2. ## Re: Please someone could you mark and correct my solutions if I made a mistake

2a) You calculated the gradient of the curve only at the given point, and assumed it (the gradient) was the same for all other points on the curve.

The question asked you to find the function which has gradient function ${dy}/{dx} = \cos 2x$ for all values of $x$ (in the domain), and which also passes through $(\pi/2,\, 1)$.

The linear equation you gave for your answer has gradient function ${dy}/{dx} = \cos 2x$ at $(\pi/2,\, 1)$ but not necessarily for any other point on the curve.

You should start by integrating the gradient function and then use the given point to find the value of the constant of integration. The answer should be $y = 1/2 \,\sin 2x + 1$.

2c) Between your 7th and 8th lines, note that $-1/2 \, \cos(240^\circ) \neq -1/2 \, \cos(60^\circ)$. Rather,$-1/2 \, \cos(240^\circ) = 1/2 \, \cos(60^\circ)$ . As a result, the negative sign in your 9th line should be a positive, giving $A = 9/4\, \mathrm{u}^2$

3. ## Re: Please someone could you mark and correct my solutions if I made a mistake

oki imma do every single question and send u a cheat sheet so you can cgeck with the answers

4. ## Re: Please someone could you mark and correct my solutions if I made a mistake

Originally Posted by HeroWise
oki imma do every single question and send u a cheat sheet so you can cgeck with the answers
Thank you so so much

5. ## Re: Please someone could you mark and correct my solutions if I made a mistake

https://imgur.com/a/wYrHX2J

All the solutions are here
someof them require u to rotate them,

6. ## Re: Please someone could you mark and correct my solutions if I made a mistake

Originally Posted by HeroWise
https://imgur.com/a/wYrHX2J

All the solutions are here
someof MOST of them require u to rotate them,
LOL

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