based on yesterdays help

[rachael724]

I was on yesterday with 2 similiar problems. Based on the way I was shown how to do these problems, would these be correct?

y sq. rt of x + 1 = 12; dx/dt = 8, x = 15, y = 3

Answer: This is = 3/4


and


x^2 ln y = -2 + xe^y; dx/dt = 10, x = 3, y = 1

Answer: 10e/3(3-e)

 

[jolly]

Show how you arrived at those answers.

 

[stapel]

I was on yesterday with 2 similiar problems.

In the future, please post replies within the original threads. Thank you.

y sq. rt of x + 1 = 12; dx/dt = 8, x = 15, y = 3

Answer: This is = 3/4

You have post the following as the equation:

. . . . .\(\displaystyle \large{y\,\sqrt{x}\,+\,1\,= \,12}\)

Is this what you meant? If so, then I'm getting a different answer.

x^2 ln y = -2 + xe^y; dx/dt = 10, x = 3, y = 1

Answer: 10e/3(3-e)

You have posted the following as your solution:

. . . . .\(\displaystyle \large{\frac{10e}{3}\,(3\,-\,e)}\)

Is this what you meant? If so, I'm getting a different answer.

FYI for tutors: This was originally posted here

As was requested yesterday, when you reply, please show all your work and reasoning. Thank you.

Eliz.

 

[Gene]

You are VERY close on #1
--------------
Gene

PS. There is a tutorial disagreement on whether this belongs with the old post or not. I think you done good.
New problem, new post.