MAT1512 Assignment 3 2025 - Due 18 July 2025

MAT1512
ASSIGNMENT 3 2025
DUE: 18 JULY 2025 (MEMO)

, Assignment 3, - 2025 - MAT1512


July 2, 2025
Question 1
(a) We differentiate implicitly to find y ′ . From y 2 e2x = 3y + x2 , apply the product rule on y 2 e2x
(treating y as y(x)):

d 2 2x d 2 d 2x
[y e ] = (y ) ⋅ e2x + y 2 ⋅ (e ) = (2y y ′ )e2x + y 2 (2e2x ).
​ ​ ​




dx dx dx
d
On the right, dx (3y)
​ = 3y ′ and d
dx
​(x2 ) = 2x. Equating derivatives:

2y e2x y ′ + 2y 2 e2x = 3y ′ + 2x.
Collect y ′ terms:

(2ye2x − 3)y ′ = 2x − 2y 2 e2x ,

so

2x − 2y 2 e2x
y′ = .
2ye2x − 3
​




At the point (x, y) = (0, 3), e2x = 1, so y ′ = (0 − 2 ⋅ 9)/(2 ⋅ 3 − 3) = −18/3 = −6. The
tangent line has slope m = −6 and passes through (0, 3), so its equation is y − 3 = −6(x − 0),
i.e. y = −6x + 3. The normal line is perpendicular to this tangent, so its slope is the negative
reciprocal, mnormal = 1/6
​ . Thus the normal line is y − 3 = 61 (x − 0), or y = 16 x + 3
​ ​




.

(b) Differentiate π sin y + 2xy = 2π implicitly:

π cos y y ′ + 2(x ⋅ y ′ + y ⋅ 1) = 0 ⟹ (π cos y + 2x) y ′ + 2y = 0.

Solve for y ′ :

−2y
y′ = .
π cos y + 2x
​