Math / Calculus

Question2. (a) A curve is defined by the function $y=x^{3}-3 x-5$
Find i) the gradient of the curve at the point where $x=2$ ii) the coordinates of the maximum and minimum turning point (b) Obtain from first principles the derivative of the function $y=2 x^{2}+4 x-3$ (c) Use integration by substitution to find the given integral $\int_{0}^{1}\left(3 x^{2}+2\right)^{4} d x$ (d) Use the product rule to find $\frac{d y}{d x}$ given i. $y=\left(3 x^{2}+1\right)(5 t-8)$ ii. $y=t^{2} \sin t$

Studdy Solution

STEP 1

Assumptions
1. The function for the curve is $y = x^3 - 3x - 5$ .
2. We need to find the gradient of the curve at $x = 2$ .
3. We need to find the coordinates of the maximum and minimum turning points of the curve.
4. For part (b), we need to derive the function $y = 2x^2 + 4x - 3$ from first principles.
5. For part (c), we need to use integration by substitution to evaluate the integral $\int_{0}^{1} (3x^2 + 2)^4 \, dx$ .
6. For part (d), we need to use the product rule to find $\frac{dy}{dx}$ for the given functions.

STEP 2

To find the gradient of the curve at $x = 2$ , we first need to find the derivative of the function $y = x^3 - 3x - 5$ .
$\frac{dy}{dx} = \frac{d}{dx}(x^3 - 3x - 5)$

STEP 3

Differentiate the function term by term.
$\frac{dy}{dx} = 3x^2 - 3$

STEP 4

Evaluate the derivative at $x = 2$ .
$\left. \frac{dy}{dx} \right|_{x=2} = 3(2)^2 - 3$

STEP 5

Calculate the value.
$\left. \frac{dy}{dx} \right|_{x=2} = 3(4) - 3 = 12 - 3 = 9$
The gradient of the curve at $x = 2$ is 9.

STEP 6

To find the coordinates of the maximum and minimum turning points, we need to find the critical points by setting the derivative equal to zero.
$3x^2 - 3 = 0$

STEP 7

Solve for $x$ .
$3x^2 = 3$
$x^2 = 1$
$x = \pm 1$

STEP 8

Determine the y-coordinates by substituting $x = 1$ and $x = -1$ into the original function $y = x^3 - 3x - 5$ .
For $x = 1$ :
$y = 1^3 - 3(1) - 5 = 1 - 3 - 5 = -7$
For $x = -1$ :
$y = (-1)^3 - 3(-1) - 5 = -1 + 3 - 5 = -3$

STEP 9

The coordinates of the turning points are $(1, -7)$ and $(-1, -3)$ .

STEP 10

To determine whether these points are maxima or minima, we need to find the second derivative of the function.
$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 3)$

STEP 11

Differentiate the first derivative.
$\frac{d^2y}{dx^2} = 6x$

STEP 12

Evaluate the second derivative at $x = 1$ and $x = -1$ .
For $x = 1$ :
$\left. \frac{d^2y}{dx^2} \right|_{x=1} = 6(1) = 6$
For $x = -1$ :
$\left. \frac{d^2y}{dx^2} \right|_{x=-1} = 6(-1) = -6$

STEP 13

Since $\frac{d^2y}{dx^2}$ is positive at $x = 1$ , the point $(1, -7)$ is a minimum. Since $\frac{d^2y}{dx^2}$ is negative at $x = -1$ , the point $(-1, -3)$ is a maximum.

STEP 14

For part (b), we need to find the derivative of $y = 2x^2 + 4x - 3$ from first principles.
The definition of the derivative from first principles is:
$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

STEP 15

Substitute $y = 2x^2 + 4x - 3$ into the definition.
$f(x+h) = 2(x+h)^2 + 4(x+h) - 3$

STEP 16

Expand $f(x+h)$ .
$f(x+h) = 2(x^2 + 2xh + h^2) + 4x + 4h - 3$
$f(x+h) = 2x^2 + 4xh + 2h^2 + 4x + 4h - 3$

STEP 17

Subtract $f(x)$ from $f(x+h)$ .
$f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 4x + 4h - 3) - (2x^2 + 4x - 3)$
$f(x+h) - f(x) = 4xh + 2h^2 + 4h$

STEP 18

Divide by $h$ .
$\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 4h}{h} = 4x + 2h + 4$

STEP 19

Take the limit as $h$ approaches 0.
$\lim_{h \to 0} (4x + 2h + 4) = 4x + 4$
The derivative of $y = 2x^2 + 4x - 3$ is $\frac{dy}{dx} = 4x + 4$ .

STEP 20

For part (c), we need to use integration by substitution to evaluate $\int_{0}^{1} (3x^2 + 2)^4 \, dx$ .
Let $u = 3x^2 + 2$ .

