Differentiation

To find the length of the tangent at the point P(‌

π

4

) on the curve x2∕3+y2∕3=22∕3, we can proceed as follows:
The given equation of the curve can be parameterized as:
x=2cos3θ,‌‌y=2sin‌3θ
For the point P, we substitute θ=‌

π

4

:
x=2cos3‌

π

4

,‌‌y=2sin‌3‌

π

4

Calculating the coordinates:
x=2(‌

1

√2

)3=‌

2

2√2

=‌

1

√2

,‌‌y=2(‌

1

√2

)3=‌

2

2√2

=‌

1

√2

So the point is (‌

1

√2

,‌

1

√2

).
Next, differentiate the curve equation x2∕3+y2∕3=22∕3 with respect to x :
‌

2

3

x−1∕3+‌

2

3

y−1∕3‌

dy

dx

=0
Solving for ‌

dy

dx

:
‌

dy

dx

=−(‌

y

x

)1∕3
At the point (‌

1

√2

,‌

1

√2

),
‌

dy

dx

=−1
The formula to find the length of the tangent at a point is:
L=y√1+(‌

dx

dy

)2
Given ‌

dx

dy

=−1, the length of the tangent at the point is calculated as:
L=‌

1

√2

√1+(−1)2=‌

1

√2

×√2=1
Thus, the length of the tangent is 1 .