Do polar coordinates have to be converted to rectangular coordinates For example, there is a problem, r = a (1-cos θ), to find the tangent equation at (R, θ) = a certain point I know it's a polar equation, x = RCOs θ y = rsin θ Why can't r = a (1-cos θ) be directly regarded as y = a (1-cos x) and find the tangent equation at a certain point (y, x)? Why is this wrong?

Do polar coordinates have to be converted to rectangular coordinates For example, there is a problem, r = a (1-cos θ), to find the tangent equation at (R, θ) = a certain point I know it's a polar equation, x = RCOs θ y = rsin θ Why can't r = a (1-cos θ) be directly regarded as y = a (1-cos x) and find the tangent equation at a certain point (y, x)? Why is this wrong?

How can r = a (1-cos θ) be directly regarded as y = a (1-cos x), which is totally wrong

How to transform polar coordinate ρ = 2 into rectangular coordinate

x=p*cosa y=p*sina
x^2+y^2=p^2
p=√(x^2+y^2)
=2
Then x ^ 2 + y ^ 2 = 4

How to transform polar coordinates into rectangular coordinates?

Let a polar coordinate be (a, P), then its corresponding rectangular coordinate is (pcosa, psina), which can be obtained as a right triangle

Conjecture induction: 1 {2x-y = 1,2x + y = 1}; 2 {2x-y = 1,4x-2y = 1}, 3 {2x-y = 1,6x-3y = 3};
For binary linear equations {a1x + BIY = C1, a2x + b2y = C2} (a1.a2.b1.b2 are not zero)
(1) When is there a unique solution? (2) when is there no solution?
(3) When will there be innumerable solutions?
(4) Use your induction to solve the equations and judge the solution of the equations {y = KX + B, y = (3K-1) x + 2} about X and y

From a1x + BIY = c1a2x + b2y = C2, we get: a1x + BIY = C1 (b2-b1 * A2 / A1) * y = c2-c1 * A2 / A1 (1) when b2-b1 * A2 / A1 ≠ 0, that is, A1B2 ≠ b1a2. (2) when b2-b1 * A2 / A1 = 0 and c2-c1 * A2 / A1 ≠ 0, there is no solution. (3) when b2-b1 * A2 / A1 = 0 and c2-c1 * A2 / A1 = 0, there is no solution

The law of series circuit voltage and parallel circuit voltage
one
2. Conjecture or hypothesis
3. Design experiment
4. Experiment (table)
5. Conclusion
I'm sorry to trouble you

1. Ask questions
In series circuit and parallel circuit, what is the relationship between the voltage of each electrical appliance and the total voltage?
2. Conjecture or hypothesis
Series circuit: u = U1 + U2
Parallel circuit: u = U1 = U2
3. Design experiment
Draw the series and parallel circuits of two small bulbs
4. Experiment (table)
5. Conclusion
In a series circuit, the sum of the voltages of the consumers is equal to the total voltage;
In parallel circuit, the voltage of each consumer is the same as that of the power supply

Is 1 / m of 2 a fraction?
The m-th power of 2 is in, and M is the denominator. It contains letters, right?

It's not a fraction, because no matter how much power 2 is, it's still a constant

Two white papers of "220 V, 60 W" are connected in series in the home circuit, and the total power of the lamp is 0

If the influence of temperature on the resistance is not considered, the total resistance of two identical lamps in series is twice that of one lamp, so the total power of two lamps in series is equal to half of the rated power of one lamp, that is, the required power is p = 60 / 2 = 30 watts

When a, B and C satisfy what kind of relationship, 2 ^ 6A = 3 ^ 3B = 6 ^ 2C

Let 2 ^ 6A = 3 ^ 3B = 6 ^ 2C = k, then 6A = log (2, K) = 1 / log (k, 2). 3B = log (3, K) = 1 / log (k, 3) 2C = log (6, K) = 1 / log (K, 6) = 1 / [log (k, 2) + log (k, 3)] = 1 / [1 / 6A + 1 / 3B], that is, 1 / 2C = 1 / 6A + 1 / 3b, then 3ab-2ac-bc = 0 is obtained

The current of an electrical appliance with a resistance value of 10 Ω is 0.3 a when it works normally. How to connect it to a circuit with a resistance value of 0.8 a?
Not only to solve the problem process, focusing on the analysis process, it is best to add similar problems

The current of the circuit is 0.8A ＞ 0.3A when the user works normally, so it has to be shunted,
Shunt can be done by paralleling a resistor,
Because it is in parallel, the voltage on both sides of the resistor is equal to the voltage on both sides of the consumer, i.e. u = IR = 0.3x10 = 3V,
If I '= 0.8-0.3-0.5a, then R' = u / I '= 3 / 0.5 = 6 Ω
That is, a 6 Ω resistor is connected in parallel

As shown in the picture, there is a 2.1m-high wooden column with a bottom circumference of 40cm. When preparing for the new year's party, in order to create a festive atmosphere, the teacher asked Xiao Ming to wind a ribbon evenly from the bottom column to the top of the column for 7 circles until it is directly above the starting point. The length of the ribbon Xiao Ming needs to prepare is at least ()
A. 70457cmB. 350cmC. 2803cmD. 300cm

If the surface of the cylinder is cut and unfolded into a rectangle, the length of the helix is the diagonal length of the rectangle after seven rectangles are arranged side by side, ∵ the height of the cylinder is 2.1 m, the perimeter of the bottom surface is 0.4 m, X2 = (40 × 7) 2 + 2102 = 122500, the solution is x = 350, so the length of the ribbon is at least 350 cm