- #1
hwill205
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Hello All,
I need help in my Calc 3 class and I decided to come here for homework help. What I'm looking for is someone to just check my work for a couple of homework problems. I've already done the problems, I would just like my work checked. Anyone who helps, your kindness is greatly appreciated.
Question 1:
Find the equations of the tangent plane and normal line to the given surface at the given point.
z=Log2 (x(y^2)+(x^2)y). So the base of the logarithm function is 2. The point is (1,-2,1)
I rewrote the function as ln(x(y^2) + (x^2)y))/ln(2)
The derivative with respect to x is (y+2x)/((ln2)(xy+x^2))
The derivative with respect to y is (x+2y)/((ln2)(xy+y^2))
The derivative with respect to z is just -1
When you plug in the point (1,-2,1), you get fx=0, fy= -3/((ln2)(2)) and fz=-1
So the equation of the tangent plan is fy(y+2)-(z-1)=0 (it was just easier to use fy instead of the whole expression).
The equation of the normal line is (y+2)/fy=(z-1)/-1, x=1
This is the symmetric equation.
For the second question:
Find the points on the hyperboloid x^2-2y^2-z^2=-2 at which the tangent plane is parallel to the plane 2x-3y+2z+7=0
F(x,y,z)= x^2-2y^2-z^2
The gradient vector for F(x,y,z) is a normal vector for the surface and thus, a normal vector for the tangent plane and for the parallel plane. The gradient vector is:
<2x,-4y,-2z>
so <2x,-4y,-2z>=k<2,-3,3> since <2,-3,3> is a normal vector for the parallel plane.
2x=2k, x=k
-4y=-3k, y=(3/4)k
-2z=2k, z=-k
Plug these values for x,y,z back into the equation for the hyperboloid:
k^2-2((3/4)k)^2-(-k)^2=-2
k^2-(9/8)k^2-k^2=-2
-(9/8)k^2=-2
k=(4/3)
x=4/3
y=1
z=-4/3
The point is (4/3, 1, -4/3)
Can someone tell me here I'm going wrong. I don't think either of these answers are right. Thanks for your help.
I need help in my Calc 3 class and I decided to come here for homework help. What I'm looking for is someone to just check my work for a couple of homework problems. I've already done the problems, I would just like my work checked. Anyone who helps, your kindness is greatly appreciated.
Question 1:
Find the equations of the tangent plane and normal line to the given surface at the given point.
z=Log2 (x(y^2)+(x^2)y). So the base of the logarithm function is 2. The point is (1,-2,1)
I rewrote the function as ln(x(y^2) + (x^2)y))/ln(2)
The derivative with respect to x is (y+2x)/((ln2)(xy+x^2))
The derivative with respect to y is (x+2y)/((ln2)(xy+y^2))
The derivative with respect to z is just -1
When you plug in the point (1,-2,1), you get fx=0, fy= -3/((ln2)(2)) and fz=-1
So the equation of the tangent plan is fy(y+2)-(z-1)=0 (it was just easier to use fy instead of the whole expression).
The equation of the normal line is (y+2)/fy=(z-1)/-1, x=1
This is the symmetric equation.
For the second question:
Find the points on the hyperboloid x^2-2y^2-z^2=-2 at which the tangent plane is parallel to the plane 2x-3y+2z+7=0
F(x,y,z)= x^2-2y^2-z^2
The gradient vector for F(x,y,z) is a normal vector for the surface and thus, a normal vector for the tangent plane and for the parallel plane. The gradient vector is:
<2x,-4y,-2z>
so <2x,-4y,-2z>=k<2,-3,3> since <2,-3,3> is a normal vector for the parallel plane.
2x=2k, x=k
-4y=-3k, y=(3/4)k
-2z=2k, z=-k
Plug these values for x,y,z back into the equation for the hyperboloid:
k^2-2((3/4)k)^2-(-k)^2=-2
k^2-(9/8)k^2-k^2=-2
-(9/8)k^2=-2
k=(4/3)
x=4/3
y=1
z=-4/3
The point is (4/3, 1, -4/3)
Can someone tell me here I'm going wrong. I don't think either of these answers are right. Thanks for your help.