Consider the curve given by the equation ln(y + x + 3) = xy + y3. Find an equation of the tangent line at the point (−4, 2)
ln(x+y+3) = xy + y^3
(1+y')/(y+x+3) = y + xy' + 3y^2 y'
Collect terms and solve for y':
y' =
1-y(x+y+3)
--------------------------
(x+y+3)(x+3y^2)-1
so, at (-4,2), y' = -1/7
So, now you have a point and a slope, and the line is
y-2 = -1/7 (x+4)
I will assume that y3 is y^3
1/(y+x+3) * (dy/dx + 1) = x dy/dx + y + 3y^2 dy/dx
plug in the point (-4,2) :
1/(2-4+3) * (dy/dx + 1) = -4dy/dx + 2 + 12 dy/dx
1/ln1 * (dy/dx + 1) = 2 + 8dy/dx
arggghhh , 1/ln1 = 1/0, which is undefined
so we have a vertical tangent at (-4,2)
equation is x = -4
messed up in 4th step.
why is there a y' on the 3y^2 ?
@ steve
because y is a function of x. You must use the chain rule.
d/dx (3y^2) = d/dy (3y^2) * dy/dx
To find the equation of the tangent line at a given point on a curve, we need to find the slope of the tangent line and the point of tangency.
Step 1: Find the slope of the tangent line
To find the slope of the tangent line, we need to find the derivative of the curve equation with respect to x.
Differentiating both sides of the equation ln(y + x + 3) = xy + y^3 with respect to x, we get:
1 / (y + x + 3) * (dy/dx + 1) = y + x * dy/dx + 3y^2 * dy/dx
Rearranging the equation to solve for dy/dx, we get:
dy/dx = (y + x + 3 - 3y^2) / (1 - y - xy)
Step 2: Find the point of tangency
Given that the point of tangency is (-4, 2), we substitute these values into the curve equation to solve for y:
ln(y - 4 + 3) = -4y + y^3 + 2^3
ln(y - 1) = -4y + y^3 + 8
Now we solve this equation to find the value of y.
Step 3: Substitute the values into the slope formula
Now that we have the slope of the tangent line from step 1 and the y-value of the point of tangency from step 2, we substitute them into the formula for the equation of the tangent line: y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope.
Finally, we substitute the values obtained into the equation above to find the equation of the tangent line.