Mathematics Question Thread

[georgebanis]

Hey guys, just need a hand with part B of both of these applications of calculus questions.

Thanks

[RuiAce]

Hey guys, just need a hand with part B of both of these applications of calculus questions.

Thanks

These questions involve related rates which are only a part of the MX1 course. It would be great if you could repost this to our MX1 Question Thread, just so we can do the proper solution and it isn't confusing for any 2U students looking through this later. Here is the link!

Jamon Edit: Added link!

[benneale]

Hey!

Having trouble working out the indefinite integral of e^x^2...
Thanks!

[jamonwindeyer]

Hey!

Having trouble working out the indefinite integral of e^x^2...
Thanks!

Hey! That isn't an integral you can perform, at least not nicely so I don't blame you for having trouble!! You don't need to be able to do it

[benneale]

Help!
Find the exact area enclosed between the curve y=e^2x and the lines y=1 and x=2

[fun_jirachi]

From the reference sheet, we know that


Noting this, graph everything. It'll make it easier to visualise. Essentially now, we have an enclosed area between the line y=1 and y=e^2x, so this becomes a top curve minus bottom curve sort of question.

Your area should be something similar to the below, work from there to get your answer!


Hope this helps

[emmajb37]

Having a struggle please help!

[RuiAce]

Having a struggle please help!

This is just plugging into the Simpson's rule formula the usual way. Noting that \( \frac\pi8 = \frac{4\pi}{32}\) and \(\frac\pi4 = \frac{8\pi}{32}\), the required function values are \( \frac{4\pi}{32}\), \(\frac{5\pi}{32}\), \( \frac{6\pi}{32}\), \(\frac{7\pi}{32}\), \( \frac{8\pi}{32}\). If you have any further queries, please specify exactly where they are.

[emmajb37]

This is just plugging into the Simpson's rule formula the usual way. Noting that \( \frac\pi8 = \frac{4\pi}{32}\) and \(\frac\pi4 = \frac{8\pi}{32}\), the required function values are \( \frac{4\pi}{32}\), \(\frac{5\pi}{32}\), \( \frac{6\pi}{32}\), \(\frac{7\pi}{32}\), \( \frac{8\pi}{32}\). If you have any further queries, please specify exactly where they are.

Thanks so much

[emmajb37]

No idea where to start, pls help!
The area of the sector of a circle is 4π untis² and the length of the arc bounded by this sector is π/8 units. Find the radius of the circle and the angle that is subtended at the centre.

[fun_jirachi]

Work from there to find your radius, and then find your angle by substituting all the given information back into one of the two formulas (easier to do it with the length of an arc (less computations)).

Hope this helps

[emmajb37]

Somehow I got 2.18 when the answer is 2.04 
Use the trapezoidal rule with 4 subintervals to find, correct to 3 decimal places, an approximation to the volume of the solid formed by rotating the curve y = sin x  about the x-axis from x=0.2  to x = 0.6.

[emmajb37]

Tried this algebraically and graphically but still getting the wong values.
Find all the points of inflexion on the curve y = 3cos(2x +π/4)  for 0 ≤ x ≤ 2π

[fun_jirachi]

Somehow I got 2.18 when the answer is 2.04 
Use the trapezoidal rule with 4 subintervals to find, correct to 3 decimal places, an approximation to the volume of the solid formed by rotating the curve y = sin x  about the x-axis from x=0.2  to x = 0.6.

That's odd, the answer you've given is wrong? It's a magnitude of 10 off; I got roughly 0.204.
Basically, remember that for volumes you use the curve squared instead of the curve itself. ie.

When using the trapezoidal rule, you're approximating the integral of y squared, so you need to sub your values into y squared, not y. Given your boundaries are 0.2 and 0.6, with four sections you have the x values 0.2, 0.3, 0.4, 0.5 and 0.6. Sub these into (sin x)^2 to find your y values (your side lengths of the trapezium if you will). Remember that the trapezoidal rule is height/2 multiplied by (first + last + 2(everything else)), ie.

Tried this algebraically and graphically but still getting the wong values.
Find all the points of inflexion on the curve y = 3cos(2x +π/4)  for 0 ≤ x ≤ 2π

For this one, note that the cosine graph isn't actually shifted up or down. If you know your trigonometric graphs well enough, you know that the sine and cosine waves change concavity on the x-axis, and since the graph isn't shifted up or down, the inflexion points should still lie on the x-axis. You're essentially solving for 3cos(2x +π/4) = 0, if that helps.

Hope this helps

[emmajb37]

That's odd, the answer you've given is wrong? It's a magnitude of 10 off; I got roughly 0.204.
Basically, remember that for volumes you use the curve squared instead of the curve itself. ie.

When using the trapezoidal rule, you're approximating the integral of y squared, so you need to sub your values into y squared, not y. Given your boundaries are 0.2 and 0.6, with four sections you have the x values 0.2, 0.3, 0.4, 0.5 and 0.6. Sub these into (sin x)^2 to find your y values (your side lengths of the trapezium if you will). Remember that the trapezoidal rule is height/2 multiplied by (first + last + 2(everything else)), ie.


For this one, note that the cosine graph isn't actually shifted up or down. If you know your trigonometric graphs well enough, you know that the sine and cosine waves change concavity on the x-axis, and since the graph isn't shifted up or down, the inflexion points should still lie on the x-axis. You're essentially solving for 3cos(2x +π/4) = 0, if that helps.

Hope this helps

Unfrtunately I am still getting the wrong values for that second question. I am getting 0, π/2, π, 3π/2 and 2π
However the answers are π/8, 5π/8, 9π/8 and 13π/8