The Student Room Group

C3 Solomon A help!

Hey guys I'm just going through the uploaded question. I've managed part A, but on part B I'm unsure of what the question is asking. So far I have differentiated the function y and tried to work out the gradient when x has a value of pi/4 but I'm getting a weird decimal answer :s-smilie: I think the question is asking me to work out the equation of the tangent to the curve and then sub in the values of when x=0 to work out the y intercept? Is this correct, and if not what am I doing wrong :frown: Thanks!
Reply 1
Original post by Bugsy
Hey guys I'm just going through the uploaded question. I've managed part A, but on part B I'm unsure of what the question is asking. So far I have differentiated the function y and tried to work out the gradient when x has a value of pi/4 but I'm getting a weird decimal answer :s-smilie: I think the question is asking me to work out the equation of the tangent to the curve and then sub in the values of when x=0 to work out the y intercept? Is this correct, and if not what am I doing wrong :frown: Thanks!


Yes that is what it is asking, what have you got for the equation of the tangent?
Reply 2
Original post by Bugsy
Hey guys I'm just going through the uploaded question. I've managed part A, but on part B I'm unsure of what the question is asking. So far I have differentiated the function y and tried to work out the gradient when x has a value of pi/4 but I'm getting a weird decimal answer :s-smilie: I think the question is asking me to work out the equation of the tangent to the curve and then sub in the values of when x=0 to work out the y intercept? Is this correct, and if not what am I doing wrong :frown: Thanks!


Find the equation of the tangent at x=π4 x = \dfrac{\pi}4

Then sub in x=0 to get the answer.
Reply 3
Original post by TheJ0ker
Yes that is what it is asking, what have you got for the equation of the tangent?


I haven't managed to get an equation for the tangent yet, because I can't work out a value for m :s-smilie: I tried to sub in pi/4 into my dy/dx but like I said I get a long decimal which I don't think is correct.. Once I get a M value I can use the y-y1=m(x-x1) to find the equation for the tangent..

Original post by raheem94
Find the equation of the tangent at x=π4 x = \dfrac{\pi}4

Then sub in x=0 to get the answer.


Yeah this is what I'm having problems with.. For my dy/dx I got 2x.Sec^2 x + 2tanx
Reply 4
Original post by Bugsy
I haven't managed to get an equation for the tangent yet, because I can't work out a value for m :s-smilie: I tried to sub in pi/4 into my dy/dx but like I said I get a long decimal which I don't think is correct.. Once I get a M value I can use the y-y1=m(x-x1) to find the equation for the tangent..



Yeah this is what I'm having problems with.. For my dy/dx I got 2x.Sec^2 x + 2tanx


Don't just plug the numbers into a calculator then. If you do it using your grey matter you will get a simple gradient involving an integer and pi.
Reply 5
Original post by Bugsy

Yeah this is what I'm having problems with.. For my dy/dx I got 2x.Sec^2 x + 2tanx


Sub in x=π4 x = \dfrac{\pi}4

dydx=2×π4×sec2π4+2tanπ4=π2×2+2=2+π \displaystyle \frac{dy}{dx} = 2 \times \frac{\pi}4 \times sec^2\frac{\pi}{4} + 2 tan \frac{\pi}4 = \frac{\pi}2 \times 2 + 2 = 2 + \pi

It is an exact value.
Reply 6
Original post by TheJ0ker
Don't just plug the numbers into a calculator then. If you do it using your grey matter you will get a simple gradient involving an integer and pi.


Original post by raheem94
Sub in x=π4 x = \dfrac{\pi}4

dydx=2×π4×sec2π4+2tanπ4=π2×2+2=2+π \displaystyle \frac{dy}{dx} = 2 \times \frac{\pi}4 \times sec^2\frac{\pi}{4} + 2 tan \frac{\pi}4 = \frac{\pi}2 \times 2 + 2 = 2 + \pi

It is an exact value.


Ah okay, great I get the answer now! Sorry for being so silly, I should have tried to do it without a calculator to get an exact value.. Thanks for your help :smile:

Quick Reply

Latest