fine the slope intercept equation of the tangent line to the curve by the relation 5x^3+10y^2=8x^5+2y at points (1,1)
To find the slope-intercept equation of the tangent line to a curve at a given point, you need to follow these steps:
1. Differentiate the equation of the curve to find its derivative.
2. Evaluate the derivative at the given point to find the slope of the tangent line.
3. Use the point-slope form (y - y1 = m(x - x1)) or the slope-intercept form (y = mx + b) to write the equation of the tangent line, substituting the slope and the coordinates of the given point.
Let's go through these steps in detail to find the slope-intercept equation of the tangent line to the curve at the point (1, 1).
Step 1: Differentiate the equation of the curve
Differentiate the equation 5x^3 + 10y^2 = 8x^5 + 2y with respect to x, treating y as a function of x.
To differentiate, apply the chain rule to any terms involving y:
d/dx (5x^3 + 10y^2) = d/dx (8x^5 + 2y)
Differentiating each term:
15x^2 + 20y(dy/dx) = 40x^4 + 2(dy/dx)
Step 2: Evaluate the derivative at the given point
Substitute the x and y values of the given point (1, 1) into the derived equation to determine the slope of the tangent line.
At (x = 1, y = 1):
15(1)^2 + 20(1)(dy/dx) = 40(1)^4 + 2(dy/dx)
15 + 20(dy/dx) = 40 + 2(dy/dx)
Rearranging the equation:
18(dy/dx) = 25
Divide both sides by 18:
dy/dx = 25/18
So, the slope of the tangent line at the point (1, 1) is 25/18.
Step 3: Write the equation of the tangent line
Now that we have the slope (m = 25/18) and the point (x1 = 1, y1 = 1), we can use the point-slope form or the slope-intercept form to write the equation of the tangent line.
Using the slope-intercept form (y = mx + b), substitute the slope, point coordinates, and solve for b:
y = mx + b
1 = (25/18)(1) + b
Simplifying:
1 = 25/18 + b
18/18 - 25/18 = b
(18 - 25)/18 = b
-7/18 = b
The equation of the tangent line is:
y = (25/18)x - 7/18
Therefore, the slope-intercept equation of the tangent line to the curve at the point (1, 1) is y = (25/18)x - 7/18.