E = 2A + sqrt(2)*C
F = 2B + sqrt(2)*C

Well you would get area J is approx. area H for rectangles which look like a square. Not so much though if one side was much smaller than the other.

This is because, if you draw lines from the bottom-right point of your rectangle to the top-right and bottom-left vertices of the triangle you get isosceles triangles and the area is approximate but not exact because of the little bit left over in both J and H.

In your 2nd diagram above the difference between J and H would be very small (~2 or 3 %)

Hmm ok.

So are you saying

E = 2A + sqrt(2)*C
F = 2B + sqrt(2)*C

does in fact produce the shortest D or are you saying there is a better one? We are kind of splitting hairs at this point because I think the E & F we are getting now are going to produce a good value of D regardless (one that is close enough to the true solution). I would be interested still to know if this is the exact answer and how to solve such a problem, just for my own interest.

It is an ok approximation so long as C is short. An even better approximation should be:

E = 2A + C
F = 2B + C

Well... somewhere in between 2A + (1/sqrt(2))C and C. So maybe 2A + 0.85C is the best approximation

I am assuming that C is short and that the rectangle is approx. a square. (2:1 max)

Hi, guys.

I'm having trouble with a problem and am hoping someone on here can help me out.

I need to find the derivative of the inverse function at the point a= -2 assuming x > 0

The function given is f(x)= e^3x + 2e^x - 5


I tried solving for the inverse and failed.
I also tried to solve for x when f(x) = -2 and failed.
Any help is appreciated.

thanks

tilted, are you sure you have the equation correct? What level of math are you in?

I started out making the substitution z=e^x and had

f(z)=z^3 +2z -5, but that doesn't factor.

let u = e^x. You get a cubic in u. Good luck from there?

I'm taking calc II and the equation is written correctly

I get u^3 + 2u - 5, but don't really know what to do from there.

Neither do I. You can solve it using numerical methods but I can't think of anything else.

Let g be the inverse function. We know
g(f(x)) = x.
Differentiating both sides and using the chain rule gives
g'(f(x)) * f'(x) = 1.
Solving for g' implies
g'(f(x)) = 1 / f'(x).
So while this doesn't tell us the analytic form of g'(a), it does let us compute it for specific values of a.

I'm a little unclear on what, exactly, you're looking for. I think you want to find g'(a) for a = -2. To do this, we need to find the corresponding value of x. We have
-2 = e^{3x} + 2 e^x - 5.
Letting z = e^x to clean things up, as some have suggested, gives
z^3 + 2 z - 3 = 0
This factors as
(z-1)(z^2 + z + 3) = 0.
The only real solution is z = 1, which corresponds to x = 0. So I don't know what you mean when you say "a = -2 assuming x > 0", because if I've interpreted your statement correctly, such a point doesn't exist.

So let's assume we want the derivative of the inverse at x = 0. Then we have
g'(f(x)) = 1 / f'(x) = 1 / (3 e^{3x} + 2 e^x)
Plugging in for x = 0 gives
g'(-2) = 1/5.

Jewbinson's method is correct. You have incorrectly calculated m. It should be about .8, which gives E as roughly 67, F as roughly 54 and D as roughly 86 by my calcs.

I'm going to try to describe this without drawing a picture, which may be difficult.

I rotated the triangle so that the right angle is at the origin, and the legs are the x- and y-axes. Stripped to its essentials, we have a point P at (x,y) (in your diagram, this is the endpoint of the line that separates H and J). We want to find the shortest line segment passing through P with ends at the x- and y-axes.

Let's suppose this line segment makes an angle phi with the x-axis (angle ED, in your picture). The length of the line segment is then
y/sin(phi) + x/cos(phi).
We want to find the value of phi that minimizes this, so we differentiate w.r.t. phi and set equal to zero. This gives
tan^3(phi) = y/x.
This line intersects the x-axis at x-coordinate
E = x + y/tan(phi) = x^{1/3} (x^{2/3} + y^{2/3})
from the origin, and it intersects the y-axis at y coordinate
F = y + x*tan(phi) = y^{1/3} (x^{2/3} + y^{2/3}).
Note that the total length of the line segment in question is
(E^2 + F^2)^{1/2} = (x^{2/3} + y^{2/3})^{3/2}.
Now just take
x = A + C / sqrt(2)
y = B + C / sqrt(2)
and you're done.

