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zimovet [89]
3 years ago
10

Which two transformations must be applied to the graph of y = ln(x) to result in the graph of y = –ln(x) + 64

Mathematics
1 answer:
Neporo4naja [7]3 years ago
6 0

Answer:

A) reflection over the x-axis, plus a vertical translation

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Consider F and C below. F(x, y, z) = 2xz + y2 i + 2xy j + x2 + 9z2 k C: x = t2, y = t + 3, z = 3t − 1, 0 ≤ t ≤ 1 (a) Find a func
Ray Of Light [21]

Answer:

a)∇f = 2y + 2x + 18z

b) \int\limits^._C {F} \, dr =108

Step-by-step explanation:

Given:

f (x,y,z ) = (2xz+ y^{2})i + (2xy) j +(x^{2} + 9z^{2})k

The curve C :

x=t^{2} ,\\y= t+3\\z= 3t-1

where 0 ≤ t ≤ 1

Required:

(a) F = ∇f =? (F is a vector here)

(b) \int\limits^._C {F} \, dr =?

Solution

First we will find the directional derivative F = ∇f

for that , we will use the formula :

∇f = F_{x}i+ F_{y} j+F_{z}k

Fx= δf/δx = δ/δx (2xz+ y^{2})i = 2z i

Fy= δf/δy=δ/δy (2xy)j = 2x j

Fz= δf/δz=δ/δz(x^{2} + 9z^{2})k = 18z k

∇f = (2z) i .i + (2x) j.j + (18z) k.k

∇f = 2z + 2x + 18z

<em>For part b):</em>

<em>we will use line integral formula:</em>

\int\limits^._C {F} \, dr

to calculate dr, we will need the curve C:

r = x(t)+y(t)+z(t)

r=(t^{2})i + (t+3) j +(3t-1) k

\frac{dr}{dt}=\frac{dx}{dt} +\frac{dy}{dt} + \frac{dz}{dt}

\frac{dx}{dt} = 2t

\frac{dy}{dt} = 1

\frac{dz}{dt} = 3

\int\limits^._C {F} \, dr = \int\limits^1_0 {F_{x} } \, dx+ F_{y} dy +F_{z} dz

= \int\limits^1_0 {(2z(2t) + 2x(1) + 18z (3)} \, )

put values of y, x and z

= \int\limits^1_0 {2(3t-1) + 2(t^{2}) +18 (3) (3t-1)} \,

={2t^{2} + 6t+162t -54-2}\, |^1_0

={ 2t^{2}+ 168t - 56} \,|^1_0               (Note : f(1)-f(0))

=2(1)+162(1)+2(0)+162(0)-56

= 2+162 -56

\int\limits^._C {F} \, dr =108

3 0
3 years ago
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