-
Best of luck to the class of 2024 for their HSC exams. You got this! Let us know your thoughts on the HSC exams here -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
MATH1251 Questions HELP (2 Viewers)
- Thread starter 1008
- Start date
You can try testing whether it's right by checking whether the given point (4,2,1) lies on your plane. Subbing it in, we see the plane equation isn't satisfied, so the answer's not right.Also, is this right?
You did everything right up till the "Equation of tangent" part. The equation is just meant to be what you did without the f(x_0) (i.e. 8(x-4) + 4(y-2) + 2(z-1) = 0, which you can simplify by dividing through by 2).
Thanks, but why isn't the f(x_0) part included?
The equation of a plane with normal n = (n1,n2,n3) and through the point a = (a,b,c) is just n•(x - a) = 0, i.e. n1*(x-a) + n2*(y-b) + n3*(z-c) = 0.
Ohh right thanks!
Part (c) (only) for this one please:
Thanks. Yeah figured out the integral bit lol The sub-intervals used are $I_{n} = [n,n+1]$ (so spacing is $h=1$), for $n=50,51,52,\ldots$. Note $f''(x) = 6x^{-4}$. Since $f''$ is decreasing on each $I_{n}$, the maximum of $|f''(x)|$ on $I_{n}$, say $M_{n}$, occurs on the left endpoint of $I_{n}$, i.e. at $x=n$. So $M_{n} = |f''(n)| = 6n^{-4}$. So the absolute error on $I_{n}$ is bounded above by $\frac{1^3 \times 6n^{-4}}{12} = \frac{1}{2}n^{-4}$. Now the overall absolute error is the absolute error in the sum of the estimates on each interval, which is smaller than the sum of the absolute errors (by the triangle inequality). Hence the overall error is less than $\sum_{n=50}^{\infty} \frac{1}{2}n^{-4}$, as required. I'll leave it for you to show this sum is smaller than the given integral (draw a picture etc.), and the last part is just evaluating that integral. $)
Had another question:
I found the tangent to be:
2p(x-p) + 2q(x-q) - (p^2+q^2)(z+p^2+q^2) = 0
Now for the bit where it asks to find where it cuts the z-axis, do I set x = y = 0? I did that and I get z= 0, so the only place where the axis is cut is the origin? Is that right?
The coefficient of the (z + (p^2 + q^2)) part should just be 1. But yeah, set x = y = 0 to find the z-intercept.Thanks. Yeah figured out the integral bit lol
Had another question:
I found the tangent to be:
2p(x-p) + 2q(x-q) - (p^2+q^2)(z+p^2+q^2) = 0
Now for the bit where it asks to find where it cuts the z-axis, do I set x = y = 0? I did that and I get z= 0, so the only place where the axis is cut is the origin? Is that right?
Yeah got it thanks.
For this one I get sum of intercepts of the plane as
seanieg89
Well-Known Member
- Joined
- Aug 8, 2006
- Messages
- 2,662
- Gender
- Male
- HSC
- 2007
You should make a habit of checking dimensions to check that your answer makes sense. An intercept of plane is a real number (the ordinate at which the plane cuts the axis), the thing you wrote is a vector.
This is how you would do the question posted:
=\sqrt{x}+\sqrt{y}+\sqrt{z}-\sqrt{a}\\ \\ \Rightarrow \nabla F(x_0,y_0,z_0)=\frac{1}{2}(x_0^{-1/2},y_0^{-1/2},z_0^{-1/2})\\ \\ \Rightarrow T_P = \{(x,y,z): (x-x_0,y-y_0,z-z_0)\cdot (x_0^{-1/2},y_0^{-1/2},z_0^{-1/2})=0\}\\ \\ $Setting $y=z=0$, we get$\\ \\ x=x_0+\sqrt{x_0y_0}+\sqrt{x_0z_0}.\\ \\ $Hence by symmetry the sum of intercepts is given by$\\ \\ x_0+y_0+z_0+2(\sqrt{x_0y_0}+\sqrt{x_0z_0}+\sqrt{y_0z_0})=(\sqrt{x_0}+\sqrt{y_0}+\sqrt{z_0})^2=a\\ \\ $which is independent of $ P.)
