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A gasoline storage tank is in the shape of a righ…

Problem 41 Easy Difficulty

A rectangular solar cell panel has a perimeter $p$ and a width $w$ Express the area $A$ of the panel in terms of $p$ and $w$ and evaluate the area for $p=250 \mathrm{cm}$ and $w=55 \mathrm{cm} .$ See Fig. 29.6.

Answer

$A=\frac{p w-2 w^{2}}{2}, 3850 \mathrm{cm}^{2}$

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Basic Technical Mathematics with Calculus

Chapter 29

Partial Derivatives and Double Integrals

Section 1

Functions of Two Variables

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Video Transcript

31 is talking about some solar panels, and I is the intensity of the sun rays and A is the area of the solar panel. Part A says Fine. The magnitude of I. So we're gonna crank that out. So we're gonna say square root of negative 0.2 squared, plus negative point no. One squared. Keep going there and you'll get square root of point. Oh 05 Which ends up being 0.0 to to And then it says, What does that mean? This is the intensity of the sun's rays that's measured in watts per centimeter squared the intensity of the sun's rays. Okay, the next part was magnitude of a so square root of 300 squared plus 400 square. That's a 345 triangle. Um, so it should be 500. Uh, if you crank it all out, you will get 500. And that represents the area of the solar panel that's measured in centimeters squared. Part B says to find W, which is the absolute value of I dotted with A. So we're gonna do the absolute value of my dog with a so to be negative 0.2 Time 300 plus negative point. +01 times 400. So crank out the math, you get negative six plus negative four inside the absolute value. So that becomes negative. 10 in the absolute value. So you get to and this means 10 watts of energy collected. Okay, Parsi says, When is a maximum energy collected? That happens when the solar panels are facing the sun when panels face the sun and when I and A are parallel, okay?

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Problem

A gasoline storage tank is in the shape of a righ…

Problem 41 Easy Difficulty

A rectangular solar cell panel has a perimeter $p$ and a width $w$ Express the area $A$ of the panel in terms of $p$ and $w$ and evaluate the area for $p=250 \mathrm{cm}$ and $w=55 \mathrm{cm} .$ See Fig. 29.6.

Answer

$A=\frac{p w-2 w^{2}}{2}, 3850 \mathrm{cm}^{2}$

Related Courses

Basic Technical Mathematics with Calculus

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Watch More Solved Questions in Chapter 29

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Caleb E.

Kristen K.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

$A=\frac{p w-2 w^{2}}{2}, 3850 \mathrm{cm}^{2}$

Basic Technical Mathematics with Calculus

Caleb E.

Kristen K.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Caleb E.

Kristen K.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus