So say you bought $50.00 worth of groceries at the food store. And now you have to add tax which is 8.5%. My question is, who does not understand the use of 50 * 1.085 and why do you not get it?

So my current signature asks to take the derivative of 2x²(sin(2ln(x(e^x)))). So correct me if I'm wrong:
dy/dx=(2x²)(cos(2ln(x(e^x))))(2)(1/(x(e^x)))(x(e^x))(e^x)+4x(sin(2ln(x(e^x))))
dy/dx=(4x³e^(2x)cos(2ln(x(e^x))))/(x(e^x))+4xsin(2ln(x(e^x)))
dy/dx=4x²(e^x)cos(2ln(x(e^x)))+4xsin(2ln(x(e^x)))
dy/dx=(2x²)(cos(2ln(x(e^x))))(2)(1/(x(e^x)))(x(e^x))(e^x)+4x(sin(2ln(x(e^x))))
dy/dx=(4x³e^(2x)cos(2ln(x(e^x))))/(x(e^x))+4xsin(2ln(x(e^x)))
dy/dx=4x²(e^x)cos(2ln(x(e^x)))+4xsin(2ln(x(e^x)))
seanlego23 wrote:
Could anyone explain to me how to find the derivative of the inverse to f(x)=2x^3 + x^2  1 at x=2?
[img]https://wikimedia.org/api/rest_v1/media/math/render/svg/f95e8b8db53babeadfae565759a5d9b5607efea8[/img]
f'(2)=28, so [f⁻¹]'(2)=1/28
jonbush wrote:
f'(2)=28, so [f⁻¹]'(2)=1/28
I should have said I already know that. And I know how to solve it that way.
My instructions are:
 Find the inverse by switching x and y.
 Implicitly differentiate the equation.
 Solve for dy/dx. It will be in terms of y.
 Replace x and b in your inverse function and solve for y.
 Plug the value of y into dy/dx.
b being x=b or 2 in this case.
Does anyone understand how to solve it this way?
seanlego23 wrote:
Updated the Math website. Two meetings in. This is really fun. We talked about Row Echelon form today. In the calculator, that's denoted as rref() found by this 2nd[matrix]>Alpha[B]. It is used in finding the point where two or more lines intersect. Hint Hint...Systems of Equations.
Say you have a 2 equations: 2x + 2y = 4 and 3x + 5y = 6. Type this into a 2 x 3 matrix.
Code:
rref([matrix name]
Click enter and it will output this:
Code:
x = .5
y = 1.5
If a problem you have returns this:
Code:
Then there is no solution, as long as you plugged it in correctly, so make sure of that first. You will use this in algebra and up; but you won't learn it, if you do, until PreCalculus.
Say you have a 2 equations: 2x + 2y = 4 and 3x + 5y = 6. Type this into a 2 x 3 matrix.
Code:
2 2 4
3 5 6
rref([matrix name]
Click enter and it will output this:
Code:
1 0 .5
0 1 1.5
x = .5
y = 1.5
If a problem you have returns this:
Code:
1 1 0
0 0 1
Then there is no solution, as long as you plugged it in correctly, so make sure of that first. You will use this in algebra and up; but you won't learn it, if you do, until PreCalculus.
Actually, what you used (RREF()) is Reduced RowEchelon Form, which is obtained through the use of Gaussian Jordan Elimination. RowEchelon Form(REF()) is obtained through the use of just Gaussian Elimination.
These methods can be used to solve a system of equations of any order. Not just where two or more lines intersect; it isn't limited to just two variables, as you implied.
As for the last part, you're iffy; you should explain it further. If you obtain a row composed of zeroes, but the last number in the row is nonzero, then there is no solution. If the row is entirely zero, then it is probably a system with infinite solutions and will need a parameterized solution. Else, you should be homefree.
 ProgrammerNerd
 Power User (Posts: 375)
 Re: Math Help/Fun/Strategies/Calculator/etc
 19 Sep 2016 09:12:18 pm Permalink
seanlego23 wrote:
Code:
for(i=96; i==100; i++) {
int odd;
odd += i;
odd += i+1;
}
You never initialize odd. I do not quite understand this. Is odd supposed to be out of the for loop? Or did you mean to just do odd = 0;. Remember if you don't initialize the variable it could be anything.
I am sure that if enough people run this program someone will get a value that is not zero:
Code:
#include <stdio.h>
int main(void){
int uninitialized;
printf("This variable may not have a value of zero: %x\n",uninitialized);
return 0;
}
seanlego23 wrote:
Add all the numbers between 1 and 100. What do you get? How did you get it? Try and do it without the calculator.
Is this a homework question? if not read on:
Don't do:
Code:
>>> sum(range(1,101)) # In python range is exclusive.
5050
Do:
Code:
>>> (100*(1+100))//2
5050
GTemples27: Thanks for that information. I fixed the post.
ProgrammerNerd: Fixed the code and I suggest you take a look at the post with the addition.
ProgrammerNerd: Fixed the code and I suggest you take a look at the post with the addition.
seanlego23, You are getting closer. Now check out the comparison in the for loop. The loop ends when the condition evaluates to false. This means that the loop the code inside the loop is never executed because the conditions is immediately false.
I can't figure out these problems. Knowledge of Calculus 1, Related Rates, and Geometric Formulas is required to answer these.
1. Two cars are riding on roads that meet at a 60° angle. Car A is 3 miles from the intersection traveling at 40 mph while car B is 2 miles from the intersection traveling at 50 mph. How fast is the distance between the two cars changing if a) they are both traveling away from the intersection and b) car A is traveling away from the intersection and car B is traveling towards it?
2. Beach sand is poured creating a cone whose radius is always equal to twice its height. If the sand is poured at the rate of 20 in³/sec, how fast is the height changing at the time the height is a) 2 inches b) 6 inches?
3. Water drains from a conical tank at the rate of 2 m³/sec. The tank is 16 meters high and its top radius is 4 meters. How fast does the water level fall when the water level is a) 12 meters high, b) 2 meters high?
1. Two cars are riding on roads that meet at a 60° angle. Car A is 3 miles from the intersection traveling at 40 mph while car B is 2 miles from the intersection traveling at 50 mph. How fast is the distance between the two cars changing if a) they are both traveling away from the intersection and b) car A is traveling away from the intersection and car B is traveling towards it?
2. Beach sand is poured creating a cone whose radius is always equal to twice its height. If the sand is poured at the rate of 20 in³/sec, how fast is the height changing at the time the height is a) 2 inches b) 6 inches?
3. Water drains from a conical tank at the rate of 2 m³/sec. The tank is 16 meters high and its top radius is 4 meters. How fast does the water level fall when the water level is a) 12 meters high, b) 2 meters high?
2. & 3. are effectively equivalent: the methods are identical. Here's 2.
With variables defined by the obvious letters, we have:
r=2h;
V=πr²h/3 [cone] = π(2h)²h/3 = 4πh³/3
therefore dV/dh = 4πh²;
and we are given dV/dt = 20.
By the Chain Rule, dV/dt = dV/dh * dh/dt.
so 20 = 4πh² * dh/dt
so dh/dt = 5/(πh²)
a) When h = 2, dh/dt = 5/(4π) ~= 0.398 in/sec.
b) When h = 6, dh/dt = 5/(36π) ~= 0.044 in/sec.
In 3 you should allow for the fact that the volume in the tank is decreasing, so dV/dt is negative. This will correctly give a negative for dh/dt as well, since the height is going down.
With variables defined by the obvious letters, we have:
r=2h;
V=πr²h/3 [cone] = π(2h)²h/3 = 4πh³/3
therefore dV/dh = 4πh²;
and we are given dV/dt = 20.
By the Chain Rule, dV/dt = dV/dh * dh/dt.
so 20 = 4πh² * dh/dt
so dh/dt = 5/(πh²)
a) When h = 2, dh/dt = 5/(4π) ~= 0.398 in/sec.
b) When h = 6, dh/dt = 5/(36π) ~= 0.044 in/sec.
In 3 you should allow for the fact that the volume in the tank is decreasing, so dV/dt is negative. This will correctly give a negative for dh/dt as well, since the height is going down.
I have calculated some fairly complicated function recently for a side project:
f(x) = 6059 / (101.325 * e^(x/7.28))  (59.798 / (1 + x)) (Note it only matters for x between 0 and 150)
As if this was not enough, I need to create a g(x) = g(x1) + f(x) for g(0) = 0 which is nonrecursive, or saying it in other words, a function that grows starting at (0/0) by an amount specified in f.
