โz/โx = 2xy + y^3
โz/โy = 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)
โz/โy = 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)
First, let's find the partial derivative with respect to x. Taking the derivative of ๐ฅ^2 y + x๐ฆ^3 with respect to x gives us 2xy + ๐ฆ^3.
Next, let's find the partial derivative with respect to y. We differentiate ๐ฅ^2 y + x๐ฆ^3 with respect to y, and we get ๐ฅ^2 + 3x๐ฆ^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!
First, let's find the partial derivative with respect to x:
โf/โx = 2xy + y^3
Next, let's find the partial derivative with respect to y:
โf/โy = x^2 + 3xy^2
Now, let's evaluate these derivatives at the point (x,y) = (2,1):
โf/โx = 2(2)(1) + (1)^3
= 4 + 1
= 5
โf/โy = (2)^2 + 3(2)(1)^2
= 4 + 6
= 10
So, at the point (2,1), the partial derivatives are:
โf/โx = 5
โf/โy = 10
To find the equation of the tangent plane, we can use the general form:
z - f(x0, y0) = (โf/โx)(x - x0) + (โf/โy)(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.
The gradient of a function of two variables f(x, y) is given by the vector โf(x, y) = ( โf/โx, โf/โy ), where โf/โx and โf/โy 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:
โf/โx = 2xy + y^3
โf/โy = x^2 + 3xy^2
Now we can evaluate the partial derivatives at the given point (x, y) = (2, 1):
โf/โx = 2(2)(1) + (1)^3 = 4 + 1 = 5
โf/โy = (2)^2 + 3(2)(1)^2 = 4 + 6 = 10
So, the gradient vector โf(2, 1) is (5, 10).
The equation of the tangent plane to the surface can be written as:
z - f(a, b) = โf(a, b) ยท (x - a, y - b)
where (a, b) is the given point and ยท 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) ยท (x - 2, y - 1)
Now substitute f(2, 1) into the equation:
z - (2^2 * 1 + 2 * 1^3) = (5, 10) ยท (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.