Duh!!! Happy to know a clear answer to this problem, sucks that it isn't really simply though.

Edit:

That is nice! Same answer as above too. Thanks everyone.

Well he simply wants the inverse function for the function he has given.

The inverse of e^x is ln(x) because e^ln(x) = ln(e^x) = x.

I imagine the answer is something of the form ln(...).

Stick some functions f(y) such that x = ln(f(y)) into e^3x + e^x - 5 and try to get e^3x + e^x - 5 = y. That is what we are after

Just Wikipedia "inverse function theorem"

So this might be kinda stupid but here goes

I am trying to find the equation of tangent plane at a point and due to partial y equaling 0 at the given point I get z= x+2. Assuming I'm correct this is the equation of a line so I assume it just holds true for all values of y, but how do I display that in plane form. Is it assumed that I know it's a plane because I'm using z or should I write something next to the answer or write it in some different form?

The set of points (x,y,z) that satisfy the equation z = x + 2 form a plane, because y can vary without restriction. (Yes, z = x + 2 is a line for any FIXED value of y, but in all of R^3 it's a plane). Don't be thrown by the fact that y doesn't appear; you can safely write z = x + 2 and it should be understood it's a plane.

My point was that one can actually compute the value of the derivative of the inverse function at a certain point, without actually solving for the analytic form of the inverse function. By inspection, I really don't think that function is going to have a nice inverse.

The domain of f(x) is given in the problem and the given domain is x is greater than or equal to zero.

Find the equations of the tangent plane and the normal line to the given surface at the specifed point

z=(e^x)cosy at point (0,0,1)

I understand the slopes of the tangent plane and the normal line are supposed to be reciprocal but here I get z-1=x for both because slope of x is 1 and partial y is 0 at the point. I may have forgotten to somehow take a partial for z but I'm not sure if its necessary or how to do it. partial y is 0 so how would I fit that into the symmetric equations or it doesnt matter at all. Seems impossible to be parallel and perpendicular so I must have messed something up

I'm not totally sure what the normal line is - perhaps it's the line, in the direction of the normal vector, that passes through the point (0,0,1)? If so, it is certainly perpendicular to the plane, but it's not perpendicular to the "slope" of the tangent plane (a plane doesn't even have a slope).

In any event, here's how I'd do the problem. You know that given a surface z = f(x,y), the equation of the tangent plane at the point x_0, y_0, z_0 is
z - z_0 = (f_x(x_0, y_0)) * (x - x_0) + (f_y(x_0,y_0)) * (y-y_0).

Let f(x,y) = (e^x) cos y. You take your partials to get
f_x(0,0) = 1
f_y(0,0) = 0,
so the equation of the tangent plane is
z - 1 = x,
or
x - z = -1.
The normal vector is <1,0,-1>, so you now want the equation for a line parallel to the vector <1,0,-1> and passing through the point (0,0,1), which is pretty straightforward.

ALright, so I guess I forgot how to get the normal vector or maybe I'm having trouble understanding what it actually is. We start with a surface and find the tangent plane at a point. The normal line would be orthogonal to this tangent plane. The gradient/partial derivatives determine what direction the tangent plane is going, so if the normal vector/line is perpendicular we would need or could use the reciprocal of all of these components to find the direction of the normal vector.

Do I understand this at all? And I am still not sure how to find partial z here to form the gradient or what to take partial z with respect to

Stop thinking about things being the "reciprocal." When you have two lines y = m_1 x + b_1 and y = m_2 x + b_2 in the xy-plane, they are orthogonal iff m_1 = -1 / m_2, but that doesn't generalize at all to higher dimensions and I think it's tripping you up.

It's going to be hard to explain all of these concepts in a 2+2 post. There are a few things you need to know:
*A plane P has a normal vector n that's orthogonal to it. Given the equation for P, you should be able to figure out what n is; conversely, given a normal vector n and a point (x_0, y_0, z_0) on P, you should be able to write down the equation for the plane that has normal vector n and passes through that point.
*Given an equation of a surface z = f(x,y), there's a nice formula that just gives you the equation of tangent plane at a point (x_0, y_0, z_0) on the surface (i.e. satisfying z_0 = f(x_0, y_0)). You can derive this formula if you want from orthogonalities, or you can just memorize the formula; I gave it in my last post.
*I think your question is asking you to both compute the equation of the tangent plane, and to write down the equation for the normal LINE. The idea of a "normal line" isn't one I'm familiar with, but I'm guessing it's the line that passes through the point (0,0,1) that's orthogonal to the tangent plane at that point. This line is basically an extension of the normal vector of the tangent plane.

