Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
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]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
