f(x,y)=x^2+2y^2
use the gradient vector �Þf(5, 1) to find the tangent line to the level curve f(x, y) = 27 at the point (5, 1).
answer choices:
1. 10x + 4y = 27
2. 2x + 4y = 27
3. 5x + 2y = 27
4. 10x − 4y = 27
5. 5x − 2y = 27
To find the tangent line to the level curve f(x, y) = 27 at the point (5, 1), we first need to find the gradient vector of the function f(x, y). The gradient vector gives the direction of steepest ascent at any point on the surface defined by the function.
The gradient vector of f(x, y) is denoted as ∇f(x, y) and is given by the partial derivatives of f(x, y) with respect to x and y. In this case, it is:
∇f(x, y) = [2x, 4y]
To find the gradient vector ∇f(5, 1), we substitute x = 5 and y = 1 into the equation:
∇f(5, 1) = [2(5), 4(1)] = [10, 4]
Next, we need to find the equation of the tangent line to the level curve f(x, y) = 27 at the point (5, 1). A tangent line has the same direction as the gradient vector at that point.
So, the equation of the tangent line will have the form:
a(x - 5) + b(y - 1) = 0
where a and b are constants.
To find the specific values of a and b, we substitute the coordinates (5, 1) into the equation and use the gradient vector ∇f(5, 1):
10(5 - 5) + 4(1 - 1) = 10a + 4b = 0
Solving this equation, we find that a = -2b.
Substituting this back into our equation, we have:
10(-2b) + 4b = 0
Simplifying, we get:
-20b + 4b = 0
-16b = 0
b = 0
Now we can substitute the values of a and b back into our equation:
-2b(x - 5) + b(y - 1) = 0
-2(0)(x - 5) + 0(y - 1) = 0
Simplifying, we have:
0 = 0
This means that the equation is satisfied for any values of x and y. In other words, the tangent line is a constant line.
Therefore, the correct answer choice is 5. 5x - 2y = 27, as it represents a constant line passing through the point (5, 1).