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Mathematics 12 Online
OpenStudy (anonymous):

Suppose that x=x(t) and y=y(t) are both functions of t. If y^2+xy−3x=−5, and dy/dt=−5 when x=3 and y=−4, what is dx/dt?

OpenStudy (anonymous):

@jim_thompson5910 could you help guide me through this

jimthompson5910 (jim_thompson5910):

y^2+xy−3x=−5 d/dt[ y^2+xy−3x ] = d/dt[ −5 ] 2y*dy/dt + y*dx/dt + x*dy/dt - 3x*dx/dt = 0 does that help?

OpenStudy (anonymous):

ok il see what i can do

OpenStudy (anonymous):

so i just need to derive everything, and then plug in the numbers given?

jimthompson5910 (jim_thompson5910):

exactly

jimthompson5910 (jim_thompson5910):

the derivation is already done, you just need to plug in everything and solve for dx/dt

OpenStudy (anonymous):

ok but for my other problems, also how do you know if it should be a dy/dt or a dx/dt?

jimthompson5910 (jim_thompson5910):

well in this case, they give you the value of dy/dt and they want you to solve for dx/dt

jimthompson5910 (jim_thompson5910):

so it's based on the instructions

OpenStudy (anonymous):

ok so the answer is 25/7

OpenStudy (anonymous):

this next one im going to try to figure out on my own

jimthompson5910 (jim_thompson5910):

no it's not 25/7

OpenStudy (anonymous):

i just plugged it in and it worked

jimthompson5910 (jim_thompson5910):

it's saying 25/7 is the correct answer?

jimthompson5910 (jim_thompson5910):

hmm i must be missing a sign somewhere

OpenStudy (anonymous):

ya

OpenStudy (anonymous):

Suppose that x=x(t) and y=y(t) are both functions of t. If y2+x=27, and dydt=−4 when x=2 and y=5, what is dxdt?

OpenStudy (anonymous):

so we have 2y(dy/dt)+(dx/dy)=0 correct?

jimthompson5910 (jim_thompson5910):

yes so far so good

OpenStudy (anonymous):

dx/dt=40

OpenStudy (anonymous):

Let A be the area of a circle with radius r. If drdt=4, find dAdt when r=1 so A=pi(r)^2

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

dA/dt=pi(2rdr/dt)

OpenStudy (anonymous):

dA/dt=pi*2*1*4

OpenStudy (anonymous):

Find the rate of increase (with respect to r) in the surface area (S=4πr2) of a spherical balloon when: (A) r=2 inches → Rate of increase = (B) r=4 inches → Rate of increase = (C) r=6 inches → Rate of increase =

OpenStudy (anonymous):

@jim_thompson5910 sorry to keep bothering you, but no idea how to do this one

jimthompson5910 (jim_thompson5910):

S = 4pi*r^2 dS/dt = 4pi*2*r*dr/dt

jimthompson5910 (jim_thompson5910):

you would plug in the radius (for each part), but unfortunately dr/dt is missing...hmm

jimthompson5910 (jim_thompson5910):

that info is probably buried somewhere (else)? idk

OpenStudy (anonymous):

ok thank you

jimthompson5910 (jim_thompson5910):

np

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