Continuity & Differentiability NCERT Solutions
NCERT Exercise 5.6
If `x` and `y` are connected parametrically by the equations given in Exercise-1 to 10 without eliminating the parameter, find `(dy)/dx`
Question:(1) ` x=2at^2, y=at ^4`
Solution:
Here `x= 2at^2` and `y=at^4`
Differentiating both sides w. r. t. `x`, we have
`(dx)/dt =d/dt (2at^2) and (dy)/dt=d/dt(at^4)`
`=>(dx)/dt =4at and (dy)/(dt)=4at^3`
Now `(dy)/dx =((dy)/dt)/((dx)/dt)`
=`(4at^3)/(4 at)=t^2`
Question: (2) `x=a cos theta, y=b cos theta`
Solution:
Let `x=a cos theta and y =b cos theta`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(d theta)`
` =d/(d theta) (a cos theta) and (dy)/(d theta )`
`=d/(d theta)(b cos theta)`
`=>(dx)/(d theta)=-asin theta and (dy)/(d theta)`
` =-b sin theta `
Now` (dy)/(dx) = ((dy)/(d theta))/((dx)/(d theta))`
`(-b sin theta)/(- a sin theta) = b/a`
Question: (3) `x =sin t, y= cos 2 t`
Solution:
Here `x= sin t and y= cos 2 t `
Differentiating both sides w. r. t. `x`, we have
`(dx)/(dt)= d/ dt (sin t) and (dy)/(dt)`
`=d/dt (cos 2 t)`
` => (dx)/(dt) cos t and (dy)/(dt)`
`=-2 sin 2t`
Now `(dy)/(dx) =((dy)/(dt))/((dx)/(dt))`
`(-2sin2t)/(cos t) `
=`(-2xx2 sin t cos t)/(cos t)`
`=-4sin t`
Question: (4) `x=4t, y = 4/t`
Solution:
Here `x=4 and y = 4/t`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(dt) = d/ dt (4t)and (dy)/(dt)`
`= d/(dt)(4/t)`
`=> (dx)/(dt)`
`=4 and (dy)/(dt)`
`=- 4/t^2`
Now `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))`
`=(-4/2)/(t^2/4)`
`= (-4)/(t^2)xx1/4 = -1/t^2`
Question: (5) ` x=cos theta - cos 2 theta`,
`y=sin theta -sin 2 theta`
Solution:
Here `x=cos theta - cos 2 theta and y`
`= sin theta - sin 2 theta`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(d theta)=d/(d theta) (cos theta - cos 2 theta)and (dy)/(d theta)`
` =d/(d theta)(sin theta -sin 2 theta)`
`(dx)/(d theta)`
`= - sin theta + 2 sin 2 theta and (dy)/(d theta)`
`= cos theta -2 cos 2 theta`
Now `(dy)/(d theta)=((dy)/(d theta))/((dx)/(d theta))`
`=(cos theta - 2 cos 2 theta)/(2 sin 2 theta- sin theta)`
Question: (6) `x=a (theta - sin theta),`
`y =a (1+cos theta)`
Solution:
Here `x=a(theta - sin theta) and y`
`y= a (1+ cos theta)`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(d theta)`
`= d/(d theta)[a(theta -sin theta)] and (dy)/(d theta)`
`=d/ (d theta)[a(1+ cos theta)]`
`=a (1-cos theta)`
`-a sin theta`
Now ` (dy)/(dx)`
`((dy)/(d theta))/((dx)/(d theta))`
`=(-a sin theta)/(a(1-cos alpha))`
`=(-a xx 2 sin\ theta/2 cos\ theta/ 2)/(a. 2 sin^2\ theta/2)`
`=-cot\ theta/2`
Question: (7) `x= (sin^3 t)/sqrt(cos 2t) , y= (cos^3 t)/sqrt(cos 2t)`
Solution:
Here `x= (sin^3t)/sqrt(cos 2t)and y = (cos^3t)/sqrt(cos 2t)`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(d theta)`
`= d/(dt)[(sin^3 t)/sqrt(cos 2t)]`
`=(sqrt(cos 2t). d/dt (sin^3t)-sin^3t d/dt (sqrt(cos 2t)))/((sqrt(cos 2t)))`
`=(sqrt(cos 2t).3sin^2t.cos t-sin^3t.1/(2sqrt(cos 2t))-2sin 2t)/(cos 2t)`
`=(3cos 2t.sin^2 t. cos t+ sin^3t.sin 2t)/((cos 2t)^(3//2))`
`=sin^2(3cos t cos 2t+sint.sin2t)/((cos2t)^(3/2))`
`=(sin^2t[2cost cos 2t+( cost cos 2t+ sin t sin 2t)])/((cos 2t)^(3//2))`
`=(sin^2t[2cost cos 2t+ cos(2t-t)])/((cos 2t)^(3//2))`
`=(sin^2t cost[2cos 2t+1])/((cos2t)^(3//2))`
`=(dy)/(dt)=d/(dt) [(cos^3 t)/(sqrt(cos 2t))]`