In the problem at hand, you can use a formula to just write down the equation of the tangent plane to the surface at the point (0,0,1). Then, from the equation of the plane, you can read off the normal vector. You know that the "normal line" passes through the point (0,0,1) and is in the same direction as the normal vector, and these pieces of information are enough to write down the equation of the line.

No

We can "take the gradient" of a function, not of a surface. So grad(f) makes sense when f is a scalar function, say f(x,y,z).

A "surface" is (for our purposes) the 0 set of a function. [i.e., the points (x,y,z) s.t. f(x,y,z)=0]

So, your "surface," defined by z=(e^x)cosy, is really the surface:
f(x,y,z) = z - (e^x)cos(y) = 0

Now to find the normal vector to a surface f(x,y,z) = 0 at a point (x0,y0,z0) on the surface, we just take grad(f) at (x0,y0,z0).

A tangent plane to the surface [f(x,y,z) = 0] at (x0,y0,z0) is defined by
f_x(x0,y0,z0)(x-x0) + f_y(x0,y0,z0)(y-y0) + f_z(x0,y0,z0)(z-z0) = 0
[notice this is slightly more general than the form given by slipstream, but in the case z=g(x,y), they are equivalent.]
This is a 3-D analog of the equation for a tangent line to y=f(x):
y-y0 = f '(x0)(x-x0)

<aside>
or if we think of y=f(x) as the "surface" g(x,y) = y-f(x) = 0:
g_x(x0,y0)(x-x0) + g_y(x0,y0)(y-y0) = 0, which is
-f '(x0)(x-x0) + 1*(y-y0) = 0, which is, after rearranging:
y-y0 = f '(x0)(x-x0)
</aside>

Ok, back to the tangent plane
f_x(x0,y0,z0)(x-x0) + f_y(x0,y0,z0)(y-y0) + f_z(x0,y0,z0)(z-z0) = 0

I'm going to rewrite this equation as a dot product to make it a little cleaner:
[grad(f) at (x0,y0,z0)] dot <x-x0, y-y0, z-z0> = 0

Note what this means: (x,y,z) is in the tangent plane to the surface at (x0,y0,z0) if and only if [grad(f) at (x0,y0,z0)] dot <x-x0, y-y0, z-z0> = 0.

Now, the normal vector is perpendicular to the tangent plane. That is, if I take any vector that lives in the plane and dot it with my normal, I should get 0. What does a "vector in the plane" look like? Well, it's a vector that starts and ends at a point in the plane. It suffices to just look at vectors that originate at some fixed point in the plane (all other vectors in the plane are just translates of these, and this doesn't affect orthogonality relations). So let's say all vectors that live in the tangent plane are translates of vectors that originate at (x0,y0,z0). That is, we can write all vectors in the plane as <x-x0, y-y0, z-z0> for some choice of (x,y,z) that lives in the plane.

Do you see what's happening here? The definition of (x,y,z) being in the tangent plane to the surface at (x0,y0,z0) is that:
<x-x0, y-y0, z-z0> dot [grad(f) at (x0,y0,z0)] = 0. And here we are looking for the normal, which is some vector whose dot product with <x-x0, y-y0, z-z0> is 0.

A ha! This makes it clear that grad(f)@(x0,y0,z0) is normal to this plane.

---------------------

What have we learned?

* Take a surface defined as f(x,y,z) = 0. The tangent plane to the surface as (x0,y0,z0) is:
f_x(x0,y0,z0)(x-x0) + f_y(x0,y0,z0)(y-y0) + f_z(x0,y0,z0)(z-z0) = 0

( or easier:
grad(f)@(x0,y0,z0) dot <x-x0, y-y0, z-z0> = 0 )

* Take a surface defined as f(x,y,z) = 0. The normal to the surface (and also to the tangent plane) at (x0,y0,z0) is just grad(f)@(x0,y0,z0).

I didn't say it specifically, but it's hopefully somewhat clear if you understand the above:
if you have a plane Ax+By+Cz+D=0, then <A,B,C> is normal to the plane.