Mathematics, 23.07.2019 21:00 ethangorrell67

Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 4x2 from (−1, 4) to (0, 0) (a) find a function f such that f = ∇f. f(x, y) = (b) use part (a) to evaluate c ∇f · dr along the given curve c.

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Answer from: ninaaforever

To answer this question, we need to make use of the concept of Potential Function (U)

f = grad U

The solution is:

U = x³/3 + y³/3 + K f = grad×U

b) - 54.41

f( x,y ) = x²×i + y²× j is a vectorial field that could be obtained from the gradient of an scalar, then

f( x, y ) = grad. U if that is true then:

dU/dx = x² ⇒ dU = x²× dx ⇒ Uₓ = x³/3 + g(y)

dU/dy = y² ⇒ according to that

g´(y) = y² ⇒ ∫g´(y) = y³/3 + K plugging that value in Uₓ we get the potential function U, or a family of functions according to constant (K)

U = x³/3 + y³/3 + K

b) To calculate ∫ grad.F * dr

C is the parabola y = 4×x²

we parametrize the curve as:

x = t then y = 4×t² r = ( t , 4×t² ) P ( -1 , 4) to Q ( 0 , 0) for - 1 ≤ t ≤ 0

If x = t dx = dt and dy = 8×t×dt so dr = ( 1 + 8×t ) dt

We choose K = 0 then U = x³/3 + y³/3

then U = ( t³/3 + 64×t⁶ )

∫₋₁⁰ ( t³/3 + 64×t⁶) × ( 1 + 8×t ) dt = ∫ [t³/3 + (8/3)×t⁴ + 64×t⁶ + 512×t⁷ ] ×dt

= t⁴/12 + (8/15)× t⁵ + (64/7) ×t⁷ + 512 /8× t⁸

= t⁴/12 + (8/15)× t⁵ + (64/7) ×t⁷ + 64 ×t⁸ | ₋₁⁰

= - [ 1/12 - 8/15 - 64/7 + 64 ]

= - 54.41

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Answer from: reganwade8298

We're looking for a scalar function f(x,y)

such that

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j=x^2\,\mathbf i+y^2\,\mathbf j=\mathbf f(x,y)$

We have

$\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)$
$\implies\dfrac{\partial f}{\partial y}=\dfrac\partial{\partial y}\left[\dfrac{x^3}3+g(y)\right]$
$\implies y^2=\dfrac{\mathrm dg}{\mathrm dy}$
$\implies g(y)=\dfrac{y^3}+C$

So

$f(x,y)=\dfrac{x^3+y^3}3+C$

By the fundamental theorem of calculus (for line integrals), we then have

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(0,0)-f(-1,4)=0-21=-21$

Answer from: Quest

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).