Answer to Question #104441 in Calculus for maria

Question #104441
(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 .
1
Expert's answer
2020-03-10T13:15:53-0400

"(a)(i)\n\\\\f(x,y)=cos(x^2+y^2)\n\\\\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^2cos(x^2+y^2)\\\\f'(x,y)_y=-2ysin(x^2+y^2)\n\\\\f''(x,y)_{yy}=-2sin(x^2+y^2)-2y.2ycos(x^2+y^2)=-2sin(x^2+y^2)-4y^2cos(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}\n\\\\\nf'(x,y)_y=-\\frac{x\\cos \\left(\\frac{x}{y}\\right)}{y^2}\n\\\\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\u2013x^2\u2013y^2\\ \\& \\ \\ \\ \\ ; x^2+y^2 \u22649\\\\ \nf(x,y)=10-(x^2+y^2)\n\\\\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 y3 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


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