Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Let's solve this problem step-by-step.
### i) Finding the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
Given:
1. The curve passes through the point [tex]\((1, -11)\)[/tex].
2. The gradient at any point on the curve is [tex]\(a x^2 + b\)[/tex].
3. The tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis.
Since the tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis, the gradient at [tex]\((2, -16)\)[/tex] must be [tex]\(0\)[/tex]. Therefore, we have:
[tex]\[ a \cdot 2^2 + b = 0 \][/tex]
[tex]\[ 4a + b = 0 \][/tex]
[tex]\[ b = -4a \quad \text{(Equation 1)} \][/tex]
### ii) Finding the equation of the curve
The gradient of the curve (the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex]) is given by:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
To find the equation of the curve, we need to integrate the gradient function. Integrating [tex]\(a x^2 + b\)[/tex] with respect to [tex]\(x\)[/tex], we get:
[tex]\[ y = \int (a x^2 + b) \, dx \][/tex]
[tex]\[ y = \int a x^2 \, dx + \int b \, dx \][/tex]
[tex]\[ y = a \int x^2 \, dx + b \int 1 \, dx \][/tex]
[tex]\[ y = a \left( \frac{x^3}{3} \right) + b x + C \][/tex]
[tex]\[ y = \frac{a x^3}{3} + b x + C \][/tex]
Now, we'll use the point [tex]\((1, -11)\)[/tex] to find the constant [tex]\(C\)[/tex]. Substituting [tex]\(x = 1\)[/tex] and [tex]\(y = -11\)[/tex] into the equation:
[tex]\[ -11 = \frac{a (1)^3}{3} + b (1) + C \][/tex]
[tex]\[ -11 = \frac{a}{3} + b + C \][/tex]
We already have [tex]\(b = -4a\)[/tex] from Equation 1. Substituting [tex]\(b\)[/tex] into the equation:
[tex]\[ -11 = \frac{a}{3} - 4a + C \][/tex]
[tex]\[ -11 = \frac{a - 12a}{3} + C \][/tex]
[tex]\[ -11 = \frac{-11a}{3} + C \][/tex]
[tex]\[ -11a = -33 - 3C \][/tex]
[tex]\[ 3 = 33 + 3C \][/tex]
Now we solve for [tex]\(C\)[/tex]:
[tex]\[ 3C = -22 \][/tex]
[tex]\[ C = -22/3 \][/tex]
### Summarizing the findings
1. From [tex]\(b = -4a\)[/tex]:
- a = b / (-4)
- The values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are defined parametrically:
- [tex]\(a = b/( -4 \)[/tex]
- [tex]\(b = -4a\)[/tex]
[tex]\(\therefore\)[/tex] [tex]\[ a = 1/4 \][/tex]
2. The equation of the curve is
\[ y = \frac{1}{4}x^3 - 4(-11) * x + \frac{4}{{-11}} / 13 +23/5.
### i) Finding the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
Given:
1. The curve passes through the point [tex]\((1, -11)\)[/tex].
2. The gradient at any point on the curve is [tex]\(a x^2 + b\)[/tex].
3. The tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis.
Since the tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis, the gradient at [tex]\((2, -16)\)[/tex] must be [tex]\(0\)[/tex]. Therefore, we have:
[tex]\[ a \cdot 2^2 + b = 0 \][/tex]
[tex]\[ 4a + b = 0 \][/tex]
[tex]\[ b = -4a \quad \text{(Equation 1)} \][/tex]
### ii) Finding the equation of the curve
The gradient of the curve (the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex]) is given by:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
To find the equation of the curve, we need to integrate the gradient function. Integrating [tex]\(a x^2 + b\)[/tex] with respect to [tex]\(x\)[/tex], we get:
[tex]\[ y = \int (a x^2 + b) \, dx \][/tex]
[tex]\[ y = \int a x^2 \, dx + \int b \, dx \][/tex]
[tex]\[ y = a \int x^2 \, dx + b \int 1 \, dx \][/tex]
[tex]\[ y = a \left( \frac{x^3}{3} \right) + b x + C \][/tex]
[tex]\[ y = \frac{a x^3}{3} + b x + C \][/tex]
Now, we'll use the point [tex]\((1, -11)\)[/tex] to find the constant [tex]\(C\)[/tex]. Substituting [tex]\(x = 1\)[/tex] and [tex]\(y = -11\)[/tex] into the equation:
[tex]\[ -11 = \frac{a (1)^3}{3} + b (1) + C \][/tex]
[tex]\[ -11 = \frac{a}{3} + b + C \][/tex]
We already have [tex]\(b = -4a\)[/tex] from Equation 1. Substituting [tex]\(b\)[/tex] into the equation:
[tex]\[ -11 = \frac{a}{3} - 4a + C \][/tex]
[tex]\[ -11 = \frac{a - 12a}{3} + C \][/tex]
[tex]\[ -11 = \frac{-11a}{3} + C \][/tex]
[tex]\[ -11a = -33 - 3C \][/tex]
[tex]\[ 3 = 33 + 3C \][/tex]
Now we solve for [tex]\(C\)[/tex]:
[tex]\[ 3C = -22 \][/tex]
[tex]\[ C = -22/3 \][/tex]
### Summarizing the findings
1. From [tex]\(b = -4a\)[/tex]:
- a = b / (-4)
- The values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are defined parametrically:
- [tex]\(a = b/( -4 \)[/tex]
- [tex]\(b = -4a\)[/tex]
[tex]\(\therefore\)[/tex] [tex]\[ a = 1/4 \][/tex]
2. The equation of the curve is
\[ y = \frac{1}{4}x^3 - 4(-11) * x + \frac{4}{{-11}} / 13 +23/5.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.