Problem 15 - Calculus Early Transcendentals

Question

Computing gradients Compute the gradient of the following functions and evaluate it at the given point \(P\). $$F(x, y)=e^{-x^{2}-2 y^{2}} ; P(-1,2)$$

Step-by-Step Solution

Answer

Answer: The gradient of the function at point \(P(-1,2)\) is $$\nabla F(-1, 2) = \left\langle 2e^{-17}, -8e^{-17} \right\rangle$$.

Step 1: Compute the partial derivative with respect to 'x'

To find the partial derivative of the function \(F(x, y)=e^{-x^{2}-2 y^{2}}\) with respect to 'x', treat 'y' as a constant and differentiate the function with respect to 'x'. Using the chain rule, we get: $$\frac{\partial F}{\partial x} = e^{-x^{2}-2y^{2}} \cdot (-2x) = -2xe^{-x^{2}-2y^{2}}$$

Step 2: Compute the partial derivative with respect to 'y'

Similarly, to find the partial derivative of the function \(F(x, y)=e^{-x^{2}-2 y^{2}}\) with respect to 'y', treat 'x' as a constant and differentiate the function with respect to 'y'. Again, using the chain rule, we get: $$\frac{\partial F}{\partial y} = e^{-x^{2}-2y^{2}} \cdot (-4y) = -4ye^{-x^{2}-2y^{2}}$$

Step 3: Write the gradient vector

Now that we have both partial derivatives, we can write the gradient vector by combining these two derivatives as components of the vector: $$\nabla F(x, y) = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y} \right\rangle = \left\langle -2xe^{-x^{2}-2y^{2}}, -4ye^{-x^{2}-2y^{2}} \right\rangle$$

Step 4: Evaluate the gradient vector at the given point

Now, we will evaluate the gradient vector at the given point \(P(-1, 2)\). Substitute the coordinates of the point into the gradient vector: $$\nabla F(-1, 2) = \left\langle -2(-1)e^{-(-1)^{2}-2(2)^{2}}, -4(2)e^{-(-1)^{2}-2(2)^{2}} \right\rangle$$ Evaluate the exponential function and simplify the expression: $$\nabla F(-1, 2) = \left\langle 2e^{-17}, -8e^{-17} \right\rangle$$ So, the gradient at point P is: $$\nabla F(-1, 2) = \left\langle 2e^{-17}, -8e^{-17} \right\rangle$$.