Find an equation of the tangent plane to the surface given by z = f (x,y )=𝑥^2 y + x𝑦^3 at the point (x,y)=(2,1).
∂z/∂x = 2xy + y^3
∂z/∂y = x^2 + 3xy^2
z(2,1) = 4+2 = 6
So the tangent plane at (1,2) is
z-6 = 5(x-2) + 10(y-1)
Well, let's get this plane party started! To find the equation of the tangent plane, we'll need to calculate the partial derivatives of f(x, y) with respect to x and y.
First, let's find the partial derivative with respect to x. Taking the derivative of 𝑥^2 y + x𝑦^3 with respect to x gives us 2xy + 𝑦^3.
Next, let's find the partial derivative with respect to y. We differentiate 𝑥^2 y + x𝑦^3 with respect to y, and we get 𝑥^2 + 3x𝑦^2.
Now that we have the partial derivatives, we can find the equation of the tangent plane at the point (2, 1). So, grab your math hats and let's do some calculations.
At the point (2, 1), the partial derivative with respect to x is 2(2)(1) + 1^3, which simplifies to 4 + 1, giving us 5.
Similarly, the partial derivative with respect to y is 2^2 + 3(2)(1)^2, which simplifies to 4 + 3, giving us 7.
So, the equation of the tangent plane is z - f(2, 1) = 5(x - 2) + 7(y - 1).
Now, keep in mind that I'm just a clown bot, so take my answer with a grain of humor!
To find the equation of the tangent plane to the surface at the given point, we need to calculate the partial derivatives of the function f(x, y) at that point.
First, let's find the partial derivative with respect to x:
∂f/∂x = 2xy + y^3
Next, let's find the partial derivative with respect to y:
∂f/∂y = x^2 + 3xy^2
Now, let's evaluate these derivatives at the point (x,y) = (2,1):
∂f/∂x = 2(2)(1) + (1)^3
= 4 + 1
= 5
∂f/∂y = (2)^2 + 3(2)(1)^2
= 4 + 6
= 10
So, at the point (2,1), the partial derivatives are:
∂f/∂x = 5
∂f/∂y = 10
To find the equation of the tangent plane, we can use the general form:
z - f(x0, y0) = (∂f/∂x)(x - x0) + (∂f/∂y)(y - y0)
Substituting the values we found:
z - f(2,1) = 5(x - 2) + 10(y - 1)
Expanding and simplifying:
z - (2^2)(1) + (2)(1^3) = 5x - 10 + 10y - 10
z - 2 + 2 = 5x + 10y - 20
z = 5x + 10y - 20
Therefore, the equation of the tangent plane to the surface at the point (2,1) is z = 5x + 10y - 20.
To find the equation of the tangent plane to the surface at a given point, we need to calculate the gradient of the function at that point.
The gradient of a function of two variables f(x, y) is given by the vector ∇f(x, y) = ( ∂f/∂x, ∂f/∂y ), where ∂f/∂x and ∂f/∂y represent the partial derivatives of the function with respect to x and y, respectively.
In this case, the function is f(x, y) = x^2y + xy^3. Let's find the partial derivatives first:
∂f/∂x = 2xy + y^3
∂f/∂y = x^2 + 3xy^2
Now we can evaluate the partial derivatives at the given point (x, y) = (2, 1):
∂f/∂x = 2(2)(1) + (1)^3 = 4 + 1 = 5
∂f/∂y = (2)^2 + 3(2)(1)^2 = 4 + 6 = 10
So, the gradient vector ∇f(2, 1) is (5, 10).
The equation of the tangent plane to the surface can be written as:
z - f(a, b) = ∇f(a, b) · (x - a, y - b)
where (a, b) is the given point and · represents the dot product.
Using the given point (2, 1) and the gradient vector (5, 10), the equation becomes:
z - f(2, 1) = (5, 10) · (x - 2, y - 1)
Now substitute f(2, 1) into the equation:
z - (2^2 * 1 + 2 * 1^3) = (5, 10) · (x - 2, y - 1)
Simplifying further:
z - 4 - 2 = 5(x - 2) + 10(y - 1)
z - 6 = 5x - 10 + 10y - 10
Rearranging and simplifying:
5x + 10y - z = 6
Therefore, the equation of the tangent plane to the surface at the point (2, 1) is 5x + 10y - z = 6.