Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Sure, I'll guide you through solving the two problems step-by-step.
### Problem 8: Finding [tex]\( \frac{dy}{dx} \)[/tex] using implicit differentiation
Given the equation:
[tex]\[ x^3 + 3x^2y + y^3 = 8 \][/tex]
1. Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(x^3 + 3x^2y + y^3) = \frac{d}{dx}(8) \][/tex]
The derivative of the right-hand side, a constant, is 0:
[tex]\[ \frac{d}{dx}(8) = 0 \][/tex]
2. Apply the product rule and chain rule to the left-hand side:
- The derivative of [tex]\( x^3 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 3x^2 \)[/tex].
- For [tex]\( 3x^2y \)[/tex], use the product rule:
[tex]\[ \frac{d}{dx}(3x^2y) = 3x^2\frac{dy}{dx} + 6xy \][/tex]
- The derivative of [tex]\( y^3 \)[/tex] with respect to [tex]\( x \)[/tex] using the chain rule is:
[tex]\[ \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \][/tex]
3. Combine these derivatives:
[tex]\[ 3x^2 + 3x^2\frac{dy}{dx} + 6xy + 3y^2\frac{dy}{dx} = 0 \][/tex]
4. Collect the terms involving [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 3x^2 + 6xy + \left(3x^2 + 3y^2\right) \frac{dy}{dx} = 0 \][/tex]
5. Isolate [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 6xy + 3x^2 = - \left(3x^2 + 3y^2\right) \frac{dy}{dx} \][/tex]
[tex]\[ 6xy + 3x^2 = - 3(x^2 + y^2) \frac{dy}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{-(6xy + 3x^2)}{- 3(x^2 + y^2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{6xy + 3x^2}{3(x^2 + y^2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{x^2 + 2xy}{x^2 + y^2} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\frac{x^2 + 2xy}{x^2 + y^2}} \][/tex]
Neither of the choices A or B match the calculation exactly, so double-checking and correcting one of the computations or choices might be necessary.
### Problem 9: Find the linearization [tex]\( L(x) \)[/tex] of [tex]\( f(x) \)[/tex] at [tex]\( x=a \)[/tex]
Given:
[tex]\[ f(x) = 2x^2 + 5x + 5 \][/tex]
[tex]\[ a = 3 \][/tex]
1. Find [tex]\( f'(x) \)[/tex], the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(2x^2 + 5x + 5) \][/tex]
[tex]\[ f'(x) = 4x + 5 \][/tex]
2. Evaluate [tex]\( f(a) \)[/tex] and [tex]\( f'(a) \)[/tex]:
[tex]\[ f(3) = 2(3)^2 + 5(3) + 5 \][/tex]
[tex]\[ f(3) = 2 \cdot 9 + 15 + 5 \][/tex]
[tex]\[ f(3) = 18 + 15 + 5 \][/tex]
[tex]\[ f(3) = 38 \][/tex]
[tex]\[ f'(3) = 4(3) + 5 \][/tex]
[tex]\[ f'(3) = 12 + 5 \][/tex]
[tex]\[ f'(3) = 17 \][/tex]
3. Form the linearization [tex]\( L(x) \)[/tex]:
The linearization at [tex]\( x = a \)[/tex] is given by:
[tex]\[ L(x) = f(a) + f'(a)(x - a) \][/tex]
[tex]\[ L(x) = 38 + 17(x - 3) \][/tex]
4. Simplify [tex]\( L(x) \)[/tex]:
[tex]\[ L(x) = 38 + 17x - 51 \][/tex]
[tex]\[ L(x) = 17x - 13 \][/tex]
So, the linearization [tex]\( L(x) \)[/tex] at [tex]\( x = 3 \)[/tex] is:
[tex]\[ \boxed{L(x) = 38} \][/tex]
If should you not simplify it.
### Problem 8: Finding [tex]\( \frac{dy}{dx} \)[/tex] using implicit differentiation
Given the equation:
[tex]\[ x^3 + 3x^2y + y^3 = 8 \][/tex]
1. Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(x^3 + 3x^2y + y^3) = \frac{d}{dx}(8) \][/tex]
The derivative of the right-hand side, a constant, is 0:
[tex]\[ \frac{d}{dx}(8) = 0 \][/tex]
2. Apply the product rule and chain rule to the left-hand side:
- The derivative of [tex]\( x^3 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 3x^2 \)[/tex].
- For [tex]\( 3x^2y \)[/tex], use the product rule:
[tex]\[ \frac{d}{dx}(3x^2y) = 3x^2\frac{dy}{dx} + 6xy \][/tex]
- The derivative of [tex]\( y^3 \)[/tex] with respect to [tex]\( x \)[/tex] using the chain rule is:
[tex]\[ \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \][/tex]
3. Combine these derivatives:
[tex]\[ 3x^2 + 3x^2\frac{dy}{dx} + 6xy + 3y^2\frac{dy}{dx} = 0 \][/tex]
4. Collect the terms involving [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 3x^2 + 6xy + \left(3x^2 + 3y^2\right) \frac{dy}{dx} = 0 \][/tex]
5. Isolate [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 6xy + 3x^2 = - \left(3x^2 + 3y^2\right) \frac{dy}{dx} \][/tex]
[tex]\[ 6xy + 3x^2 = - 3(x^2 + y^2) \frac{dy}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{-(6xy + 3x^2)}{- 3(x^2 + y^2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{6xy + 3x^2}{3(x^2 + y^2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{x^2 + 2xy}{x^2 + y^2} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\frac{x^2 + 2xy}{x^2 + y^2}} \][/tex]
Neither of the choices A or B match the calculation exactly, so double-checking and correcting one of the computations or choices might be necessary.
### Problem 9: Find the linearization [tex]\( L(x) \)[/tex] of [tex]\( f(x) \)[/tex] at [tex]\( x=a \)[/tex]
Given:
[tex]\[ f(x) = 2x^2 + 5x + 5 \][/tex]
[tex]\[ a = 3 \][/tex]
1. Find [tex]\( f'(x) \)[/tex], the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(2x^2 + 5x + 5) \][/tex]
[tex]\[ f'(x) = 4x + 5 \][/tex]
2. Evaluate [tex]\( f(a) \)[/tex] and [tex]\( f'(a) \)[/tex]:
[tex]\[ f(3) = 2(3)^2 + 5(3) + 5 \][/tex]
[tex]\[ f(3) = 2 \cdot 9 + 15 + 5 \][/tex]
[tex]\[ f(3) = 18 + 15 + 5 \][/tex]
[tex]\[ f(3) = 38 \][/tex]
[tex]\[ f'(3) = 4(3) + 5 \][/tex]
[tex]\[ f'(3) = 12 + 5 \][/tex]
[tex]\[ f'(3) = 17 \][/tex]
3. Form the linearization [tex]\( L(x) \)[/tex]:
The linearization at [tex]\( x = a \)[/tex] is given by:
[tex]\[ L(x) = f(a) + f'(a)(x - a) \][/tex]
[tex]\[ L(x) = 38 + 17(x - 3) \][/tex]
4. Simplify [tex]\( L(x) \)[/tex]:
[tex]\[ L(x) = 38 + 17x - 51 \][/tex]
[tex]\[ L(x) = 17x - 13 \][/tex]
So, the linearization [tex]\( L(x) \)[/tex] at [tex]\( x = 3 \)[/tex] is:
[tex]\[ \boxed{L(x) = 38} \][/tex]
If should you not simplify it.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
