Differentiation (My answer is wrong)?

Question

A viscous liquid is poured onto a flat surface. It forms a circular patch which grows at a steady rate of 8 cm^2/s. Find, in terms of pi,

(a) the radius of the patch 25 s after pouring has commenced;
(b) the rate at which the radius is increasing at this instant.

For (a), I get the radius as 10(sqrt(2/pi)) cm (which is the correct answer.)

But then, for (b), the answer is dr/dt = (sqrt(2pi)) / 5pi. I don't get that...

This is what I get:

dr / dt = ( 1 / 2 π r ) * (8 cm² / s ) = 4 / π r

If I substitute the r into 4 / π r, I got 40(sqrt2π) / (π^2 * r), which is not the correct answer...

Help me out? Thanks.

Answer 1

Givens:

1. A = pi r^2

2. dA/dr = 2pir

3. dA/dt = dA/dr(dr/dt) = 8 cm^2/s

4. at t=0, r=0



from 2. and 3. we have that:

8 cm^2/s = 2pir(dr/dt)

so that

dr/dt = 4/pir = (4/pi)r^-1cm/s (when r is measured in cm's)



(a) You can integrate this to get r(t) and use 4. to determine the constant of integration, but it's easier to use 3.



At t = 25 the area is 8(25) = 200 cm^2 from 1. we have

200 = pi r^2

r = sqrt(200/pi) = 10(sqrt(2/pi))



(b) Since dr/dt = 4/pir then dr/dt when r = 10(sqrt(2/pi)) is



4/pi(10(sqrt(2/pi))) = 2/5pi(sqrt(2/pi)) (sqrt(2/pi)/(sqrt(2/pi)

= 2sqrt(2/pi)/(5pi)(2/pi) = 2sqrt(2/pi)/10 = sqrt(2/pi)/5



(I think you put r in the numerator of 4/pir when substituting rather than the denominator)



That's your solution.

Hope this makes sense.

Answer 2

a) This one is easy. The area is just 8cm^2/s * 25s = 200cm^2. You are right the radius at that time is 10sqrt(2/pi) using algebra.



b) This one is more difficult. dA/dt = dA/dr*dr/dt = 8 = (2pi(r))(dr/dt). Just plug in the correct value for 'r' and use algebra to find dr/dt. This represents the instantaneous rate of change of r at that time.



It sounds like you may have made an algebraic error for part b.

Answer 3

A=pi*r^2

dA/dt=2*pi*r*dr/dt

dA/dt=8

r=10(sqrt(2/pi))

8/(2*pi*10sqrt(2/pi))=dr/dt

dr/dt=2/(5*pi*sqrt(2/pi))

Answer 4

i'm in actuality follwing what you probably did, I copied it, and then point out the place you went incorrect. i'm getting a distinctive answer than the single you assert is right nevertheless. y = (2ax^2 + bx) / (bx^3 - cx) dy/dx = [(bx^3 - cx )* (4ax + b) - (2ax^2 + bx) * (3bx^2 - c)] / (bx^3 - cx)^2 (2ax^2 + bx) * (3bx^2 - c)] = 6abx^4 - 2acx^2 + 3b^2x^3 - bcx dy/dx = [4bax^4 + b^2x^3 - 4cax^2 - bcx - 6abx^4 + 2acx^2 - 3b^2x^3 + bcx] / (bx^3 - cx)^2 in the final equation you have the signs and symptoms on those 2 words incorrect: + 3b^2x^3 - bcx Making this correction you get: dy/dx = [-2abx^4 - 2b^2x^3 - 2acx^2 ] / (bx^3 - cx)^2 element out an x^2 on the precise and the minus sign: -x^2[2abx^2 + 2b^2x + 2ac] / (bx^3 - cx)^2 element an x out of the backside and get rid of the x^2: -[2abx^2 + 2b^2x + 2ac] / (bx^2 - c)^2 Are you beneficial you typed in the splendid answer. i'm getting it with an x factored out of the denominator. easily you will discover this by utilizing taking an x out of the unique fact: y = (2ax + b) / (bx^2 - c) that's greater handy to tell apart: [(bx^2 - c)(2a) - (2ax + b)(2bx)] / [bx^2 - c]^2 [2abx^2 - 2ac - 4abx^2 - 2b^2x] / [bx^2 - c]^2 [-2abx^2 - 2b^2x - 2ac] / [bx^2 - c]^2 -[2abx^2 + 2b^2x + 2ac] / (bx^2 - c)^2

Answer 5

A = π r^2 = 8t

r = [8t/π]^(1/2) = 2[2/π]^(1/2)t^(1/2)

dr/dt = [2/π]^(1/2)t^(-1/2) = [2/π]^(1/2)(25)^(-1/2) = [2π]^(1/2)/(5π)