Skip to main content
Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.8.9
Chapter 3, Problem 3.8.9

If L = √(x² + y²), dx/dt = –1, and dy/dt = 3, find dL/dt when x = 5 and y = 12.

Verified step by step guidance
1
First, recognize that L is a function of x and y, specifically L = √(x² + y²). This is the formula for the length of the hypotenuse in a right triangle with legs x and y.
To find dL/dt, use implicit differentiation with respect to time t. Start by differentiating both sides of the equation L = √(x² + y²) with respect to t.
Apply the chain rule to differentiate L = √(x² + y²). The derivative of L with respect to t is (1/2) * (1/√(x² + y²)) * (2x * dx/dt + 2y * dy/dt).
Substitute the given values into the differentiated equation: dx/dt = -1, dy/dt = 3, x = 5, and y = 12.
Simplify the expression to find dL/dt. Remember to substitute the values into the expression and simplify step by step, ensuring each part of the equation is correctly calculated.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that if a variable z depends on y, which in turn depends on x, then the derivative of z with respect to x can be found by multiplying the derivative of z with respect to y by the derivative of y with respect to x. In this problem, it helps in finding dL/dt by relating the derivatives of L with respect to x and y.
Recommended video:
05:02
Intro to the Chain Rule

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. It is particularly useful when dealing with equations involving multiple variables. In this scenario, implicit differentiation allows us to differentiate L = √(x² + y²) with respect to time, t, by considering the derivatives dx/dt and dy/dt.
Recommended video:
05:14
Finding The Implicit Derivative

Pythagorean Theorem

The Pythagorean theorem is a geometric principle that relates the lengths of the sides of a right triangle. It states that the square of the hypotenuse (L) is equal to the sum of the squares of the other two sides (x and y). In this problem, L = √(x² + y²) represents the hypotenuse, and understanding this relationship is crucial for setting up the equation to find dL/dt.
Recommended video:
06:11
Fundamental Theorem of Calculus Part 1
Related Practice
Textbook Question

Derivatives in Differential Form


In Exercises 17–28, find dy.


y = (2√x)/(3(1 + √x))

239
views
Textbook Question

Slopes and Tangent Lines


In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points P₁ and P₂.

226
views
Textbook Question

Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form

P = (nRT / (V − nb)) − (an² / V²),

in which a, b, n, and R are constants. Find dP/dV. (See accompanying figure.)


<IMAGE>

184
views
Textbook Question

Find the derivatives of the functions in Exercises 19–40.


g(t) = (1 + sin(3t) / (3 − 2t))⁻¹

253
views
Textbook Question

Finding Linearizations


In Exercises 1–5, find the linearization L(x) of f(x) at x = a.


f(x) = ∛x, a = −8

204
views
Textbook Question

Derivatives in Differential Form


In Exercises 17–28, find dy.


xy² − 4x³/² − y = 0

186
views