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# Find a normal vector of the curve 6x^2+2y^2+z^2 = 225 at the given point P : (5; 5; 5).

Find a normal vector of the curve 6x^2+2y^2+z^2 = 225 at the given point P : (5; 5; 5).

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### calculus - Finding a normal vector of a curve - Mathematics ...

Find a normal vector of the curve \$ 6x^2+2y^2+z^2 = 225 \$ at the given point \$ P : (5 ... Finding a normal vector of a curve. ... normal vector of the curve \$ 6x^2+2y ...

### calculus - Finding a normal vector of a curve - Mathematics ...

Find a normal vector of the curve \$ 6x^2+2y^2+z^2 = 225 \$ at the given point \$ P : (5, 5, 5) ... Find a normal vector of the curve \$ 6x^2+2y^2+z^2 = 225 \$ at the ...

### 1) Find A Normal Vector Of The Curve Or Surface At ... - Chegg

Answer to 1) Find a normal vector of the curve or surface at the given point p .make a sketch 6x^2+2y^2+z^2=225, p:(5,5,5) 2)Direc... Textbook Solutions ...

### Exam 2 Sample Solutions - Whitman People

... At a given point on a curve (x(t 0);y(t 0);z 0 ... z(t 0)) with normal vector is B ... and normal line to the given surface at the speci ed point: x2 + 2y2 3z2 ...

### Find the slope of the tangent line to the curve 0x^2&#8722 ...

The point P(4, −2)lies on the curve y ... find an equation of the tangent line to the curve at the given point. graph ... =x^2+2y^2 use the gradient vector Þf(5 ...

### The Gradient and the Level Curve - Whitman College

The Gradient and the Level Curve ... z = x2 + y2 at (1;3). Fix the level curve k = f(x;y) ... Now consider the gradient vector at the point (a;b), ...

### how to find equation of horizontal tangent line of 2y^3+6x^2y ...

... 2 = 3x + 53 +2 +2 ... algebra 1. 6x+2y=8 2. y=-2 3. x=5 4. 4x-3y=11 Tell whether ... Find the distance between P (–2, 5) ... (4,5) 3) the curve has aturning ...

### Exam Review Name Find the arc length of the given curve. 1)

Find the arc length of the given curve. 1)x =2 ... The plane through the point P(-6, 6, -7) and normal ... The line through the point P(3, 4, -5) parallel to the ...

## Suggested Questions And Answer :

### Find a normal vector of the curve 6x^2+2y^2+z^2 = 225 at the given point P : (5; 5; 5).

Find a normal vector of the curve 6x^2+2y^2+z^2 = 225 at the given point P : (5; 5; 5).​ The normal vector at a point on a surface is given by (1) where and are partial derivatives. and (x0, y0) = (5,5) we have z = (225 - 6x^2 - 2y^2)^(1/2) = f(x,y), giving us fx = -6x(225 - 6x^2 - 2y^2)^(-1/2),     and   fx(5,5) = -30(225 - 150 - 25)^(-1/2) = -30(25)^(-1/2) = -6 fy = -2y(225 - 6x^2 - 2y^2)^(-1/2),     and   fx(5,5) = -10(225 - 150 - 25)^(-1/2) = -10(25)^(-1/2) = -2 Therefore, N = | -6 |                         | -2 |                         | -1 |

### Find an equation for the line through point (3, -2, 1) parallel to the line x=1+2t, y=2-t, z=3t

Given two parametric lines defined by p = p1 + t(A) q = q1 + s(B) these lines will be parallel if the unit direction vectors A and B are the same. Let us define p as p = p1 + t(p2 - p1) where p1 = (1, 2, 0) and p2 - p1 = (2, -1, 3) taken from x = 1 + 2t, y = 2 - t, z = 3t Let A = (p2-p1)/norm(p2-p1) be the normalized direction vector of p2-p1 where norm(p2-p1) = sqrt(4 + 1 + 9) = sqrt(14) Then A = (2/sqrt(14), -1/sqrt(14), 3/sqrt(14)) which is the normalized direction vector for line p so that p = p1 + t(A) If we set B = A, then the parametric line q passing through the point q1 = (3, -2, 1) is given by q = q1 + s(B) Because B = A, line q is parallel to the line p given above. Note that when the direction vector is normalized, the parameters "s" and "t" for lines q and p represent the Euclidian distance from the points q1 and p1 respectively.

