Related rates word problems

[Guest]

Find the equation of the tangent to the graph y^3-3xy-5=0 at the point (2,-1).

y^3-3xy-5=0
3y^2dy/dx-3y+xdy/dx=0
dy/dx=3y/3y^2+x
m=-3/5

y=mx+b
-1=-3/5(2)+b
b=1/5

y=mx+b
y=-3/5x+1/5
5y=-3x+1

ans is SUPPOSE TO BE x-y-3=0 WHAT DID I DO WRONG?!?!?
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The postition of an object moving along a straight line is described by the function
s(t)=3t^2-10 for t greater than or equal to 0. Is the object moving away from or towards its starting position when t=3?

what do u have to find here? the acceleration or velocity? I tend to get confused between the two, or do u jsut plug in 3 right away..

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Sand is being poured onto a conical pile at the rate of 9m^3/h. Friction forces in the sand are such that the slope of the sides of the conical pile is always 2/3.

a) How fast is the altitude increasing when the radius of the base of the pile is 6m?

dv/dt= 9m^3
h=10m
dh/dt=?
Whats the 2/3? I dont really understand that part..its the side..but why do we need it..

Thanks for the help

 

[skeeter]

Find the equation of the tangent to the graph y^3-3xy-5=0 at the point (2,-1).

y^3-3xy-5=0
3y^2dy/dx-3y+xdy/dx=0 should be -3x(dy/dx)
dy/dx=3y/3y^2+x
m=-3/5

y=mx+b
-1=-3/5(2)+b
b=1/5

y=mx+b
y=-3/5x+1/5
5y=-3x+1

ans is SUPPOSE TO BE x-y-3=0 WHAT DID I DO WRONG?!?!?
-------------------------------------------------------------------------------
The postition of an object moving along a straight line is described by the function
s(t)=3t^2-10 for t greater than or equal to 0. Is the object moving away from or towards its starting position when t=3?
look at the sign of s'(3) = v(3) ... sign of velocity gives one the direction of motion

what do u have to find here? the acceleration or velocity? I tend to get confused between the two, or do u jsut plug in 3 right away..

-----------------------------------------------------------------
Sand is being poured onto a conical pile at the rate of 9m^3/h. Friction forces in the sand are such that the slope of the sides of the conical pile is always 2/3.

a) How fast is the altitude increasing when the radius of the base of the pile is 6m?

dv/dt= 9m^3
h=10m
dh/dt=?
Whats the 2/3? I dont really understand that part..its the side..but why do we need it..
volume of a cone is
V = (pi/3)r^2*h
h/r = 2/3 is the slope of the cone's side ... 2r = 3h, or r = (3/2)h
V = (pi/3)(3h/2)^2*h
V = (3pi/4)h^3
now find the derivative of the above equation w/r to time ... substitute your known values (you'll have to solve for h, they gave you r) and solve for dh/dt, since that's what they're asking you to find

Thanks for the help

 

[soroban]

Hello, bittersweet!

Find the equation of the tangent to the graph \(\displaystyle y^3\,-\,3xy\,-\,5\:=\:0\) at the point (2,-1).

\(\displaystyle y^3\,-\,3xy\,-\,5\:=\:0\)

\(\displaystyle 3y^2\left(\frac{dy}{dx}\right)\,-\,3y\,+\,x\left(\frac{dy}{dx}\right)\:=\:0\;\;\) ← here!

It should be: \(\displaystyle \:3y^2\left(\frac{dy}{dx}\right)\,-\,3y\,-\,3x\left(\frac{dy}{dx}\right)\;=\;0\)

The postition of an object moving along a straight line is described by the function: \(\displaystyle \,s(t)\:=\:3t^2\,-\,10\) for \(\displaystyle t\,\geq\,0\)
Is the object moving away from or towards its starting position when t=3?

The velocity gives us both the speed and the direction of the object.

We have: \(\displaystyle \,v(t)\;=\;s'(t)\;=\;6t\)

The starting position is: \(\displaystyle \,s(0)\:=\:3\cdot0^2\,-\,10\:=\:-10\;\) ... left of the origin.
When \(\displaystyle t\,=\,3\), its position is: \(\displaystyle \,s(3)\:=\:3\cdot3^2\,-\,10\:=\:+17\;\) ... right of the origin.

When \(\displaystyle t\,=\,3\) we have: \(\displaystyle \,v(3)\,=\,6(3)\,=\,+18\)
It is moving to the right ... away from the starting position.

Sand is being poured onto a conical pile at the rate of 9m^3/h.
Friction forces in the sand are such that the slope of the sides of the conical pile is always 2/3.

a) How fast is the altitude increasing when the radius of the base of the pile is 6m?

dv/dt = 9 m^3/h
h = 10 m ?
dh/dt = ?

                *
              / | \
            /   |   \
          /     |2k   \
        /       |       \
      * - - - - + - - - - *
                    3k

"The slope of the sides is 2/3" means "rise over run" = 2/3.
So the ratio of the height to the radius is 2/3.
If the height is \(\displaystyle 2k\), then the radius is \(\displaystyle 3k\).

We have: \(\displaystyle \,h\,=\,2k,\;\;r\,=\,3k\;\;\Rightarrow\;\;r\,=\,\frac{3}{2}h\;\) [1]


Volume formula: \(\displaystyle \L\:V\:=\:\frac{1}{3}\pi r^2h\)

Substitute [1]: \(\displaystyle \L\:V\:=\:\frac{1}{3}\pi\left(\frac{3}{2}h\right)^2h \:=\:\frac{3}{4}\pi h^3\)

Differentiate with respect to time: \(\displaystyle \L\:\frac{dV}{dt}\:=\:\frac{9}{4}\pi h^2\left(\frac{dh}{dt}\right)\)


We are given: \(\displaystyle \,\frac{dV}{dt}\,=\,9,\;\;r\,=\,6\;\;\Rightarrow\;\;h\,=\,4\;\;\) from [1]

So we have: \(\displaystyle \L\:9\;=\;\frac{9}{4}\pi(4^2)\left(\frac{dh}{dt}\right)\)

Therefore: \(\displaystyle \L\:\frac{dh}{dt}\,=\,\frac{1}{4\pi}\) m/h