Question

if x=a (cost+tsint), y=a (sint-tcost) then find dy/dx​

Answer 1

GIVEN :

If x=a (cost+tsint), y=a (sint-tcost) then find $\frac{dy}{dx}$

TO FIND :

The value of  $\frac{dy}{dx}$

SOLUTION :

Given that the values are  x=a (cost+tsint), y=a (sint-tcost)

Since the given values of x and y are in terms of t we have that

$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}$

$\frac{dy}{dx}=\frac{dy}{dt}.\frac{1}{\frac{dx}{dt}}$

y=a (sint-tcost)

Differentiating y with respect to t we get

$\frac{dy}{dt}=a[cost-d(tcost)]$

By using the differentiation formulae  :

i) d(cost)=-sint

ii) d(sint)=cost

iii) $d(uv)=uv^{\prime}+u^{\prime}v$

 

$=a[cost-(t(-sint)+cost(1))]$

$=a[cost+tsint-cost]$

$=atsint$

$\frac{dy}{dt}=atsint$

x=a (cost+tsint)

Differentiating x with respect to t we get

$\frac{dx}{dt}=a[sint+d(tsint)]$

By using the differentiation formulae  :

i) d(cost)=-sint

ii) d(sint)=cost

iii) $d(uv)=uv^{\prime}+u^{\prime}v$

 

$=a[sint+(t(cost)+sint(1))]$

$=a[sint+tcost-sint]$

$=atcost$

$\frac{dx}{dt}=atcost$

Now $\frac{dy}{dx}=atsint(\frac{1}{\atcost})$

$=\frac{sint}{cost}$

By using the trignometric formula :

$\frac{sinx}{cosx}=tanx$

$=tant$

$\frac{dy}{dx}=tant$

∴ the value is $\frac{dy}{dx}=tant$