Calculus 3: Equation of a plane and partial derivative chain rule

[lbros55]

So I need to find the equation of a plane that has the point (-1,4,2) and the line l=(1,1,0)+t(0,2,1)

I found the vector between the two points which is (-2,3,2) and then cross multiplied with (0,2,1). I ended up getting (1,-2,4)(x+1,y-4, z-2)=0. I this correct?


For another hw problem, I needed to find dz/du when given : z=e^(x^(2)y), x=sqrt(uv), and y=1/v

After doing it out, I got 2xye^(x^(2)y)v/(2sqrt(uv)). I this also correct?

 

[j-astron]

Not touching the first question...

For another hw problem, I needed to find dz/du when given : z=e^(x^(2)y), x=sqrt(uv), and y=1/v

After doing it out, I got 2xye^(x^(2)y)v/(2sqrt(uv)). I this also correct?

Am I right in thinking that the functions you've been given are:

\(\displaystyle \displaystyle z = e^{x^2 y}\)
\(\displaystyle \displaystyle x = \sqrt{uv}\)
\(\displaystyle \displaystyle y = \frac{1}{v}\)

?

If so, substituting the expressions for x and y in terms of u and v into the equation causes it to simplify to something quite basic...

 

[lbros55]

it would just be e^(u) right? I need to show it using partial derivatives, so I may just be having a hard time doing the simplification.

 

[Steven G]

So I need to find the equation of a plane that has the point (-1,4,2) and the line l=(1,1,0)+t(0,2,1)

I found the vector between the two points which is (-2,3,2) and then cross multiplied with (0,2,1). I ended up getting (1,-2,4)(x+1,y-4, z-2)=0. I this correct?


For another hw problem, I needed to find dz/du when given : z=e^(x^(2)y), x=sqrt(uv), and y=1/v
After doing it out, I got 2xye^(x^(2)y)v/(2sqrt(uv)). I this also correct?

The equation of the plane will be in the form a(x+1) + b(y-4) + c(z-2)=0.

One direction in the plane is u =<0, 2, 1>

If t=0, then (1, 1, 0) is a point in the plane and so is v = <1, 1, 0> -<-1, 4, 2> = <2, -3, -2>.

Now find uxv which will give you a, b and c.

 

[Steven G]

For another hw problem, I needed to find dz/du when given : z=e^(x^(2)y), x=sqrt(uv), and y=1/v

After doing it out, I got 2xye^(x^(2)y)v/(2sqrt(uv)). I this also correct?

After doing it out, I got 2xye^(x^(2)y)v/(2sqrt(uv). But this is not a solution as it does not say what it is a solution to! Is it suppose to be a solution to dz/du? If so, then why not say that.

This could not be any easier to compute. If x=sqrt(uv), and y=1/v, then what does x²y equal in terms of u and v. Just square x and multiply it by y and you'll have x²y in terms of u and v. Then z = e ^{the answer you just found}

Continue from here

 

[lbros55]

After doing it out, I got 2xye^(x^(2)y)v/(2sqrt(uv). But this is not a solution as it does not say what it is a solution to! Is it suppose to be a solution to dz/du? If so, then why not say that.

This could not be any easier to compute. If x=sqrt(uv), and y=1/v, then what does x²y equal in terms of u and v. Just square x and multiply it by y and you'll have x²y in terms of u and v. Then z = e ^{the answer you just found}

Continue from here

Solved, thank you!