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. 

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

Step-by-Step Solution

Answer

The directional derivative of the given

function is $\frac{-18}{13}\sqrt{26}$

Step 1: Given data

$f(x,y)=x^2-y^2$

$P=\left(x_0,y_0\right)=(3,3)\text{ and }y=(-1,5)$

Step 2: solution

Therefore

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

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

Directional derivation function at  $P$point is given by

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

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

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

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

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

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

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