`=(sqrt(cos 2t)\.d//(dt)(cos^3t)-cos^3t\ d//dt(sqrt(cos 2t)))/((sqrt(cos 2t)^2))`
`=(sqrt(cos 2t).3 cos^2t.(-sin t)-cos^3t\.1/(2sqrt(cos 2t)) xx - 2 sin 2t)/(sqrt(cos 2t)^2`
`=(-3cos 2t cos^2 t sin t + cos^3t sin 2t)/((cos 2t)^(3//2))`
`=(cos^2t[-3sint cos 2t+ cost sin 2t])/((cos 2t)^(3//2))`
`=(cos ^2t[-2sint cos 2t+ cost sin 2t- sin t cos 2t])/((cos 2t)^(3//2))`
`= (cos ^2t[-2 sin t cos 2t+ sint( 2t-t)])/((cos 2t)^(3//2))`
`= (cos^2 t sin t [1-2 cos 2t])/((cos 2t)^(3//2))`
Now `(dy)/(dx)=((dy)//(dt))/((dx)//(dt))`
`=((cos^2 t sin t [1-2 cos 2t])/((cos 2t)^(3//2)))/((sin^2t cost[2cos 2t+1])/((cos2t)^(3//2)))`
`=(cos^2 t sin t [1-2 cos 2t])/(sin^2t cost[2cos 2t+1])`
`=(cost(1-2 cos 2t))/(sint(2cos 2t+1))`
Question: (8) `x= a( cos t + log tan\ t/2), y= a sin t.`
Solution:
Here `x= a( cos t + log tan\ t/2) and y= a sin t`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(dt)`
`= d/(dt)[a(cos t + log tan\ t/2)]`
`=a[-sin t+ 1/(tan\ t//2) d/(dt)(tan\ t/2)]`
`=a[-sin t+ 1/(tan\ t//2) sec^2\ t/2 .d/(dt)(t/2)]`
`= a[-sin t+ (sec^2\ t/2)/(tan\ t/2) xx 1/2]`
`=a [-sin t+\ 1/(2sin\ t/2 cos\ t/2)]`
`=a[-sin t+ 1/(sin t)]`
`=a [(-sin^2 t+1)/(sin t)]`
`= (a cos^2 t)/(sin t)`
`(dy)/(dt)= d/(dt)[a sin t] =a cos t`
now ` (dy)/(dx)=((dy)/(dt))/((dx)/(dt))`
`=(a cos t)/(a(cos^2t)//(sin t))`
`costxx (sin t)/(cos ^2t)= tan t. `
Question: (9) `x = a sec theta, y=b tan theta`
Solution:
Here `x= a sec theta and y= b tan theta`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(d theta)= d/(d theta) (a sec theta)`
`= a sec theta tan theta`
`= (dy)/(d theta)= d/(d theta) (b tan theta)`
`= b sec theta`
Now ` (dy)/(dx)= (dy//d theta)/(dx//(d theta)`
`=(b sec^2 theta)/(a sec theta tan theta)`
`=b/a (sec theta)/(tan theta)`
`= b/a cosec theta`
Question: (10) `x= a(cos theta + sec theta), y`
`a(sin theta -theta cos theta)`
Solution:
Here `x= a(cos theta + sin theta) and y`
` a( sin theta -theta cos theta)`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(d theta)` ` = d/(d theta) [a(cos theta + sec theta)]`
`= a[-sin theta + ( theta\ d/(d theta) sin theta + sin theta\ d/(d theta) (theta))]`
`= a[-sin theta + theta cos theta + sin theta ]= a theta cos theta`
` (dy)/(d theta) = d/ (d theta) [a(sin theta - theta cos theta)]`
`=a[cos theta - ( theta\ d/(d theta)cos theta + cos theta\ d/(d theta)(theta))]`
`= [cos theta -(-theta sin theta + cos theta)]`
`= a[cos theta + theta sin theta - cos theta]`
Now `(dy)/(dx)= (dy//d theta)/(dx//(d theta)`
`= (a theta sin theta)/(a theta cos theta)= tan theta`
Question: (11) `x= sqrt(a^(sin-1)t), y= sqrt(a^(cos-1)t)` , show that `(dy)/(dx)= - y/x`
Solution:
Here `x= sqrt(a^(sin-1)t), y= sqrt(a^(cos-1)t)`
Differentiating both sides w. r. t. `x`, we have
`(dx)/(dt)` ` = d/(dt) [sqrt(a^(sin-1)t)]`
`=1/(2sqrt(a^(sin-1)t)) . d/ (dt) (a^(sin-1)t)`
`=1/(2sqrt(a^(sin-1)t)) a^(sin-1)t.log a . d/(dt) (sin ^(-1)t)`
`=1/(2sqrt(a^(sin-1)t)) a^(sin-1)t.log a. 1/(sqrt(1-t^2)`
`=(sqrt(a^(sin-1)t).log a)/(2sqrt(1-t^2)`
`=(x log a)/(2sqrt(1-t^2))`
`(dy)/(dt)` ` = d/(dt) [sqrt(a^(cos-1)t)]`
`= 1/(2sqrt(a^(cos^-1t)))\ a ^(cos^(-1)t)*log a d/(dt)(cos^-1)`
`= 1/(2sqrt(a^(cos^-1t)))\ a ^(cos^(-1)t)*log a -1/sqrt(1-t^2)`
`=(-sqrt(a^(cos-1)t)log a)/(2sqrt(1-t^2)`
`=(-y loga)/(2sqrt(1-t^2))`
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