find equation of tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)
Once again I can not post my answer but in the end I get
3 y = -2 x + 5
Can other teachers post all right ?
no luck - hope someone can help you.
the idea is to do f(d+dx) - f(x)
where f(x+dx) = f(x) + (dy/dx) dx
I've sent your posts to Bob and Leo. I'll send this one too.
Thanks - I can not really answer this because it will not let me write out derivative method.
Bob knows. He has cleaned out the banned list, and said he'd check more specifically later.
I've texted Leo ... I'm surprised there's been no answer.
=(
(1/3)x^(-2/3) + (1/2)y^(-1/2) dy/dx = 0
at (1,1)
(1/3)(1) + (1/2)(1)dy/dx =0
dy/dx =-(1/2) / (1/3) = -3/2
so y-1 = (-3/2)(x-1)
2y - 2 = -3x + 3
2y = -3x + 5
y = (-3/2)x + 5/2
Damon, I had the same problem, now it seems to work
tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)
-----------------------------------
(1/3) / x^(2/3) + (1/2) dy/dx /y^(1/2)=0
so
(1/2)dy/dx = -(1/3) y^(1/2)/x^(2/3)
at (1.1)
dy/dx = -(2/3)
then
y = -2x/3 + b
1 = -2/3 + b
b = 5/3
so
y = -2x/3 + 5/3
3 y = -2 x + 5
To find the equation of the tangent line to the curve at a given point, we need to calculate the derivative of the function and evaluate it at the given point.
First, let's differentiate the equation x^(1/3) + y^(1/2) = 2 with respect to x.
Differentiating x^(1/3) with respect to x using the power rule, we get:
(1/3) * x^(-2/3)
Differentiating y^(1/2) with respect to x using the chain rule since y is a function of x, we get:
(1/2) * 2 * y^(-1/2) * dy/dx
Since we are finding the derivative with respect to x, we need to use dy/dx instead of just y, and chain rule is necessary.
Since the equation is x^(1/3) + y^(1/2) = 2, we can express y in terms of x:
y^(1/2) = 2 - x^(1/3)
y = (2 - x^(1/3))^2
Now let's differentiate y = (2 - x^(1/3))^2 with respect to x.
Using the chain rule, we get:
dy/dx = 2 * (2 - x^(1/3))^2 * (-1/3) * x^(-2/3)
Now we have the derivative of y with respect to x. Next, we substitute the x and y values of the given point (1,1) into the derivative to find the slope of the tangent line at that point.
Slope (m) = dy/dx = 2 * (2 - 1^(1/3))^2 * (-1/3) * 1^(-2/3)
Simplifying the expression we get:
m = 2 * (2 - 1)^2 * (-1/3) * 1
m = -2/3
The slope of the tangent line at the point (1,1) is -2/3.
Now we can use the point-slope form to find the equation of the tangent line. The point-slope form is given by:
(y - y1) = m * (x - x1), where (x1, y1) is the given point and m is the slope.
Substituting the values into the equation, we get:
(y - 1) = (-2/3) * (x - 1)
Simplifying the equation further, we get the equation of the tangent line:
3y - 3 = -2(x - 1)
Expanding and rearranging, we get the final equation of the tangent line:
3y - 3 = -2x + 2
Therefore, the equation of the tangent line to the curve x^(1/3) + y^(1/2) = 2 at the point (1,1) is 2x + 3y = 5.