Partial Derivative with Respect to Time

Q 1.
Answer
T = 24e−0.5t cos (0.4x) + te−0.2t sin x
1.
Partial derivative with respect to time
∂T
∂t = cos (0.4x) ∂
∂t (24e−0.5t) + sin x ∂
∂t (te−0.2t)
[Applying Product rule where needed]
= cos (0.4x)(−12e−0.5t) + sin x(e(−0.2t)(1) + t(−0.2e−0.2t)
= −12e−0.5t cos (0.4x) + sin xe−0.2t(1 − 0.2t)
At x = 0 and t = 1.9
∂T
∂t
∣
∣
∣x=0,t=1.9 = −12e−0.5×1.9 cos (0) + sin (0)e−0.2×1.9(1 − 0.2 × 1.9)
= −12e−0.95 = −4.641
2.
Partial derivative with respect to time represents the temporal rate of change
of temperature at a location. x = 0 is a boundary location. The solution
for T (x, t) indicates that temperature at the boundary x = 0 is function of
1

time and the partial derivative of T (w.r.t. time) indicates that boundary
temperature is decreasing with time.
Answer
1.
y dy
dx = e− 2x
19 + cos ( 0.1x
19
)
y
Taking all the y terms to the left side and all the x terms to the right
y2dy = e− 2x
19 + cos
( 0.1x
19
)
dx
[Integrating both sides]
∫
y2dy =
∫
e− 2x
19 + cos
( 0.1x
19
)
dx
y3
3 = − 19
2 e− 2x
19 + 190 sin
( 0.1x
19
)
+ C′
∴ y = 3
√
−28.5e− 2x
19 + 570 sin
( 0.1x
19
)
+ C
2.
2