---
title: "Directional derivative of xy²+yz³"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:directional-derivative-xy2-yz3"
source: https://www.bctengineer.com/q/semester_2/mathematics/directional-derivative-xy2-yz3
site: BCT Engineering
---

# Directional derivative of xy²+yz³

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2074 Bhadra, 5 marks; 2073 Bhadra, 5 marks; 2072 Magh, 5 marks (asked 3 times)
- Video: [f(x,y,z)=xy2+yz3 Directional Derivative at point (2,-1,1) GATE 2019 Q-44](https://www.youtube.com/watch?v=ZVpT84LDNTI) — Agricultural Engineering Online
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/directional-derivative-xy2-yz3
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Find the directional derivative of φ(x,y,z) = xy² + yz³ at the point (2,−1,1) in the direction of vector i+2j+2k.

## Short answer

Gradient ∇φ = (y², 2xy+z³, 3yz²) yields (1, -3, -3) at (2,-1,1). Unit direction d̂ = (i+2j+2k)/3. Directional derivative = ∇φ·d̂ = -11/3.

## Answer

The directional derivative of φ at a point in the direction of a vector d is ∇φ · d̂, where d̂ is the unit vector along d. It is the rate of change of φ per unit distance in that direction.

#### Gradient
φ = xy² + yz³

```text
∂φ/∂x = y²
∂φ/∂y = 2xy + z³
∂φ/∂z = 3yz²
```
At (2, −1, 1):

```text
∂φ/∂x = 1
∂φ/∂y = 2(2)(−1) + 1 = −3
∂φ/∂z = 3(−1)(1) = −3
∇φ = i − 3j − 3k
```

#### Unit vector
d = i + 2j + 2k, |d| = √(1 + 4 + 4) = 3, so d̂ = (i + 2j + 2k)/3.

#### Directional derivative
```text
∇φ · d̂ = (1·1 + (−3)(2) + (−3)(2))/3 = (1 − 6 − 6)/3 = −11/3
```

**Answer: the directional derivative is −11/3 ≈ −3.67** (φ decreases in that direction).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The directional derivative along d is:
   A. ∇φ · d
   B. ∇φ · d̂ with d̂ = d/|d|
   C. ∇φ × d̂
   D. |∇φ| |d|
   Answer: B. The unit vector is needed for rate per unit distance. (Key point: Formula)

2. ∇φ at (2, −1, 1) for φ = xy^2 + yz^3 is:
   A. i − 3j − 3k
   B. i + 3j − 3k
   C. i − 3j + 3k
   D. −i − 3j − 3k
   Answer: A. φy = 2xy + z^3 = −4 + 1 = −3, φz = 3yz^2 = −3, φx = y^2 = 1. (Key point: Gradient)

3. For d = i + 2j + 2k, |d| equals:
   A. 9
   B. 5
   C. √5
   D. 3
   Answer: D. sqrt(1+4+4) = 3. (Key point: Unit vector)

4. The directional derivative is:
   A. 11/3
   B. −11
   C. −11/3
   D. −3
   Answer: C. (1 − 6 − 6)/3 = −11/3; φ decreases in that direction. (Key point: Result)

## Close variants set in other exams

- [Directional derivative of x²yz+4xz²](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Adirectional-derivative-x2yz-4xz2) — Mathematics, chapter 5, asked 4×

## Related in Mathematics

- [Reciprocal system of 2i−3j+k, i+2j−k, 3i−j+2k](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Areciprocal-system-2i-3j-k) — Mathematics, chapter 5, asked 4×
- [Reciprocal system of 2i+3j−k, i−j−2k, −i+2j+2k](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Areciprocal-system-2i-3j-k-2) — Mathematics, chapter 5, asked 2×
- [Condition for constant direction of vector](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acondition-constant-direction-vector) — Mathematics, chapter 5, asked 6×

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