Seperable Differentiation. Help, please. =\

In summary, we discussed the velocity of a skydiver after her parachute opens, which can be expressed using the differential equation dv/dt = -2v-32. By using separation of variables and integrating with corresponding limits, we obtained an expression for v in terms of t, which is v(t) = -34e^-2t - 16. This can also be written as -2t = ln((-v-16)/34), which can then be solved for v(t).
  • #1
mikiritenshi
5
0
I blanked out. Any help is greatly appreciated.

Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>0. After her parachute opens, her velocity satisfies the differential equation: dv/dt= -2v-32, with initial condition v(0)=-50.

(a) Use separation of variables to find an expression for v in terms of t, where t is measured in seconds.

I got up to this far:

integral [dv/(-2v-32)]= integral (dt)
u= -2v-32; du= -2dv.

Help. :cry:
 
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  • #2
Good.May i suggest integrating with corresponding limits...?

[tex] \int_{-50}^{v} \frac{dv}{-2v-32} = \int_{0}^{t} \ dt [/tex]

The LHS integral is doable.With a substitution that u already did...So carry on.Pay attention to the way the limits are transformed.


Daniel.
 
  • #3
Using the substitution that I did, I obtained:

(-1/2) integral(1/u)du = integral (dt)

= (-1/2) ln|u| + C = t+ C

I'm not sure on what to do next. =\
 
  • #4
substitue the original variables back for the u substitutions. then use the initial conditon to find out what C is.
 
  • #5
I know you plug in 0 for t, but what'll happen to the -50?
And do I have to exponentiate both sides?
 
  • #6
From the integration i get

[tex] v(t)=-34e^{-2t}-16 [/tex]


Daniel.
 
  • #7
That is the answer, but I don't know how to get it.
 
  • #8
The RHS is 't',okay??

Now,the LHS can be written

[tex]-\frac{1}{2} \int _{-50}^{v} \frac{d(-v-16)}{-v-16} = -\frac{1}{2}\left[\ln (-v-16)\right]\left |_{-50}^{v}\right = -\frac{1}{2} \ln \left(\frac{-v-16}{34}\right) [/tex]

Therefore

[tex] -2t=\ln\left(\frac{-v-16}{34}\right)\Rightarrow v(t)=-34e^{-2t}-16 [/tex]

Daniel.
 
Last edited:
  • #9
Wow, thanks.
 

1. What is separable differentiation?

Separable differentiation is a method used to find the derivatives of functions that can be written in the form of the product of two simpler functions.

2. How is separable differentiation different from other methods of differentiation?

Unlike other methods of differentiation, separable differentiation is specifically used for functions that can be broken down into simpler functions. This method involves finding the derivatives of each individual function and then combining them to find the overall derivative.

3. What types of functions can be differentiated using the separable differentiation method?

Functions that can be expressed as a product of simpler functions, such as exponential, logarithmic, and trigonometric functions, can be differentiated using the separable differentiation method.

4. What are the steps involved in separable differentiation?

The steps involved in separable differentiation are as follows: 1) Write the given function in the form of the product of two simpler functions, 2) Find the derivatives of each individual function, 3) Combine the derivatives using the product rule, and 4) Simplify the resulting expression to get the final derivative.

5. Why is separable differentiation useful in scientific research?

Separable differentiation is a useful tool in scientific research as it allows for the quick and efficient calculation of derivatives for functions that can be expressed as a product of simpler functions. This can be particularly helpful in fields such as physics, engineering, and economics where these types of functions are commonly used.

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