Applied Calculus: UNIT II: Functions of Several Variables
Partial derivatives: Problems under type-1
Explanation, Formula, Equation, Example and Solved Problems - Partial derivatives: Problems under type-1
PROBLEMS UNDER TYPE 1
Let u= f(x,y).
If x and y are functions of one independent variable t, then
du/dt = (∂u/∂x).(dx/dt) + (∂u/∂y).(dy/dt)
Example
20. Find dz/dt if z = x2 +
y2, x = at2, y = 2at
Solution:
Given z = x2 + y2
∂z∂x =2x;
∂z∂y =2y;
Also given x = at2;
y = 2at
dx/dt
= 2at;
dy/dt
= 2a;
dz/dt = (∂z/∂x).(dx/dt) + (∂z/∂y).(dy/dt)
= 2x 2at + 2y 2a
= 4a[xt+ya]
= 4a[(at2)t + (2at)a]
dz/dt = 4a[at3
+ 2a2t]
Example
21. If u = x3+ y3, where x = a cost, y = a sint,
find du/dt and verify the result.
Solution:
du/dt = (∂u/∂x).(dx/dt) + (∂u/∂y).(dy/dt)
= 3x2(‒a sin t) + 3y2a cos t
= 3(a cost)2(‒a
sin t) + 3(a sin t)2 (a cost)
= ‒3a3 cos2t
sint + 3a3 sin2t cost
du/dt = 3a3
sint cost (sin t ‒ cost)
To verify the result
u = x3 + y3
= a3cos3t + a3sin3t
du/dt = a33cos2t (‒sint) + a33sin2t cost
du/dt
= 3a3 cost sint (sint‒cost)
Hence the result is
verified.
Example
22. If u = log(x + y + z), where x = e‒t, y =
sint, z = cost, then Find du/dt.
Solution:
Example
23. Find du/dt if u = exy, where x = √[a2‒t2], y = sin3t
Solution:
Given u = exy
Example
24. If u = xy + yz + zx, where x = et, y = e‒t,
z= 1/t find du/dt.
Solution:
Example
25. Given u = sin(x/y), x = et
and y = t2.
Find
du/dt as a function of t. Verify your result by direct substitution.
Solution:
Applied Calculus: UNIT II: Functions of Several Variables : Tag: Applied Calculus : - Partial derivatives: Problems under type-1
Applied Calculus: UNIT II: Functions of Several Variables
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Applied Calculus
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