Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) i

MATH200 December 2013

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Question 01 (b) i
Find the equation of the tangent plane to the surface $\displaystyle x^{2}z^{3}+y\sin(\pi x)=-y^{2}$ at the point $\displaystyle P=(1,1,-1)$.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Remember the formula for the tangent plane to a function $\displaystyle z=f(x,y)$ at the point $(x_{0},y_{0})$ is given by $\displaystyle (z-z_{0})=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$ where $z_{0}=f(x_{0},y_{0})$.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The equation of a tangent plane at $P(x_{0},y_{0},z_{0})$ is: $\displaystyle z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$ We need to find $\displaystyle f_{x}(x_{0},y_{0})$ and $\displaystyle f_{y}(x_{0},y_{0})$, but because z is defined implicitly, we need to be careful finding these partial derivatives. First, let's keep y constant and differentiate implicitly with respect to x to find $f_{x}$ (or ${\frac {\partial z}{\partial x}}$): $2xz^{3}+3x^{2}z^{2}{\frac {\partial z}{\partial x}}+\pi y\cos(\pi x)=0$ (Be careful differentiating the first term of the given equation - you must use the product rule!) Rearranging to solve for the partial derivative, you get: ${\frac {\partial z}{\partial x}}=-{\frac {2xz^{3}+\pi y\cos(\pi x)}{3x^{2}z^{2}}}$ Therefore, at P(1,1,-1), you get: ${\frac {\partial z}{\partial x}}=-{\frac {2(1)(-1)^{3}+\pi (1)\cos(\pi (1))}{3(1)^{2}(-1)^{2}}}={\frac {2+\pi }{3}}$ Similarly, for $f_{y}$ (or ${\frac {\partial z}{\partial y}}$), we get: $\displaystyle 3x^{2}z^{2}{\frac {\partial z}{\partial y}}+\sin(\pi x)=-2y$ Rearranging: $\displaystyle {\frac {\partial z}{\partial y}}=-{\frac {2y+\sin(\pi x)}{3x^{2}z^{2}}}$ At P(1,1,-1): $\displaystyle {\frac {\partial z}{\partial y}}=-{\frac {2(1)+\sin(\pi (1))}{3(1)^{2}(-1)^{2}}}=-{\frac {2}{3}}$ Now, plugging these values into the tangent plane formula, we get: $\displaystyle z-(-1)={\frac {2+\pi }{3}}(x-1)+-{\frac {2}{3}}(y-1)$ Making things nice and beautiful, the final answer is: ${\color {blue}z+1={\frac {2+\pi }{3}}(x-1)-{\frac {2}{3}}(y-1)}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Note: There is another, perhaps easier, way of solving for the partial derivatives in the previous solution. If we move all the variables onto one side of the equation and leave the other as zero, we get: $\displaystyle x^{2}z^{3}+y\sin(\pi x)+y^{2}=0$ Let's define a new function of three variables F(x,y,z), $\displaystyle F(x,y,z)=x^{2}z^{3}+y\sin(\pi x)+y^{2},$ where the level surface is given by F(x,y,z)=0. Then, the partial derivatives we found earlier in this question can be expressed simply as: ${\begin{aligned}{\frac {\partial z}{\partial x}}&=-{\frac {F_{x}}{F_{z}}}\\{\frac {\partial z}{\partial y}}&=-{\frac {F_{y}}{F_{z}}}\end{aligned}}$ Let's test this out. If we find the partial derivatives of F with respect to x and to z, we get: ${\begin{aligned}{\frac {\partial F}{\partial x}}&=2xz^{3}+\pi y\cos(\pi x)\\{\frac {\partial F}{\partial z}}&=3x^{2}z^{2}\end{aligned}}$ Therefore, the partial derivative of z with respect to x is: ${\frac {\partial z}{\partial x}}=-{\frac {F_{x}}{F_{z}}}=-{\frac {2xz^{3}+\pi y\cos(\pi x)}{3x^{2}z^{2}}}$ This is the same equation as the one we found earlier using implicit differentiation. Try out the other partial derivative on your own to make sure both work. Using the result obtained for ${\frac {\partial z}{\partial y}}$ and using the formula for the tangent plane and plugging the values of x,y, and z at the point P gives ${\color {blue}z+1={\frac {2+\pi }{3}}(x-1)-{\frac {2}{3}}(y-1)}.$

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