Find an equation of the tangent plane to the given parametric surface at the specified point. x=u2+1,y=v3+1,z=u+v;(5,2,3)

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joshyoung05M · Accepted Answer

We can write r(u,v)=(u2+1)i+(v3+1)j+(u+v)k We first compute the tangent vectors ru=∂x∂ui+∂y∂uj+∂z∂uk=2ui+0j+k rv=∂x∂vi+∂y∂vj+∂z∂vk=0i+3v2j+k Thus the normal vector to the tangent plane is ru×rv=|ijk2u0103v21| Expand along Row 1 i|013v21|−j|2u101|+k|2u003v2| =−3v2i−2uj+6uv2k Notice that the point (5,2,3) corresponds to the parameter values u=2 and v=1, So the normal vector there is =−3(1)2i−2(2)j+6(2)(1)2k =−3i−4j+12k If a is the position vector of a point on the plane. And n is a vector normal to the plane. Then the equation of the plane is (r−a)×n=0 (x−5)×(−3)+(y−2)×(−4)+(z−3)×(12)=0 −3x+15−4y+8+12z−36=0 −3x−4y+12z−13=0 3x+4y−12z+13=0 Result: 3x+4y−12z+13=0

alenahelenash · Accepted Answer

To find the equation of the tangent plane to the given parametric surface at the specified point (5,2,3), we need to calculate the partial derivatives and evaluate them at the point of interest.Let the surface be defined by the parametric equations:x=u2+1,y=v3+1,z=u+v.The partial derivatives with respect to u and v are given by:∂x∂u=2u,∂x∂v=0,∂y∂u=0,∂y∂v=3v2,∂z∂u=1,∂z∂v=1.Evaluating these derivatives at the point (5,2,3), we get:∂x∂u=2(5)=10,∂x∂v=0,∂y∂u=0,∂y∂v=3(2)2=12,∂z∂u=1,∂z∂v=1.The equation of the tangent plane can be expressed as:∂z∂u(u−u0)+∂z∂v(v−v0)=z−z0,where (u0,v0,z0) represents the coordinates of the point of interest. In this case, (u0,v0,z0)=(5,2,3).Substituting the values into the equation, we have:1(u−5)+1(v−2)=z−3.Simplifying the equation, we obtain the equation of the tangent plane as:u+v−z+4=0.

star233 · Accepted Answer

Answer:x+12y−z−26=0Explanation:Let&#039;s start by finding the partial derivatives of x, y, and z with respect to u and v:∂x∂u=ddu(u2+1)=2u∂x∂v=ddv(u2+1)=0∂y∂u=ddu(v3+1)=0∂y∂v=ddv(v3+1)=3v2∂z∂u=ddu(u+v)=1∂z∂v=ddv(u+v)=1Next, we evaluate these partial derivatives at the point (5,2,3). Substituting u=5 and v=2 into the above expressions, we get:∂x∂u=2(5)=10∂x∂v=0∂y∂u=0∂y∂v=3(22)=12∂z∂u=1∂z∂v=1Now we can use these partial derivatives to construct the equation of the tangent plane. The equation of a plane can be written as:(x−x0)∂z∂u+(y−y0)∂z∂v=(z−z0)Substituting the values of x0=5, y0=2, z0=3, and the partial derivatives we obtained, the equation of the tangent plane becomes:(x−5)(1)+(y−2)(12)=(z−3)Simplifying further:x+12y−5−24=z−3x+12y−z−26=0Therefore, the equation of the tangent plane to the given parametric surface at the point (5,2,3) is x+12y−z−26=0.

karton · Accepted Answer

Step 1:Given the parametric equations of the surface:x=u2+1,y=v3+1,z=u+v.Step 2:We&#039;ll start by finding the partial derivatives:∂x∂u=ddu(u2+1)=2u,∂x∂v=0,∂y∂u=0,∂y∂v=ddv(v3+1)=3v2,∂z∂u=1,∂z∂v=1.Step 3:Next, we evaluate these partial derivatives at the point (5,2,3):∂x∂u(5,2,3)=2·5=10,∂x∂v(5,2,3)=0,∂y∂u(5,2,3)=0,∂y∂v(5,2,3)=3·22=12,∂z∂u(5,2,3)=1,∂z∂v(5,2,3)=1.Step 4:Now we have the directional vectors for the tangent plane at the point (5,2,3), which are ∂𝐫∂u(5,2,3)=(10,0,1) and ∂𝐫∂v(5,2,3)=(0,12,1).The equation of the tangent plane can be written as:(x−x0)·∂x∂u(x0,y0,z0)+(y−y0)·∂y∂u(x0,y0,z0)+(z−z0)·∂z∂u(x0,y0,z0)=0,where (x0,y0,z0) is the point on the surface. Substituting the known values:(x−5)·10+(y−2)·0+(z−3)·1=0,10x−50+z−3=0,10x+z−53=0.Thus, the equation of the tangent plane to the parametric surface at the point (5,2,3) is 10x+z−53=0.

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