Find the equation for the plane tangent to each surface z=f(x,y) at the indicated point. (a)z=x3+y3−2xy,at the point(1,2,−5) (b)z=(cos⁡(x))(cos⁡(y)),at the point(0,π2,0) (c)z=(cos⁡(x))(sin⁡(y)),at the point(0,π2,1)

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Jeremy Merritt · Accepted Answer

Part (a): We have: Surface z=x3+y3−2xy,at the point(1,2,−5) (x0,y0,z0)=(1,2,−5) Used formula in equation: An equation of the tangent plane to the surface z=f(x,y) at the point P(x0,y0,z0) is: z−z0=fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0) Differentiating partially equation of surface with respect to x and y:  fx=∂z∂x=3x2−2y  ∂z∂x=3(1)2−2(2)=−1  fy=∂z∂x=3y2−2x  ∂z∂x=3(2)2−2(1)=10  Now, put the valuesin formula: z−z0=fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0) z−(−5)=(−1)(x−1)+(10)(y−2) x−10y+z+24=0

temnimam2 · Accepted Answer

Part (b):Given: Surface z=cos⁡x×cos⁡y,at the point(0,π2,0)(x0,y0,z0)=(0,π2,0) Differentiating partially equation of surface with respect to x and y: fx=∂z∂x=(−sin⁡x)(cos⁡y) ∂z∂x=(−sin⁡0)(cos⁡π2)=0 fy=∂z∂x=(cos⁡x)(−sin⁡y) ∂z∂x=(cos⁡0)(−sin∣π2)=−1 Now, put the values in formula: z−z0=fx(x0,y0)(x−x0)+fy(x0,y0) z−(0)=(0)(x−0)+(−1)(y−π2) y+zπ2=0

nick1337 · Accepted Answer

Part (c):Given: Surface z=(cos⁡x)(cos⁡y), at the point (0,π2,1) (x0,y0,z0)=(0,π2,1) Differentiating partially equation of surface with respect to x and y: fx=∂z∂x=(−sin⁡x)(sin⁡y) ∂z∂x=(−sin⁡0)(sin⁡π2)=0 fy∂z∂x=(cos⁡x)(cos⁡y) ∂z∂x=(cos⁡0)(cos⁡π2)=0 Now, put the values in formula: z−z0=fx(x0,y0)(x0,y0)+fy(x0,y0)(y−y0) z−(1)=(0)(x−0)+(0)(y−π2) z−1=0

Find the equation for the plane tangent to each surface

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