For f(x, y) = xe2y, find the partial derivative ∂f/∂y at the point (1, 0.5).
Multivariable calculus worksheet with answers
1.Easy
f(x, y) = xe2yP = (1, 0.5)
fy(1, 0.5) =
2.Easy
Evaluate I = ∫25 ∫12 (2x·ey) dy dx over the rectangle 2 ≤ x ≤ 5, 1 ≤ y ≤ 2.
f = 2x·ey2 ≤ x ≤ 51 ≤ y ≤ 2
I =
3.Medium
The tangent plane to z = y·e3xy at the point where (x, y) = (0.5, 0.5) can be written as z = px + qy + c. Find c.
f(x, y) = y·e3xy(a, b) = (0.5, 0.5)
c =
4.Medium
z = −xy + x³ + 5y³ with x = t − 2 and y = 2t² − 2. Use the chain rule to find dz/dt at t = 0.
z = −xy + x³ + 5y³x = t − 2y = 2t² − 2t = 0
dz/dt =
5.Hard
The marked points are the critical points of f(x, y) = x³ − 3x + y². Click the saddle point.
f(x, y) = x³ − 3x + y²
- 1A (1, 0)
- 2B (−1, 0)
6.Hard
Use polar coordinates to evaluate I = ∬R e−(x² + y²) dA, where R is the quarter of the disc x² + y² ≤ 4 in the first quadrant.
f = e−(x² + y²)0 ≤ r ≤ 20 ≤ θ ≤ π/2
I =
Answer key
- 1. fy(1, 0.5) = 5.437
- 2. I = 98.09
- 3. c = −1.588
- 4. dz/dt = 14
- 5. B (−1, 0)
- 6. I = 0.771
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