Find the equation of the tangent plane at the given point.
1)x^2+y^2-z=1 at the point (1,3,9)
2)x^2*y^2+z-40=0 at x=2, y=3
3) x^2+ln(xy)+z=6 at point (4,0.25,2)
For the first one I got the partial fraction in terms of x, y and z and my final answer was -2x+6y-z+2. I don't know if that is right.
For the second one, the back of the book says the answer is -36x-24y+148.
I didn't get the partial fraction with respect to z, I just plugged in x=2 and y=3 for z=-x+2y^2+40 and got 4. Then, my final answer was 36x+24y-140. I don't know how to get negative 4.
I haven't done prob. 3 because I'm not sure what is the right procedure.
#1:
∇f = 2x,2y,-1
∇f(1,3,9) = 2,6,-1
plane: 2(x-1)+6(y-3)-1(z-9) = 0
2x+6y-z = 11
#2:
at x=2,y=3, z=4
∇f = 2xy^2,2x^2y,1
∇f(2,3,4) = 36,24,1
plane: 36(x-2)+24(y-2)+1(z-4) = 0
36x+24y+z = 124
#3:
∇f = 2x+1/x,1/y,1
∇f(4,1/4,2) = 33/4,4,1
plane: 33/4 (x-4) + 4(y-1/4) + 1(z-2) = 0
33x+16y+4z = 144
O' course, check my arithmetic
To find the equation of the tangent plane at a given point in 3-dimensional space, you need to follow a specific procedure. Let's go through each problem one by one.
1) Equation: x^2 + y^2 - z = 1 at the point (1, 3, 9)
To find the equation of the tangent plane, you need to use the gradient vector. First, rewrite the equation in the form: f(x, y, z) = 0. So, the equation becomes: x^2 + y^2 - z - 1 = 0.
Next, calculate the gradient vector (∇f) of the function f(x, y, z) = x^2 + y^2 - z - 1. The gradient vector is a vector pointing in the direction of the steepest increase of the function. It is calculated as follows:
∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z), where ∂f represents the partial derivative of f.
Here, ∂f/∂x = 2x, ∂f/∂y = 2y, and ∂f/∂z = -1. So, the gradient vector (∇f) is (2x, 2y, -1).
Now, evaluate (∇f) at the given point (1, 3, 9). This gives you (∇f) = (2, 6, -1).
Finally, use the equation of the tangent plane, which is given by:
(2, 6, -1) · (x - 1, y - 3, z - 9) = 0, where · represents the dot product.
Expanding this equation gives: 2(x - 1) + 6(y - 3) - 1(z - 9) = 0.
Simplifying further, you get the equation of the tangent plane: 2x + 6y - z + 12 = 0.
So, the correct equation of the tangent plane is 2x + 6y - z + 12 = 0. It seems that the expression you got is incorrect.
2) Equation: x^2 * y^2 + z - 40 = 0, at x = 2, y = 3
A similar approach is followed to find the equation of the tangent plane.
First, rewrite the equation as: f(x, y, z) = x^2 * y^2 + z - 40 = 0.
Now, calculate the partial derivatives of f(x, y, z) with respect to x, y, and z:
∂f/∂x = 2xy^2,
∂f/∂y = 2x^2y,
∂f/∂z = 1.
The gradient vector (∇f) is (∂f/∂x, ∂f/∂y, ∂f/∂z):
(2xy^2, 2x^2y, 1).
Substituting the given values x = 2 and y = 3, we get (∇f) = (36, 24, 1).
The equation of the tangent plane is given by: (36, 24, 1) · (x - 2, y - 3, z - f(2,3)) = 0.
Expanding and simplifying: 36(x - 2) + 24(y - 3) + (z - f(2,3)) = 0.
Since z - f(2,3) = 0 (since the function evaluates to 0 at the point), we have: 36(x - 2) + 24(y - 3) = 0.
Simplifying, we get the equation of the tangent plane: 36x + 24y - 132 = 0.
So, the correct equation of the tangent plane is 36x + 24y - 132 = 0. The expression you got seems to be incorrect.
For problem 3, please provide the equation and the point at which the tangent plane needs to be found, and I'll guide you through the process.