### let f(x)=x^105+x^53+x^27+x^13+x^3+3x+1.if g(x) is inverse of function f(x),then g'(x) is

let f(x)=x^105+x^53+x^27+x^13+x^3+3x+1.if g(x) is inverse of function f(x),then g'(x) is ?   We have f(x) = y = x^105 + x^53 + x^27 + x^13 + x^3 + 3x + 1   Normally, the inverse function is given by f^(-1)(x) = g(x), where g(x) is gotten by solving the equation y = x^105 + x^53 + x^27 + x^13 + x^3 + 3x + 1 for x, giving x = g(y). We then write y = g(x) and give the inverse function as f^(-1)(x) = g(x). However, we cannot solve y = x^105 + x^53 + x^27 + x^13 + x^3 + 3x + 1, for x. Instead we get the inverse by writing x = y^105+ y^53 + y^27 + y^13 + y^3 + 3y + 1 = h(y). If we plot the function x = h(y), we will be plotting the inverse function f^(-1)(x) = y. If we were to plot both f(x) and h(y) (i.e. f^(-1)(x)), we would see that they were reflections of each other about the line y = x. We would also be able to see what intersection points, if any, there were between the two curves. The question asks us to find g’(x), the slope of the inverse function, which happens to be:  h(y) = x = y^105+ y^53 + y^27 + y^13 + y^3 + 3y + 1 Differentiating, dx/dy = 105y^104+ 53y^52 + 27y^26 + 13y^12 + 3y^2 + 3 Therefore, g’(x) = dy/dx = 1/(dx/dy) = 1/(105y^104+ 53y^52 + 27y^26 + 13y^12 + 3y^2 + 3) The solution options given for the slope are numerical values, therefore we must be getting asked for the slope at a particular point, which is some value for x and some value for y. The options for the slope are 3, 1/3 and -1/3. Looking at the expression for g’(x) = 1/(105y^104+ 53y^52 + 27y^26 + 13y^12 + 3y^2 + 3) The only viable option is to choose y = 0, giving g’(x) = 1/3 Answer: g’(x) = 1/3   Here is an image of f(x) and h(y) plotted on the same graph.     The blue line is: f(x) = y = x^105 + x^53 + x^27 + x^13 + x^3 + 3x + 1 The red line is: h(y) = x = y^105 + y^53 + y^27 + y^13 + y^3 + 3y + 1 The white line is: y = x. You can see that both curves are reflections of each other about the line y = x. They have a point of intersection at approx. (-0.5, -0.5) The intersection of h(y) on the y-axis is at about y = -1/3. The intersection of h(y) on the x-axis is at x = 1. These values give an approximate slope, in that area, of 1/3. This would confirm the value of 1/3 found above for g’(x) at y = 0.

### find n if position vectors 3i-2j-k,2k+3j-4k,-i+j+2k,4i+5j+nk line on the same plane

The second point should be 2i+3j-4k? By taking pairs of points we can work out three line vectors. By taking two such lines, we can use the vector cross-product to find the normal orthogonal vector. The third vector line will be orthogonal to the normal and therefore its dot product with the normal vector will be zero. Call the points P1(3,-2,-1), P2(2,3,-4), P3(-1,1,2), P4(4,5,n) and vectors V1=P2-P1, V2=P3-P1, V3=P4-P1. Note that P1 is taken to be the reference point from which the three vector lines radiate. V1=-i+5j-3k, V2=-4i+3j+3k, V3=i+7j+(n+1)k X-product of unit vectors i j k i 0 k -j j -k 0 i k j -i 0 V4=V2 X V3=-28k+4(n+1)j-3k+3(n+1)i+3j-21i=3(n-6)i+(4n+7)j-31k. V1 . V4=0=3(6-n)+5(4n+7)+93 18-3n+20n+35+93=0 17n+146=0, n=-146/17

### Find the point on the curve y=x^2 which is nearest to the point A(0,3)

To find the nearest point on the curve we're looking for the line that is perpendicular to the tangent (normal). First we find the gradient by differentiation: 2x.  The normal has a gradient of -1/(2x) because the product of the slope of the normal and the slope of the tangent (gradient) is -1 and 2x*(-1/(2x))=-1. The gradient of the normal varies according to the position on the curve. The equation of the normal must pass through the point A. So y=mx+b must be satisfied by x=0 and y=3: so b=3 and the equation is y=mx+3, where m is to be determined. This line meets the curve when x^2=mx+3. If we put m=-1/(2x), we get x^2=-1/2+3=5/2 and x=+sqrt(2.5) and y=2.5. Therefore the nearest point is either (sqrt(2.5),2.5) or (-sqrt(2.5),2,5). Since the parabola is symmetrical the points are equidistant from A because A is on the axis of symmetry. The normal at (-sqrt(2.5),2.5) is the reflection of the normal at (sqrt(2.5),2.5). [The equations of the two normals are y=3-x/sqrt(10) (right-hand) and y=3+x/sqrt(10) (left-hand). The distance from A=sqrt(2.75).]

