Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
To solve for the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], we will follow a few steps involving the given conditions. Let's break it down:
### Step 1: Understanding the Gradient at a Point
The gradient of the curve at any point [tex]\(x\)[/tex] is given as:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
### Step 2: Condition at [tex]\( (2, -16) \)[/tex]
The tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis. This implies that the gradient at [tex]\(x = 2\)[/tex] is 0. Therefore:
[tex]\[ \left. \frac{dy}{dx} \right|_{x = 2} = 0 \][/tex]
Substituting [tex]\(x = 2\)[/tex] into the gradient expression:
[tex]\[ a (2)^2 + b = 0 \][/tex]
[tex]\[ 4a + b = 0 \][/tex]
So, we have our first equation:
[tex]\[ b = -4a \][/tex]
### Step 3: Integrating the Gradient
To find the curve [tex]\(y\)[/tex] which passes through the given points, we need to integrate the gradient function:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
Integrate with respect to [tex]\(x\)[/tex]:
[tex]\[ y = \int (a x^2 + b) \, dx \][/tex]
[tex]\[ y = \int a x^2 \, dx + \int b \, dx \][/tex]
[tex]\[ y = \frac{a}{3} x^3 + bx + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
### Step 4: Applying the Given Point [tex]\((1, -11)\)[/tex]
The curve passes through the point [tex]\((1, -11)\)[/tex], so we substitute [tex]\(x = 1\)[/tex] and [tex]\(y = -11\)[/tex] into the equation of the curve:
[tex]\[ -11 = \frac{a}{3} (1)^3 + b(1) + C \][/tex]
[tex]\[ -11 = \frac{a}{3} + b + C \][/tex]
Using [tex]\( b = -4a \)[/tex], substitute [tex]\(b\)[/tex] into the equation:
[tex]\[ -11 = \frac{a}{3} - 4a + C \][/tex]
[tex]\[ -11 = \frac{a}{3} - \frac{12a}{3} + C \][/tex]
[tex]\[ -11 = -\frac{11a}{3} + C \][/tex]
Multiplying through by 3:
[tex]\[ -33 = -11a + 3C \][/tex]
[tex]\[ 3C = -33 + 11a \][/tex]
[tex]\[ C = -11 + \frac{11a}{3} \][/tex]
[tex]\[ C = -11 + \frac{11a}{3} \][/tex]
### Step 5: Applying the Given Point [tex]\((2, -16)\)[/tex]
The curve also passes through the point [tex]\((2, -16)\)[/tex], so we substitute [tex]\(x = 2\)[/tex] and [tex]\(y = -16\)[/tex] into the equation of the curve:
[tex]\[ -16 = \frac{a}{3} (2)^3 + b(2) + C \][/tex]
[tex]\[ -16 = \frac{8a}{3} + 2b + C \][/tex]
Using [tex]\( b = -4a \)[/tex] and [tex]\( C = -11 + \frac{11a}{3}\)[/tex], substitute [tex]\(b\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ -16 = \frac{8a}{3} + 2(-4a) + \left( -11 + \frac{11a}{3} \right) \][/tex]
[tex]\[ -16 = \frac{8a}{3} - 8a - 11 + \frac{11a}{3} \][/tex]
[tex]\[ -16 = \frac{8a + 11a}{3} - 8a - 11 \][/tex]
[tex]\[ -16 = \frac{19a}{3} - 8a - 11 \][/tex]
Combine like terms:
[tex]\[ -16 = \frac{19a}{3} - \frac{24a}{3} - 11 \][/tex]
[tex]\[ -16 = -\frac{5a}{3} - 11 \][/tex]
[tex]\[ -16 + 11 = -\frac{5a}{3} \][/tex]
[tex]\[ -5 = -\frac{5a}{3} \][/tex]
[tex]\[ 5 = \frac{5a}{3} \][/tex]
Multiply both sides by 3:
[tex]\[ 15 = 5a \][/tex]
[tex]\[ a = 3 \][/tex]
Now substitute [tex]\(a = 3\)[/tex] back into the equation for [tex]\(b\)[/tex]:
[tex]\[ b = -4a \][/tex]
[tex]\[ b = -4(3) = -12 \][/tex]