This is how you would do the question posted:
Last edited:
Thanks for clarifying that seanieg89. I misinterpreted that by intercepts they meant the actual points (in vector form) where the plane cuts the x, y and z axes. But yeah I realised that now thanks to your responseYou should make a habit of checking dimensions to check that your answer makes sense. An intercept of plane is a real number (the ordinate at which the plane cuts the axis), the thing you wrote is a vector.
This is how you would do the question posted:
You can do these fairly algorithmically, just tedious.
The fact that the regions are closed and bounded (compact) and the functions are continuous tells us there is a maximum and a minimum. The functions are differentiable, so the extrema occur either at stationary points in the interior of the domains or on the boundaries. So find the stationary points by setting f_x and f_y equal to 0 and solving for x and y. For the boundaries, you can parameterise them in terms of a single parameter (say t), so that on the boundary, f is a function of a single variable, which makes it easy to optimise there, so you can find the extreme values on each boundary. Then you'll have a list of function values (extreme values on its boundary lines and from the interior). The greatest one is the maximum and least one is the minimum.
The fact that the regions are closed and bounded (compact) and the functions are continuous tells us there is a maximum and a minimum. The functions are differentiable, so the extrema occur either at stationary points in the interior of the domains or on the boundaries. So find the stationary points by setting f_x and f_y equal to 0 and solving for x and y. For the boundaries, you can parameterise them in terms of a single parameter (say t), so that on the boundary, f is a function of a single variable, which makes it easy to optimise there, so you can find the extreme values on each boundary. Then you'll have a list of function values (extreme values on its boundary lines and from the interior). The greatest one is the maximum and least one is the minimum.
Last edited:
seanieg89
Well-Known Member
- Joined
- Aug 8, 2006
- Messages
- 2,662
- Gender
- Male
- HSC
- 2007
Integrand's explanation above is the general procedure. Find the maximum and minimum values of }f(x,y) = x-x^2+y^2\text{ on the rectangle }\\0 \leq x \leq 3, 0 \leq y \leq 2.\\\text{(b) Find the maximum and minimum values of }f(x,y) = xy(1-x-y)\text{ on the triangle with vertices (0,0), (2,0), (0,2).})
There are often shortcuts you can take by noting the structure of the inequalities and the geometry of the domains.
a) Since the first domain is a rectangle and the function is a sum a(x)+b(y), we can find maxima/minima by maximising/minimising a and b separately.
b) Since we note the presence of xy and x+y we can use AM-GM. Noting the geometry of our domain, we can then maximise/minimise f on the line segments {x+y=c & x,y >=0} by noting that sqrt(xy) =< (x+y)/2 for non-negative x,y with equality at x=y (being careful because there is a sign change of f when c exceeds 1).
This reduces the optimisation problem to a single variable optimisation problem in the variable c in [0,2] which you can treat using calculus or whatever you like.
You can often make your life easier in these kinds of problems by choosing coordinates/parametrisations wisely.
Oh yeah, forgot to mention there are sometimes shortcuts. Was going to edit it in with a sample answer but ended up being too lazy then to type up the answer (it would be fairly long) and left the post unedited.
Last edited:
Thanks!
Could you please demonstrate how one would go about doing these by solving these using the shortcuts you've provided?
(a) Using the notation in the suggestion provided, I get a'(x) = 1-2x and b'(x) = 2y.
If I then set these to equal 0, I get x = 1/2 and y = 0. I also used parts of my original approach to determine the other critical points. However, none of these provide the maximum value that is provided in the answers, at the point (1/2, 2) = 17/4, while the maximum I get is at (0,2) = 4. However, my minimas match.
(b) Sorry, but I have no idea how to approach this one...