Find the rate at which the depth of the wine in the glass is increasing when the depth is 4cm.

A wine glass is being filled with wine at a rate of 8 cm^3/s. The volume, V cm^3, of wine in the glass when the depth of the wine in the glass is x cm is given by V = 4x^3/2.

I dont understand how to form the related rate equation, for instance my dx/dt equation gave me 4/(3x^1/2) when i did my working out. This is obviously wrong because dx/dt should have t as the variable, if i am not mistaken? I am really confused!!!

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We are given that the volume of the wine glass is `V=4x^(3/2)` ; the rate that the glass is being filled is 8`"cm"^3/s"sec"` .

We wish to find the rate of the change in depth when the depth is 4cm.

`(dV)/(dt)=(3/2)(4)(x^(1/2))(dx)/(dt)=6x^(1/2) (dx)/(dt)` Use the chain rule

We know `(dV)/(dt)=8,x=4` so

`8=6(4)^(1/2)(dx)/(dt)`

`(dx)/(dt)=8/12`

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**The depth is changing at a rate of `2/3 "cm"/"sec"` when the depth is 4cm**

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Since the volume of the wine class is given by `V=4x^{3/2}` , which means when we differentiate it with respect to time we get:

`{dV}/{dt}=4 3/2 x^{1/2} {dx}/{dt}` simplify

`=6x^{1/2} {dx}/{dt}`

In your question, you have forgotton to multiply the derivative of V by the derivative of x with respect to time. This is why we use Leibniz notation.

To find the rate at which the glass is filling, we now substitute the value of `x=4` for the depth, and `{dV}/{dt}=8` , which means we can solve for the rate of the depth changing:

`{dx}/{dt}={{dV}/{dt}}/{6x^{1/2}}` sub values

`=8/{6 sqrt4}`

`=8/12`

`=2/3`

**The rate at which is depth is increasing is 2/3 cm/s.**

We are given that the volume of the wine glass is ; the rate that the glass is being filled is 8 .

We wish to find the rate of the change in depth when the depth is 4cm.

Use the chain rule

We know so

The depth is changing at a rate of when the depth is 4cm

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