Mathematics Question Thread

[emmajb37]

Hey, I got the answer 1 + e for this question but the answer is actually 2e. I have no idea how to get that though, please help!

Find the exact gradient of the tangent to the curve y = e^x+lnx at the point where x = 1.

Thank you!!

[emmajb37]

Please Help!!
I got 0.18 somehow 

[MB_]

Hey, I got the answer 1 + e for this question but the answer is actually 2e. I have no idea how to get that though, please help!

Find the exact gradient of the tangent to the curve y = e^x+lnx at the point where x = 1.

Thank you!!

Please Help!!
I got 0.18 somehow 

Can you show your working so we can identify where you may have made a mistake?

[jamonwindeyer]

I need some help!

The curve y=2^x is rotated about the x-axis between x=1 and x=2. Use Simpson's value with 3 function values to find an approximation of the volume of the solid formed to 2 s.f.

So I calculated the area and squared it and multiplied by pi to get 26.2, but answers say 27.2 Typo or ?

and: Find the exact area bounded by the parabola y=x^2 and the line y=4-x.

I found the x-intercepts to be (-1+-sqr17)/2. So when I sub into the integral, I have to expand those big fractions?? Am I doing this right/is there an easier method

Hey! Sorry for the late reply - But yep you are over-complicating it a tad! They've told you where the bounds are for the volume, so no need to calculate x-intercepts. For 3 function values, you'll use \(x=1\), \(x=1.5\), and \(x=2\). The Simpson's Rule for volume is:



Notice it is just the normal rule with a \(\pi\) out the front and all function values squared, give that a go!

[jamonwindeyer]

Hey, I got the answer 1 + e for this question but the answer is actually 2e. I have no idea how to get that though, please help!

Find the exact gradient of the tangent to the curve y = e^x+lnx at the point where x = 1.

Thank you!!

Hey! As a quick check, is your derivative:

Please Help!!
I got 0.18 somehow 

And equally, is your integral here...



Always good to check the Calculus bits first! If these match, then as MB_ said I reckon we could spot the mistake if you post your working!

[emmajb37]

Hi there, struggling with this question please help:
 Find the exact area enclosed by the curve y = sinx and the line y = 1/2 for 0 ≤ x ≤ 2pi

[fun_jirachi]

Have you tried drawing out the graphs of these two functions over the domain provided? It really helps you visualise what you're actually looking for. If you haven't that's okay, draw it up as you answer the question

Notice that when you do draw the graph, it results in only one area. You can find the intersection of the two curves by solving sinx=1/2, which should yield two solutions ie. pi/6 and 5pi/6. These solutions will provide your lower bound and upper bound respectively. From here, use the fact that the area is equal to the integral of the top curve minus the integral of the bottom curve ie. sinx-1/2. Work from there to get your answer! You should get sqrt3-pi/3.

Hope this helps

[emmajb37]

Have you tried drawing out the graphs of these two functions over the domain provided? It really helps you visualise what you're actually looking for. If you haven't that's okay, draw it up as you answer the question

Notice that when you do draw the graph, it results in only one area. You can find the intersection of the two curves by solving sinx=1/2, which should yield two solutions ie. pi/6 and 5pi/6. These solutions will provide your lower bound and upper bound respectively. From here, use the fact that the area is equal to the integral of the top curve minus the integral of the bottom curve ie. sinx-1/2. Work from there to get your answer! You should get sqrt3-pi/3.

Hope this helps

Thank you so much!!

[emmajb37]

Hey, I dont understand with this question, which function would be the top and which would be the bottom because they keep switching. Would I need to do it in several parts?
Thanks

Find the exact area bounded by the curves y = sinx and  y = cos x in the domain 0 < x < 2pi

[fun_jirachi]

Similar thing here with the question above; graph the functions over the same domain! There's only one area (between pi/4 and 5pi/4, found by solving sinx=cosx), and the sine wave is above the cosine wave here. You should get 2sqrt2.

Hope this helps

[emmajb37]

Please Help!!
Solve cos2x = 2x/3 for 0<x<2pi

[benneale]

heyo could you please help with this:

find the equation of the normal to the curve y=e^x at the point where x=3, in exact form  ...?

thanks!

[RuiAce]

Please Help!!
Solve cos2x = 2x/3 for 0<x<2pi

You cannot do this algebraically. (As in, you physically cannot - there is no mathematical method in existence to do it.)

If the question was given to you like that without any more information, it is a faulty question. If you were told some other information (e.g. use a graph), please provide it.

heyo could you please help with this:

find the equation of the normal to the curve y=e^x at the point where x=3, in exact form  ...?

thanks!

This looks like the usual method. You should be able to check that the gradient of the tangent at 3 is just \(e^3\), so the normal has gradient \( -e^{-3}\), and then sub into \(y-y_1=m(x-x_1)\).

[benneale]

You cannot do this algebraically. (As in, you physically cannot - there is no mathematical method in existence to do it.)

If the question was given to you like that without any more information, it is a faulty question. If you were told some other information (e.g. use a graph), please provide it.
This looks like the usual method. You should be able to check that the gradient of the tangent at 3 is just \(e^3\), so the normal has gradient \( -e^{-3}\), and then sub into \(y-y_1=m(x-x_1)\).

did all this and still couldn't get to the answer. The book says the answer is x+e^3 y-3-e^6 = 0... how would I be able to get to that? thanks

[RuiAce]

did all this and still couldn't get to the answer. The book says the answer is x+e^3 y-3-e^6 = 0... how would I be able to get to that? thanks

\begin{align*} y - e^3 &=- \frac{1}{e^3} (x-3)\\ e^3(y - e^3) &= -(x - 3)\\ e^3 y - e^6 &= -x + 3\\ x + e^3 y - 3 - e^6 &= 0 \end{align*}