Answer to Question #104429 in Differential Equations for Ajay

Question

(a)Calculate all the four second-order partial derivatives of the following functions:
(i) f(x,y)=cos(x^2+y^2)
(ii) f(x,y)=sin(x/y)

(b) Find the range of the function f defined by f(x,y) 10–x^2–y^2 for all (x,y)
for which x^2+y^2 ≤9 Sketch two of its level curves.
(c)Check whether the following functions are differentiable at the point given
against them:
(i)f(x,y)=|x=1|at (1,0)
(ii) f(x,y) =y^3+ ysin2x +e^(x+y) at (1,–1)

(d)Find dw/dt
w = xy + x,z = cos t, y = sint , z=1 at t= 0 .

Answer 1

$(a)(i) f(x,y)=cos(x ^ 2 +y^ 2 )$

$f ′ (x,y)_ x ​ =−2xsin(x ^2 +y^ 2 )$

$f ′′ (x,y)_ {xx} ​ =−2sin(x^ 2 +y ^ 2 )−2x.2xcos(x^ 2 +y ^ 2 )=−2sin(x^ 2 +y ^ 2 )$ $−4x^ 2 cos(x^ 2 +y ^ 2 )$

$f ′ (x,y)_ y ​ =−2ysin(x ^ 2 +y ^ 2 )$

$f ′′ (x,y)_ {yy} ​ =−2sin(x^ 2 +y ^ 2 )−2y.2ycos(x^ 2 +y ^ 2 )=−2sin(x^ 2 +y ^ 2 )$ $−4y^ 2 cos(x^ 2 +y ^ 2 )$

$(ii) \ f(x,y)=sin(x/y)\\f'(x,y)_x=\dfrac{cos(x/y)}{y}\\f''(x,y)_{xx}=-\dfrac{sin(x/y)}{y^2} \\ f'(x,y)_y=-\frac{x\cos \left(\frac{x}{y}\right)}{y^2} \\f''(x,y)_{yy}=\mathrm{\:}x\frac{\frac{x\sin \left(\frac{x}{y}\right)}{y^2}y^2-2y\cos \left(\frac{x}{y}\right)}{\left(y^2\right)^2}=\frac{x\left(x\sin \left(\frac{x}{y}\right)-2y\cos \left(\frac{x}{y}\right)\right)}{y^4}$

$(b) f(x,y) =10–x^2–y^2\ \& \ \ \ \ ; x^2+y^2 ≤9\\ f(x,y)=10-(x^2+y^2) \\maximum \ value\ of x^2+y^2 \ is\ 9\ and \ minimum\ value\ is \ 0$

$f(x,y) \ ranges \ between \ 1 \ to\ 10$

$(c)f(x,y)=|x-1|\ at\ (0,1)\\$

From the graph we can see that there is sharp turn at x=1 which means the function is not differentiable at x=1

$(ii) y^{3}+y\sin2x+e^{x+y}=0$

In the above function no subpart of the function which is not differentiable at any point

sin , exponential and y^3 all are continuous and differentiable at all domain

Hence the above function is differentiable

$d) w = xy + x , x = cos t , y = sint$

substituting x and y in w

$w=cos t*sin t + cos t$

$w=(sin 2t)/2 + cos t$

$dw/dt= cos 2t - sin t$

$putting t=0 dw/dt= 1$