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Original problem (third problem here)

Plane $T$ nears the surface $S$ $$S: \int_0 ^1 \frac{e^{xzt}}{x-y+z+t} dt = \ln(2)$$ in a point that is on positive $z$ -axis. Assign $T$'s equation.

So I think I need $\nabla S\times T=0$ where $T=\begin{pmatrix}x \\ y \\ z \end{pmatrix}$. So

$$\partial_x S := \partial_x \left( \int_0 ^1 \frac{e^{xzt}}{x-y+z+t} dt \right)=?$$

Now should I integrate this before differentiation or was there some rule to make this differentation-integration simpler?

Trial 4 By Leibniz rule kindly suggested by the engineer and the formula by Petersen,

$$\begin{align}\partial_x \int_0^1 f(x,t) dt \Bigg\vert_{x=0} &= \int_0^1 \partial_x f(x,t) \Big\vert_{x=0} dt \\ &=z(y-z)\log\left(\frac{1-y+z}{-y+z}\right)+z+\frac{1}{-y+z+1}-\frac{1}{-y+z} \\ &:=B \end{align}$$

Details here. Now after this, the same for $\partial_y S:=C$ and $\partial_z S :=D$ so $\nabla S = \begin{pmatrix} B \\ C \\ D\end{pmatrix}$.

But to the question, is $\nabla S = \begin{pmatrix} B \\ ... \\ ...\end{pmatrix}$ right? Or $\nabla S_{|\bar{x}=(0,0,1)} = \begin{pmatrix} -\log(2)+0.5 \\ ... \\ ...\end{pmatrix}$? Look after this I need to do $\nabla S\times T=0$ and I want to minimize the amount of terms early on because it is easy to do mistakes with long monotonous calculations.

Old trials in the chronological order

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Typo in "Leipniz". Leibniz. –  Américo Tavares Mar 26 '12 at 9:07

2 Answers 2

up vote 2 down vote accepted

Gradients of scalar functions are normal aka perpendicular to level surfaces. Planes in $\mathbb{R}^3$ are given by equations of the form $\mathbf{n}\cdot(\mathbf{x}-\mathbf{p})=0$, where $\mathbf{n}$ is a normal vector and $\mathbf{p}$ is a point on the plane.

Thus the plane you're looking for is given by

$$F(x,y,z)=\int_0^1 \frac{\exp(xzt)}{x-y+z+t}dt, \quad z_0: F(0,0,z_0)=\log2, \quad \nabla F(0,0,z_0)\cdot\big(\vec{x}-(0,0,z_0)\big)$$

The components of $\nabla F$ should all evaluate to concrete quantities. For example,

$$\int_0^1 \frac{\partial}{\partial x}\frac{\exp(xzt)}{x-y+z+t}dt=\int_0^1\left(zt-\frac{1}{x-y+z+t}\right)\frac{\exp(xzt)}{x-y+z+t}dt $$

which, after evaluation, becomes

$$\int_0^1\frac{z_0t}{z_0+t}-\frac{1}{(z_0+t)^2}dt =z_0(1-z_0\log2)-\left(\frac{1}{z_0+1}-\frac{1}{z_0}\right).$$

Also note that $z_0$ solves explicitly to $1/(e^2-1)$.

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Wolfram Alpha indicates that this integral can not be expressed by elementary functions. This is a big wrong way sign saying that you should try to solve the problem in some other way.

In your problem, you are given that the plane $T$ is tangent to the surface $S$ at a point on the $z$-axis, i.e. a point for which $x=y=0$. Start by finding this point, by solving $$ \int_0^1 \frac 1{z+t} dt = \ln 2.$$ Then find the partial derivatives at this point. Use that $$ \partial_x \int_0^1 f(x,t) dt \Bigg\vert_{x=0} = \int_0^1 \partial_x f(x,t) \Big\vert_{x=0} dt.$$

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Moved the nabla point here, trying to keep this q focused on the orig. –  hhh Mar 26 '12 at 12:09
    
Is this right here for the below equation? –  hhh Mar 26 '12 at 12:33

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