### Conclusion
The values of the constants are:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = -12 \][/tex]
### Step 1: Understanding the Gradient at a Point
The gradient of the curve at any point [tex]\(x\)[/tex] is given as:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
### Step 2: Condition at [tex]\( (2, -16) \)[/tex]
The tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis. This implies that the gradient at [tex]\(x = 2\)[/tex] is 0. Therefore:
[tex]\[ \left. \frac{dy}{dx} \right|_{x = 2} = 0 \][/tex]
Substituting [tex]\(x = 2\)[/tex] into the gradient expression:
[tex]\[ a (2)^2 + b = 0 \][/tex]
[tex]\[ 4a + b = 0 \][/tex]
So, we have our first equation:
[tex]\[ b = -4a \][/tex]
### Step 3: Integrating the Gradient
To find the curve [tex]\(y\)[/tex] which passes through the given points, we need to integrate the gradient function:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
Integrate with respect to [tex]\(x\)[/tex]:
[tex]\[ y = \int (a x^2 + b) \, dx \][/tex]
[tex]\[ y = \int a x^2 \, dx + \int b \, dx \][/tex]
[tex]\[ y = \frac{a}{3} x^3 + bx + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
### Step 4: Applying the Given Point [tex]\((1, -11)\)[/tex]
The curve passes through the point [tex]\((1, -11)\)[/tex], so we substitute [tex]\(x = 1\)[/tex] and [tex]\(y = -11\)[/tex] into the equation of the curve:
[tex]\[ -11 = \frac{a}{3} (1)^3 + b(1) + C \][/tex]
[tex]\[ -11 = \frac{a}{3} + b + C \][/tex]
Using [tex]\( b = -4a \)[/tex], substitute [tex]\(b\)[/tex] into the equation:
[tex]\[ -11 = \frac{a}{3} - 4a + C \][/tex]
[tex]\[ -11 = \frac{a}{3} - \frac{12a}{3} + C \][/tex]
[tex]\[ -11 = -\frac{11a}{3} + C \][/tex]
Multiplying through by 3:
[tex]\[ -33 = -11a + 3C \][/tex]
[tex]\[ 3C = -33 + 11a \][/tex]
[tex]\[ C = -11 + \frac{11a}{3} \][/tex]
[tex]\[ C = -11 + \frac{11a}{3} \][/tex]
### Step 5: Applying the Given Point [tex]\((2, -16)\)[/tex]
The curve also passes through the point [tex]\((2, -16)\)[/tex], so we substitute [tex]\(x = 2\)[/tex] and [tex]\(y = -16\)[/tex] into the equation of the curve:
[tex]\[ -16 = \frac{a}{3} (2)^3 + b(2) + C \][/tex]
[tex]\[ -16 = \frac{8a}{3} + 2b + C \][/tex]
Using [tex]\( b = -4a \)[/tex] and [tex]\( C = -11 + \frac{11a}{3}\)[/tex], substitute [tex]\(b\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ -16 = \frac{8a}{3} + 2(-4a) + \left( -11 + \frac{11a}{3} \right) \][/tex]
[tex]\[ -16 = \frac{8a}{3} - 8a - 11 + \frac{11a}{3} \][/tex]
[tex]\[ -16 = \frac{8a + 11a}{3} - 8a - 11 \][/tex]
[tex]\[ -16 = \frac{19a}{3} - 8a - 11 \][/tex]
Combine like terms:
[tex]\[ -16 = \frac{19a}{3} - \frac{24a}{3} - 11 \][/tex]
[tex]\[ -16 = -\frac{5a}{3} - 11 \][/tex]
[tex]\[ -16 + 11 = -\frac{5a}{3} \][/tex]
[tex]\[ -5 = -\frac{5a}{3} \][/tex]
[tex]\[ 5 = \frac{5a}{3} \][/tex]
Multiply both sides by 3:
[tex]\[ 15 = 5a \][/tex]
[tex]\[ a = 3 \][/tex]
Now substitute [tex]\(a = 3\)[/tex] back into the equation for [tex]\(b\)[/tex]:
[tex]\[ b = -4a \][/tex]
[tex]\[ b = -4(3) = -12 \][/tex]
### Conclusion
The values of the constants are:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = -12 \][/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.