### Find an equation for the line that is tangent to the curve y=3x^3-3x at point (1,0).

y=f(x)=3x³-3x ··· Eq.1 Since the given curve is continuous function, Eq.1 is differetiable at every x and the first derivative f'(x), the slope (tangent) of the original curve, is found at every point (x,y) on the curve. The 1st derivative of Eq.1 is: f'(x)=9x²-3, so the slope of the tangent at x=1 is: f'(1)=9(1)²-3=6 While, the equation of a line that passes thru a point (x1,y1), and has a slope of m is: (y-y1)=m(x-x1) ··· Eq.2  (the point- slope form of linear function) Here, m=f'(x1)=6, x1=1,y1=0   Plug these values into Eq.2.   Eq.2 can be restated and simplified as follows: y-0=6(x-1) ⇒ y=6x-6 Therefore, the tangent to the given curve at point (1,0) is: y=6x-6

### volume of a solid generated by revolving area bounded by the curves about the indicated axis?

(a) The picture shows the given curve and line. The line PQ shows the height of a cylinder where P is a point on the curve. The cylinder is the result of rotating the line PQ about the y-axis as the axis of rotation (x=0). The line and curve intersect when x=6x-x^2, that is, x(5-x)=0 at x=0 and 5. The cylinder is hollow with an infinitesimally thin wall, thickness dx. The radius of the cylinder is x, the height of the cylinder is 5x-x^2 as can be seen by the geometry, and the area of this cylindrical shell is found by "rolling out" the cylinder into a rectangular lamina. So the length of the rectangle is 2πx, the circumference of the cylinder, and the height 5x-x^2, making the area 2πx^2(5-x). The volume of the cylinder is the area multiplied by the thickness dx: 2πx^2(5-x)dx. The sum of the volumes of the cylindrical shells gives us the volume of rotation of the shape bounded by the line and curve. This sum is ∫(2πx^2(5-x)dx) between the limits x=0 to 5. The integral evaluates to 2π(5x^3/3-x^4/4) and, applying the limits: 2π(625/3-625/4)=625π/6=327.25 cu units approx. (b) This time we have a cylinder radius 4-x and height PR=2√(4-x)-4+x. The volume of the cylindrical shell is 2π(4-x)(2√(4-x)-4+x)dx and the integral ∫(2π(4-x)(2√(4-x)-4+x)dx). Since y=4-x we can replace 4-x with y and dx by -dy. The integral becomes -2π∫(y(2√y-y)dy). Since x=4-y and x=(16-y^2)/4 we can write 4-y=(16-y^2)/4 and solve for y. 16-4y=16-y^2 so y(4-y)=0 and y=0 and 4 as evidenced in the picture. The limits are 4≥y≥0. The minus on the integral is changed to plus if we reverse the limits for y. So we need to evaluate the definite integral: 2π∫((2y^(3/2)-y^2)dy)=2π[(4/5)z^(5/2)-z^3/3] for 0≤y≤4. This evaluates to 2π(128/5-64/3)=128π/15=26.81 cu units approx. (c) The intersection point between the parabola and line y=4 is given by 16=8x, so x=2 and y=4: (2,4). The parabola meets the y-axis (x=0) at the origin (0,0). For a point P(x,y) on the parabola 4-y is the radius of a disc of thickness dx and therefore the volume of the disc is π(4-y)^2dx=π(16-8y+y^2)dx. Since y^2=8x we can evaluate the integral π∫((16-16√(2x)+8x)dx) for 0≤x≤2 to find the volume of revolution around the line y=4. This evaluates to 8π[2x-4√2x^(3/2)/3+4x^2] for 0≤x≤2=8π(4-16/3+16)=352π/3=368.61 cu units.

