Question

A curve is such that dy/dx= 2Cos(2x-π/2). The curve passes through the point (π/2,3). Find the equation of the curve & the equation of the normal to the curve at the pt. where x=3π/4.

Answer 1

To get the equation of the curve you need to integrate that function with respect to x.
\[\int\limits_{}^{}2\cos(2x-\frac{\pi}{2})dx= sin(2x-\frac{\pi}{2})+c\]
Knowing that the point (pi/2, 3) is on the curve, and substituting those values in the equation we can figure out what c is:
\[3= \sin(2(\frac{\pi}{2})-\frac{\pi}{2})+c \Rightarrow 3 = \sin(\frac{\pi}{2})+c\Rightarrow 3 = 1+c\]
So c is 2, thus the equation of the curve must be:
\[f(x) = \sin(2x-\frac{\pi}{2})+ 2\]

Answer 2

You can integrate it easily.
Then you have:
\[\sin(2x-\frac{\pi}{2})+C\]
Using the point you can solve for C.
You can plug in your x value into dy/dx, get your slope. Thats the slope of the tangent. That tells you that the slope of the normal is perpendicular or the negative reciprocal of it. From there, you have a slope and a point, you can use point slope form to get a line.
PSF:
\[y-y_0=m(x-x_0)\]
:)

Answer 3

Thank you. Can you tell me the equation of the normal to the curve at the pt. where x=3π/4 ?

Answer 4

Well, plugging in 3pi/4 you get that m=-2. Taking the negative reciprocal you have m=1/2. So you have
y-3=(1/2)(x-pi/2)
y-3=(1/2)x-pi/4
y=(1/2)x-pi/4-3

Answer 5

Where do you 'plug' in 3π/4 ?

Answer 6

Into x for dy/dx
So you have:
\[\frac{dy}{dx}|_{x=\frac{3\pi}{4}}=2\cos(2(\frac{3\pi}{4})-\frac{\pi}{2})=2\cos(\pi)=2(-1)=-2\]

Answer 7

Got the answer!! Thank you so much! :)

Answer 8

No problem :)