3*(-1))/(16-8)^2] / (2(8))
Simplifying,
dy/dx = [3(64)(8) + 512] / (64) / 16
dy/dx = (1536 + 512) / 64 / 16
dy/dx = 2048 / 1024
dy/dx = 2
So, the slope (dy/dx) of the tangent line at the point (8, 8) is 2.