I've asked in multiple places, but looks like the question is very hard...
Note I mainly need it for a set of data, so I could do it iteratively, but a function would be pretty good to have...
f(x) = 6059 / (101.325 * e^(x/7.28))  (59.798 / (1 + x)) (Note it only matters for x between 0 and 150)
As if this was not enough, I need to create a g(x) = g(x1) + f(x) for g(0) = 0 which is nonrecursive, or saying it in other words, a function that grows starting at (0/0) by an amount specified in f.
I've asked in multiple places, but looks like the question is very hard...
Note I mainly need it for a set of data, so I could do it iteratively, but a function would be pretty good to have...
One way to solve 1 relies on the cosine rule, and implicit differentiation.
t hours after the situation described in a), car A will be (3+40t) miles from the junction, and car B will be (2+50t) miles from the junction. The direct distance between the cars, x, is given by the cosine rule as:
x²=(3+40t)² + (2+50t)²  2(3+40t)(2+50t)cos60°
Expanding, and using cos60°=0.5, this becomes
x²=2100t²+210t+7.
Differentiate:
2x dx/dt = 4200t +210.
At the time in question, t = 0, so we get x²=7 & 2x dx/dt = 210.
So dx/dt = 210/(2√7) = 15√7 ~= 39.7 mph.
For b) we would simply replace the (2+50t) with (250t), arriving after similar calculation at
dx/dt ~= 20.8 mph
t hours after the situation described in a), car A will be (3+40t) miles from the junction, and car B will be (2+50t) miles from the junction. The direct distance between the cars, x, is given by the cosine rule as:
x²=(3+40t)² + (2+50t)²  2(3+40t)(2+50t)cos60°
Expanding, and using cos60°=0.5, this becomes
x²=2100t²+210t+7.
Differentiate:
2x dx/dt = 4200t +210.
At the time in question, t = 0, so we get x²=7 & 2x dx/dt = 210.
So dx/dt = 210/(2√7) = 15√7 ~= 39.7 mph.
For b) we would simply replace the (2+50t) with (250t), arriving after similar calculation at
dx/dt ~= 20.8 mph
Nik wrote:
I have calculated some fairly complicated function recently for a side project:
f(x) = 6059 / (101.325 * e^(x/7.2)  (59.798 / (1 + x)) (Note it only matters for x between 0 and 150)
As if this was not enough, I need to create a g(x) = g(x1) + f(x) for g(0) = 0 which is nonrecursive, or saying it in other words, a function that grows starting at (0/0) by an amount specified in f.
I've asked in multiple places, but looks like the question is very hard...
Note I mainly need it for a set of data, so I could do it iteratively, but a function would be pretty good to have...
f(x) = 6059 / (101.325 * e^(x/7.2)  (59.798 / (1 + x)) (Note it only matters for x between 0 and 150)
As if this was not enough, I need to create a g(x) = g(x1) + f(x) for g(0) = 0 which is nonrecursive, or saying it in other words, a function that grows starting at (0/0) by an amount specified in f.
I've asked in multiple places, but looks like the question is very hard...
Note I mainly need it for a set of data, so I could do it iteratively, but a function would be pretty good to have...
Let's solve this
What is g'(x)? That is (g(x)g(x))/1 = g(x)g(x1) = g(x1)+f(x)g(x1) = f(x)! Better said: the derivatie of g is equal to f(x), so g(x) = F(x). You can calculate F(x) using WolframAlpha or something like that, and then you're done
OldMathTeacher: Thank you. Related rates in Calculus are so weird in my opinion. It will take some getting use to.
Nik:I have no idea what that is asking, so I am hoping that PT helped you out with that.
Degrees to Radians:
(Degree)(π/180)=(Radian)
Radians to Degrees:
(Radian)(180/π)=(Degree)
Nik:I have no idea what that is asking, so I am hoping that PT helped you out with that.
