5. Find the equation of the tangent plane to the surface f...

5. Find the equation of the tangent plane to the surface $f$ at the given point $P$. (a) (1 point) $f(x, y, z) = xe^{yz}$, $P(1, -1, -1)$ (b) (1 point) $f(x, y) = x\sin y$, $P(0, 0)$

Transcript

00:01 In this question we need to find equation of tangent plane.

00:17 So here we have the a part as z is equals to 3x square negative by square plus 2x where the point is 1 comma negative 2 comma 1.

00:37 So here equation of tangent plane will be equals to z negative z not will be equals to f of x x not comma y not in bracket we have x negative x not plus f of y x not here we have y negative y not so here we have f x comma y not so here we have f x comma y to be we have f x to be equals to 6x plus two and f y will be equal to negative 2 y so here if x at 1 comma negative 2 and 1 will be equals so here we have negative 12 plus 2 it will be equals to negative 10 and f y at 1 comma negative 2 comma 1 will be equals to negative 2 and here we have negative 2 so it will be equals to 4 so here we have the equation as z negative 1 it will be equals to negative 10 x negative 1 plus here we have 4 y negative negative 2 so here we get the equation as negative 10 here we have x negative 1 plus 4 here we have y plus 2 so this will be the required equation and is the required answer of it b part here we have there is equals to x cos x cos y and the point is 0 pi 0 so here we have f x comma y to be equals to x cos x cos so here f x will be equals to since we need to differentiate with respect to x only so here we have cos x plus x negative sine x and this cos y will be as it is so here f x at 0 comma pi comma 0 will be equals to here we have cos 0 plus when we put x is equal to 0 this will be 0 and here we have cos x is equal to 0 this will be 0 and here we have cos y is pi is pi so here we have negative 1 now further we need to evaluate if y so here x cos x will be as it is and the derivative of cos by will be negative sine y and f y at the values 0 comma pi comma 0 will be equals to 0 so now here the equation of tangent will be z negative 0 here we have f of x as negative 1 here we have x negative 0 plus 0 so here we have negative x here we have z so we get z plus x as the required equation of tangent c part here we have z is equals to x sine x y square at the point 0 comma 0 .0 .0 so here we first evaluate if x comma y that will be equals to so here we have if x to be equals to sign x y square plus x cos x y square and here we have y square so if x at 0 .0 .0 .0 will be equals to 0 and now we evaluate f y that will be equals to here we have x here we have cost x y square here we have 2 x y and at 0 .0 .0 if y will be equals to 0 so here we get the equation as z negative 0 is equal to 0 so here we have the equation as z is equals to 0.

06:09 D part here we have z is equals to x square plus y square and here we have x squared and here we have ln x square plus y square and we need to find it at a point 0 .1 .0 so here we have f x comma y will be equals to this now we need to find it out f x so it will be equals to this further it can be written as this now it can be evaluated at the point zero 1, 0 and it will be equals to 0 only...

Question

5. Find the equation of the tangent plane to the surface $f$ at the given point $P$. (a) (1 point) $f(x, y, z) = xe^{yz}$, $P(1, -1, -1)$ (b) (1 point) $f(x, y) = x\sin y$, $P(0, 0)$

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