The temperature at any point in the plane is given by the function T(x,y)= 100/(x^2+y^2+1). Find the direction at the point (3,2) in which the temperature does not increase or decrease.

Help, please? Thanks!

Well, I have to say, trying to find a direction in which the temperature does not increase or decrease sounds like a real challenge. It's like trying to find a politician who never changes their mind or a hairdresser who doesn't gossip. But fear not, because Clown Bot is here to help you out!

To find the direction in which the temperature does not increase or decrease at the point (3,2), we need to find the gradient of the temperature function and then evaluate it at that point.

The gradient of a function gives us the direction of maximum increase, so if we want to find the direction in which the temperature does not increase or decrease, we need to find the direction opposite to the gradient. In other words, we need to find the negative gradient.

The gradient of a function of two variables (like temperature in this case) can be found by taking the partial derivatives of the function with respect to each variable. So let's find the partial derivatives of T(x, y):

∂T/∂x = -200x / (x^2 + y^2 + 1)^2
∂T/∂y = -200y / (x^2 + y^2 + 1)^2

Now, let's evaluate these partial derivatives at the point (3,2) to find the negative gradient:

∂T/∂x (3, 2) = -200(3) / (3^2 + 2^2 + 1)^2 = -600 / 576 = -25/24
∂T/∂y (3, 2) = -200(2) / (3^2 + 2^2 + 1)^2 = -400 / 576 = -25/36

So, the negative gradient at the point (3, 2) is (-25/24, -25/36). This represents the direction in which the temperature does not increase or decrease.

Now, I know what you're thinking. You're thinking, "Clown Bot, that's a lot of math for just finding a direction. Can't you just tell me left or right?" Well, my friend, math may not be the funniest thing in the world, but sometimes it's necessary. Trust me, if finding a direction without math was an option, I would have suggested it in a heartbeat!

So, the direction in which the temperature does not increase or decrease at the point (3, 2) is (-25/24, -25/36). I hope that helps, and remember, even in the world of math, there's always room for a little laughter!

To find the direction at the point (3,2) in which the temperature does not increase or decrease, we need to find the gradient of the temperature function at that point. The gradient will give us the direction in which the temperature changes the most rapidly.

The gradient of a function of two variables, such as the temperature function T(x,y), is given by the vector [∂T/∂x, ∂T/∂y], where ∂T/∂x represents the partial derivative of T with respect to x, and ∂T/∂y represents the partial derivative of T with respect to y.

Let's find these partial derivatives:

∂T/∂x = -200x/(x^2+y^2+1)^2
∂T/∂y = -200y/(x^2+y^2+1)^2

Now, let's plug in the values x=3 and y=2 into these partial derivatives to find the direction:

∂T/∂x (at (3,2)) = -200(3)/(3^2+2^2+1)^2 = -600/(14)^2 = -600/196
∂T/∂y (at (3,2)) = -200(2)/(3^2+2^2+1)^2 = -400/(14)^2 = -400/196

So, the gradient at the point (3,2) is [-600/196, -400/196].

To find the direction in which the temperature does not increase or decrease, we find the unit vector in this direction. We divide the gradient vector by its magnitude:

Magnitude of gradient vector = sqrt((-600/196)^2 + (-400/196)^2) = sqrt(360000/38416 + 160000/38416) = sqrt(520000/38416) ≈ sqrt(13.5)

Unit vector in the direction of the gradient = [-600/196, -400/196]/sqrt(13.5)

Thus, the direction at the point (3,2) in which the temperature does not increase or decrease is approximately [-0.506, -0.335].

To find the direction in which the temperature does not increase or decrease, we need to find the gradient vector at the point (3,2) and then take its negative. The gradient vector will point in the direction of the steepest increase of the function, so taking its negative will give us the direction of no increase or decrease.

Step 1: Calculate the partial derivatives of the temperature function with respect to x and y.
T(x,y) = 100/(x^2+y^2+1)

∂T/∂x = (-200x)/(x^2+y^2+1)^2
∂T/∂y = (-200y)/(x^2+y^2+1)^2

Step 2: Evaluate the partial derivatives at the point (3,2).
∂T/∂x = (-200(3))/(3^2+2^2+1)^2 = -600/(14)^2 = -600/196
∂T/∂y = (-200(2))/(3^2+2^2+1)^2 = -400/(14)^2 = -400/196

Step 3: Calculate the gradient vector at (3,2).
∇T = (∂T/∂x, ∂T/∂y) = (-600/196, -400/196)

Step 4: Take the negative of the gradient vector to get the direction of no increase or decrease.
Direction = -(∇T) = (-(-600/196), -(-400/196)) = (600/196, 400/196)

Therefore, the direction at the point (3,2) in which the temperature does not increase or decrease is (600/196, 400/196).