Degrees to Radians:
(Degree)(π/180)=(Radian)
Radians to Degrees:
(Radian)(180/π)=(Degree)
PT_ wrote:
Nik wrote:
I have calculated some fairly complicated function recently for a side project:
f(x) = 6059 / (101.325 * e^(x/7.2)  (59.798 / (1 + x)) (Note it only matters for x between 0 and 150)
As if this was not enough, I need to create a g(x) = g(x1) + f(x) for g(0) = 0 which is nonrecursive, or saying it in other words, a function that grows starting at (0/0) by an amount specified in f.
I've asked in multiple places, but looks like the question is very hard...
Note I mainly need it for a set of data, so I could do it iteratively, but a function would be pretty good to have...
f(x) = 6059 / (101.325 * e^(x/7.2)  (59.798 / (1 + x)) (Note it only matters for x between 0 and 150)
As if this was not enough, I need to create a g(x) = g(x1) + f(x) for g(0) = 0 which is nonrecursive, or saying it in other words, a function that grows starting at (0/0) by an amount specified in f.
I've asked in multiple places, but looks like the question is very hard...
Note I mainly need it for a set of data, so I could do it iteratively, but a function would be pretty good to have...
Let's solve this
What is g'(x)? That is (g(x)g(x))/1 = g(x)g(x1) = g(x1)+f(x)g(x1) = f(x)! Better said: the derivatie of g is equal to f(x), so g(x) = F(x). You can calculate F(x) using WolframAlpha or something like that, and then you're done
Of course, that's really an approximation to what Nik wanted  a tangent gradient being used to find a chord gradient. It depends whether the approximation is adequate for Nik's purposes. If we simplify the expression for f(x) to
f(x)= 242360/4053 * e^(25x/182)  59.798/(1+x),
then it is pretty easy to differentiate or integrate without any need for Wolfram Alpha (or TINspire!). But is it what we want? If Nik has to have a g function that works specifically in step sizes of 1, then I think his iterative approach is the best.
seanlego23 wrote:
OldMathTeacher: Thank you. Related rates in Calculus are so weird in my opinion. It will take some getting use to.
The difficulty I have is not knowing where I ought to start  how much you already know, or even whether you know them by the same names as I use in England. Here are some guidelines for related rates of change.
▪Figure out how many variables you are working with  in the cone questions there are 4: r, h, V, t.
▪Find one fewer equations, giving relationships between them  here we used the 3 equations
r=2h
V=πr²h/3
dV/dt=20.
▪Differentiate any of these that are not already differential equations  here we had to substitute the first into the second, to make it differentiable.
▪Connect them to the rate you are trying to find (that was dh/dt) by constructing a suitable chain rule.
What makes the chain rule easy is that, although derivatives such as dV/dt are not strictly speaking fractions, they look like fractions and when multiplied together they cancel like fractions. So here we just had to write it in such a way that dh would cancel, in order to find the relationship between all the rates of change.
OldMathTeacher wrote:
seanlego23 wrote:
OldMathTeacher: Thank you. Related rates in Calculus are so weird in my opinion. It will take some getting use to.
The difficulty I have is not knowing where I ought to start  how much you already know, or even whether you know them by the same names as I use in England. Here are some guidelines for related rates of change.
▪Figure out how many variables you are working with  in the cone questions there are 4: r, h, V, t.
▪Find one fewer equations, giving relationships between them  here we used the 3 equations
r=2h
V=πr²h/3
dV/dt=20.
▪Differentiate any of these that are not already differential equations  here we had to substitute the first into the second, to make it differentiable.
▪Connect them to the rate you are trying to find (that was dh/dt) by constructing a suitable chain rule.
What makes the chain rule easy is that, although derivatives such as dV/dt are not strictly speaking fractions, they look like fractions and when multiplied together they cancel like fractions. So here we just had to write it in such a way that dh would cancel, in order to find the relationship between all the rates of change.
My strategy is to write what I am given and what I need, substitute a variable if I can, differentiate equation, solve for a variable I am not given, solve differential equation.
I've learned up to the trig rules, implicit differentiation, and some things regarding differentiability at certain points or intervals. This is the website my teacher uses that we get our problems and work from: www.MasterMathMentor.com.
That strategy sounds good, although with more elaborate problems (say 5 or more variables) my students have found it really helpful to have a little rule to keep track: you need n1 equations for n unknowns. Then the nth equation will be the chain rule.
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