Suppose you need to know an equation of the tangent plane to a surface S at the point P(4, 1, 3). You don't have an equation for S but you know that the curves r₁(t) = 4 + 3t, 1 − t², 3 − 5t + t²r₂(u) = 3 + u², 2u³ − 1, 2u + 1 both lie on S. Find an equation of the tangent plane at P.

Answers

Answer 1

Answer:

Answer:

30(x-4)-16(y-1)+18(z-3)

30x-16y+18z=158

Step-by-step explanation:

In order to find the tangent plan equation at point P,you know that r₁(t) and r₂(u) lie on surface S, Find the vectors B₁(t)= $(d)/(dt) \n$ r₁(t) and B₂(u)= $(d)/(du)$ r₂(u)

B₁(t)= $(d)/(dt)$ (4+3t, 1- $t^(2)$ ,3-5t+ $t^(2)$

B₁(t)=(3, -2t, -5+2t)

B₂(u)= $(d)/(du)$ (3+ $u^(2)$ , 2 $u^(3)$ -1, 2u+1)

B₂(u)=(2u, 6 $u^(2)$ , 2)

Put t=0 in r₁(t), we will get:

r₁(t)=(4, 1, 3), it means it is on point P

Put u=1 in r₂(u), we will get:

r₂(u)=(4, 1, 3), it means it is on point P

Put t=0 in B₁(t), we will get:

B₁(t)=(3,0,-5)

Put u=1 in B₂(u), we will get:

B₂(u)=(2,6,2)

So Plane Contains two vectors B₁(t) and B₂(u), For Normal Vectors to the plane Cross Product is:

B₁(t)xB₂(u)= (3,0,-5)x(2,6,2) [Cross Product]

B₁(t)xB₂(u)=(30,-16,18)

Equation will become:

30(x-4)-16(y-1)+18(z-3)

30x-16y+18z=158

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What is the quotient (125 – 8x3) ÷ (25 + 10x + 4x2)?–2x + 5
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Answers

For this case we have the following expression:

$(125-8x^3)/(25+10x+4x^2)$

Factoring the numerator we have:

$(-(2x+5)(4x^2+10x+25))/(25+10x+4x^2)$

Rewriting the denominator we have:

$(-(2x+5)(4x^2+10x+25))/(4x^2+10x+25)$

Canceling similar terms we have:

$-(2x+5)$

$-2x-5$

Answer:

The quotient of the division is given by:

$-2x-5$

If you would like to solve (125 - 8x^3) / (25 + 10x + 4x^2), you can do this using the following steps:

(125 - 8x^3) / (25 + 10x + 4x^2) = - ((2x - 5) * (4x^2 + 10x + 25)) / (4x^2 + 10x + 25) = - (2x - 5) = - 2x + 5

The correct result would be - 2x + 5.

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Answers

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