Find the slope of the tangent to the curve at the point specified.
x^3+5x^2y+2y^2=4y+11 at (1,2)
So far I simplified it to -(3x^2-10y/5x^2+4y-4) which I think then simplifies to -(8/3). What do I do from here....?
I had dy/dx = -(3x^2+10xy/5x^2+4y-4)
plugging in (1,2) we get dy/dx = -(3+20)/(5+8-4)
= -23/9
so now you have a point and the slope, so
y-2 = (-23/9)(x-1) multiply by 9
9y - 18 = -23x + 23
23x + 9y = 41 or y = -(23/9)x + 41/9
To find the slope of the tangent to the curve, you need to find the derivative of the equation with respect to x and then evaluate it at the point (1,2).
Let's differentiate both sides of the equation implicitly:
d/dx(x^3 + 5x^2y + 2y^2) = d/dx(4y + 11)
Using the product rule, we can differentiate x^2y to get:
3x^2 + 5(2x)(y) + 2y^2(dy/dx) = 0
Now, let's solve for dy/dx:
2y^2(dy/dx) = -3x^2 - 10xy
dy/dx = (-3x^2 - 10xy) / (2y^2)
Now, substitute the point (1,2) into the derivative equation:
dy/dx = (-3(1)^2 - 10(1)(2)) / (2(2)^2)
= (-3 - 20) / 8
= -23 / 8
Therefore, the slope of the tangent to the curve at the point (1,2) is -23/8.
To find the slope of the tangent to the curve at the given point (1,2), you need to find the derivative of the equation with respect to x and then evaluate it at x = 1, y = 2.
Step 1: Differentiate the equation implicitly with respect to x.
To do this, differentiate each term with respect to x while treating y as a function of x.
d/dx(x^3) + d/dx(5x^2y) + d/dx(2y^2) = d/dx(4y+11)
3x^2 + 5(2xy) + 2(2y)(dy/dx) = 0
Step 2: Solve for dy/dx to isolate the expression.
Rearrange the terms to solve for dy/dx:
2(2y)(dy/dx) = -3x^2 - 10xy
dy/dx = (-3x^2 - 10xy) / (4y)
Step 3: Substitute the coordinates of the given point (1,2) into the expression for dy/dx.
Substitute x = 1 and y = 2 into the dy/dx expression.
dy/dx = (-3(1)^2 - 10(1)(2)) / (4(2))
dy/dx = (-3 - 20) / (8)
dy/dx = -23 / 8
Therefore, the slope of the tangent to the curve at the point (1,2) is -23/8.