Q. 35 - Calculus

Question

In Exercises $35-38$, find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$.

$f(x,y)=x^2-y^2\text{ at }P=(3,3),\mathbf{v}=⟨-1,5⟩$

Step-by-Step Solution

Answer

The function of directional derivative is $\nabla f\left(P\right).u=\frac{-18}{13}\sqrt{26}$.

Step 1: Directional derivative of function.

For a given function$P=(x_0,y_0)=(3,3)$ and $v=(-1,5)$, we must find the directional derivative $f(x,y)=x^2-y^2$.

$\parallel v\parallel =\sqrt{-1^2+5^2}=\sqrt{26}$

$\therefore u=(\alpha ,\beta )=\left(\frac{-\sqrt{26}}{26},\frac{5\sqrt{26}}{26}\right)$

Step 2: Directional unit vector.

The directional derivative of a function at point $P$ with direction unit vector $u$ is computed as follows:

$\mathrm{\nabla }f\left(P\right)\cdot u=\mathrm{\nabla }f(3,3)\times u=\left({\left.\frac{df}{dx}\right|}_{(3,3)}i+{\left.\frac{df}{dy}\right|}_{(3,3)}j\right)\times \left(\frac{-\sqrt{26}}{26}i+\frac{5\sqrt{26}}{26}j\right)$

$=\left[(2x{)}_{(3,3{)}^3}i+(-2y{)}_{(3,3)}j\right]\times \left(\frac{-\sqrt{26}}{26}i+\frac{5\sqrt{26}}{26}j\right)$

$=(6i-6j)\times \left(\frac{-\sqrt{26}}{26}i+\frac{5\sqrt{26}}{26}j\right)$

$=\left(\frac{-6\sqrt{26}}{26}-\frac{30\sqrt{26}}{26}\right)$

$=\frac{-36\sqrt{26}}{26}$

$\nabla f\left(P\right).u=\frac{-18}{13}\sqrt{26}$