### area between curves Xto the fourth+ysquared=12 and x=y

The line x=y bisects the enclosed area of the other curve x^4+y^2=12. The x coords of the points of intersection are given by x^4+x^2-12=0=(x^2+4)(x^2-3), so x=±sqrt(3). The y coords are given by y=x, so the points are (sqrt(3),sqrt(3)) and (-sqrt(3),-sqrt(3)).  The picture shows how the area is bisected, so all we need to do is to find the area of the whole figure and halve it. But x^4+y^2=12 is symmetrical so to find the area of the whole figure, we just look at a quadrant and find its area, then we double the result to find the area enclosed by the line and the curve. When y=0 we have x intercept: x^4=12, so the area of a quadrant is integral of ydx between x=0 and x=12^(1/4), where y=sqrt(12-x^2) and the integrand is sqrt(12-x^2)dx. Let x=sqrt(12)sin(z); dx=sqrt(12)cos(z)dz, so the integrand becomes: sqrt(12-12sin^2(z))sqrt(12)cos(z)dz=12cos^2(z)dz. cos(2z)=2cos^2(z)-1, so 2cos^2(z)=1+cos(2z) and 12cos^2(z)=6+6cos(2z). But x=sqrt(12)sin(z), so if x=0, z=0 and if x=12^(1/4), sin(z)=12^(1/4)/12^(1/2)=12^(-1/4), z=sin^-1(12^(-1/4)). sin(z)=12^(-1/4); cos(z)=sqrt(1-12^(-1/2)). The result of integration is: 6z+3sin(2z) between the limits 0 and sin^-1(12^(-1/4)). Plugging in values we get the area of a quadrant is 3.4033 + 2.7189 = 6.1222 approx., making half the area of the figure12.2443 approx. It doesn't matter that the line x=y splits the area obliquely, it is still half the area.

### How can you get a probability of |X - 10|>1.8? given X~N (10,4)

Mean=10, SD=4. X-10>1.8, X>11.8 or 10-X>1.8, X<8.2. Convert to Z inequality: |X-10|>1.8, |X-10|/4>0.45, |Z|>0.45 means Z>0.45 or Z<-0.45. Z=(11.8-10)/4=1.8/4=0.45. This Z score corresponds to probability of 0.6736 (67.36%). But this probability is for Z<0.45, i.e., X<11.8. So we have to work out 1-0.6736=0.3264 (32.64%). Z=(8.2-10)/4=-1.8/4=-0.45. This Z score corresponds to probability of 1-0.6736=0.3264 (32.64%). So for probability of X<8.2 OR X>11.8, we add the two probabilities together: 0.6528 (65.28%). The easiest way to visualise this is to draw the normal distribution curve roughly and mark the Z=0 point which divides the curve into two equal halves. To the left of zero mark Z=-0.45 and to the right Z=0.45. The area to the right of Z=0.45 is where X>11.8, and the area to the left of Z=-0.45 is where X<8.2. These areas are the same, so we find one area and double it. P(-Z)=1-P(Z) so if Z=-0.45, we use Z=0.45 and subtract from 1. Double this to get the area (probability) expressed in the inequality.

### what is each part of the equation y=mx+b mean and who do you find them using math vocabulary

The normal meaning for this standard linear equation is that x and y are coordinates in a rectangular arrangement of axes. The y axis is North-South while the x axis is East-West. Where they cross is called the origin with coordinates (0,0), that is, x and y are both zero. The equation y=mx+b defines a straight line. It slopes at a value given by m, the slope or gradient, and m is a number which can be a whole number, a fraction, or whatever, as long as it is constant so that the line remains straight. The slope, m, is also known as the tangent, and the tangent of the angle that the line makes with the x axis has a value of m. When the straight line is at an angle of 45 degrees to the x axis, its tangent is 1 so m=1. If the line slopes backwards at 45 degrees to the x axis, it's tangent is -1 and m=-1. Forward sloping lines have a positive m, while backward sloping lines have a negative m. The value of b is also called the y intercept, because it's the point on the y axis where the straight line crosses that axis. It can have a positive or negative value. b is a constant, just like m. mx is m times x. The x axis is divided by equally spaced numbers, 0, 1, 2, 3 etc to the right, and -1, -2, -3 etc to the left of the origin. The y axis is similarly divided, postitive numbers going up and negative numbers going down. By putting numbers in the equation you can work out where points go on the line. m will have a value, like 2, for example, and b a value, say, 3, so we have y=2x+3. If we put x=0 we get y=3 which is the y intercept. So we mark that point (0,3) by going up 3 divisions on the y axis. Now put x=1, then y=5. So we move to the point (1,5), which is right 1 and up 5. We can join that point to (0,3) and continue beyond these points. What we find is that, although we have only plotted two points, other values of x and y actually fit on the line. If we look at where x=3 and go up to meet the line, then go horizontally back to the y axis, we should find it meets the point 9 on the y axis. So the line represents the relationship between x and y as given by the equation for all points including points in between our whole number divisions, like, for example, 1.5 or one and a half, halfway between 1 and 2.