find the equation of a tangent of a curve at point(1,3)
x^2.y^3+y^2.x^4+4x^2-14=0,i tried differentiate it n also applying this formula y=mx+b.bt i gt confuse
In x^2.y^3+y^2.x^4+4x^2-14=0
I will assume you are using the "." as a multiplication symbol
Usually we use the * symbol
x^2(3y^2)dy/dx + y^3(2x) + y^2(4x^3) + x^4(2y)dy/dx + 8x = 0
dy/dx(3x^2 y^2 + 2x^4 y) = -2x y^3 - 4x^3 y^2 -8x
dy/dx = (-2x y^3 - 4x^3 y^2 - 8x)/(3x^2 y^2 + 2x^4 y)
which for (1,3) becomes
dy/dx = (-54 - 36 - 8)/(27+ 6) = -98/33
so in y = mx + b , m = -98/33 and the point is (1,3)
3 = (-98/33)(1) + b
b = 197/33
equation: y = (-98/33)x + 197/33
check my arithmetic, was expecting "nicer" numbers
AHHHH, just found the problem.
One of the first things you should do in these kind of problems is to make sure that the given point is actually on the line.
IT IS NOT, so the above work is meaningless
check your typing of the equation
Ok tankz for the clue u have given me am grateful
To find the equation of a tangent to a curve at a specific point, you need to follow these steps:
1. Differentiate the equation of the curve with respect to x to find the slope of the tangent line.
2. Substitute the x-coordinate of the given point into the derivative to find the slope at that point.
3. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope you calculated, to find the equation of the tangent line.
Let's go through these steps one by one for the curve x^2y^3 + y^2x^4 + 4x^2 - 14 = 0 and the point (1,3).
Step 1: Differentiate the curve equation
Taking the derivative of the equation with respect to x, we get:
d/dx (x^2y^3 + y^2x^4 + 4x^2 - 14) = 0
Step 2: Substitute the x-coordinate into the derivative
To find the slope at the point (1,3), substitute x=1 into the derivative you just found, and solve for the slope:
m = dy/dx = [d/dx (x^2y^3 + y^2x^4 + 4x^2 - 14)] (when x=1)
Step 3: Use the point-slope form to find the equation
Now that you have the slope (m) and the point (1,3), you can plug these values into the point-slope form of the line equation:
y - 3 = m(x - 1)
So the equation of the tangent line to the curve x^2y^3 + y^2x^4 + 4x^2 - 14 = 0 at the point (1,3) is y - 3 = m(x - 1), where m is the slope you calculated in Step 2.