STEP 21

Differentiate $u$ with respect to $x$ .
$\frac{du}{dx} = 6x$
$du = 6x \, dx$

STEP 22

Rewrite the integral in terms of $u$ .
$\int_{0}^{1} (3x^2 + 2)^4 \, dx = \int_{0}^{1} u^4 \cdot \frac{du}{6x}$

STEP 23

Since $u = 3x^2 + 2$ , when $x = 0$ , $u = 2$ , and when $x = 1$ , $u = 5$ .
$\int_{2}^{5} u^4 \cdot \frac{du}{6x}$

STEP 24

Since $du = 6x \, dx$ , the integral simplifies to:
$\int_{2}^{5} \frac{u^4}{6} \, du$

STEP 25

Factor out the constant $\frac{1}{6}$ .
$\frac{1}{6} \int_{2}^{5} u^4 \, du$

STEP 26

Integrate $u^4$ .
$\frac{1}{6} \left[ \frac{u^5}{5} \right]_{2}^{5}$

STEP 27

Evaluate the definite integral.
$\frac{1}{6} \left( \frac{5^5}{5} - \frac{2^5}{5} \right)$
$\frac{1}{6} \left( 625 - \frac{32}{5} \right)$

STEP 28

Simplify the expression.
$\frac{1}{6} \left( 625 - 6.4 \right)$
$\frac{1}{6} \cdot 618.6 = 103.1$
The value of the integral is $103.1$ .

STEP 29

For part (d), we need to use the product rule to find $\frac{dy}{dx}$ for the given functions.
i. $y = (3x^2 + 1)(5t - 8)$
The product rule states:
$\frac{d}{dx} [u \cdot v] = u' \cdot v + u \cdot v'$
Let $u = 3x^2 + 1$ and $v = 5t - 8$ .

STEP 30

Differentiate $u$ and $v$ .
$u' = 6x$
$v' = 5$

STEP 31

Apply the product rule.
$\frac{dy}{dx} = (6x)(5t - 8) + (3x^2 + 1)(5)$
$\frac{dy}{dx} = 30xt - 48x + 15x^2 + 5$
ii. $y = t^2 \sin t$
Let $u = t^2$ and $v = \sin t$ .

STEP 32

Differentiate $u$ and $v$ .
$u' = 2t$
$v' = \cos t$

STEP 33

Apply the product rule.
$\frac{dy}{dx} = (2t)(\sin t) + (t^2)(\cos t)$
$\frac{dy}{dx} = 2t \sin t + t^2 \cos t$

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Studdy Solution

STEP 1

STEP 2

To find the gradient of the curve at x=2 x = 2 x=2, we first need to find the derivative of the function y=x3−3x−5 y = x^3 - 3x - 5 y=x3−3x−5.dydx=ddx(x3−3x−5) \frac{dy}{dx} = \frac{d}{dx}(x^3 - 3x - 5) dxdy​=dxd​(x3−3x−5)

STEP 3

Differentiate the function term by term.dydx=3x2−3 \frac{dy}{dx} = 3x^2 - 3 dxdy​=3x2−3

STEP 4

Evaluate the derivative at x=2 x = 2 x=2.dydx∣x=2=3(2)2−3 \left. \frac{dy}{dx} \right|_{x=2} = 3(2)^2 - 3 dxdy​∣∣​x=2​=3(2)2−3

STEP 5

Calculate the value.dydx∣x=2=3(4)−3=12−3=9 \left. \frac{dy}{dx} \right|_{x=2} = 3(4) - 3 = 12 - 3 = 9 dxdy​∣∣​x=2​=3(4)−3=12−3=9The gradient of the curve at x=2 x = 2 x=2 is 9.

STEP 6

To find the coordinates of the maximum and minimum turning points, we need to find the critical points by setting the derivative equal to zero.3x2−3=0 3x^2 - 3 = 0 3x2−3=0

STEP 7

Solve for x x x.3x2=3 3x^2 = 3 3x2=3x2=1 x^2 = 1 x2=1x=±1 x = \pm 1 x=±1

STEP 8

STEP 9

The coordinates of the turning points are (1,−7) (1, -7) (1,−7) and (−1,−3) (-1, -3) (−1,−3).

STEP 10

To determine whether these points are maxima or minima, we need to find the second derivative of the function.d2ydx2=ddx(3x2−3) \frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 3) dx2d2y​=dxd​(3x2−3)

STEP 11

Differentiate the first derivative.d2ydx2=6x \frac{d^2y}{dx^2} = 6x dx2d2y​=6x

STEP 12

STEP 13

Since d2ydx2 \frac{d^2y}{dx^2} dx2d2y​ is positive at x=1 x = 1 x=1, the point (1,−7) (1, -7) (1,−7) is a minimum. Since d2ydx2 \frac{d^2y}{dx^2} dx2d2y​ is negative at x=−1 x = -1 x=−1, the point (−1,−3) (-1, -3) (−1,−3) is a maximum.

STEP 14

For part (b), we need to find the derivative of y=2x2+4x−3 y = 2x^2 + 4x - 3 y=2x2+4x−3 from first principles.The definition of the derivative from first principles is:dydx=lim⁡h→0f(x+h)−f(x)h \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} dxdy​=h→0lim​hf(x+h)−f(x)​

STEP 15

Substitute y=2x2+4x−3 y = 2x^2 + 4x - 3 y=2x2+4x−3 into the definition.f(x+h)=2(x+h)2+4(x+h)−3 f(x+h) = 2(x+h)^2 + 4(x+h) - 3 f(x+h)=2(x+h)2+4(x+h)−3

STEP 16

Expand f(x+h) f(x+h) f(x+h).f(x+h)=2(x2+2xh+h2)+4x+4h−3 f(x+h) = 2(x^2 + 2xh + h^2) + 4x + 4h - 3 f(x+h)=2(x2+2xh+h2)+4x+4h−3f(x+h)=2x2+4xh+2h2+4x+4h−3 f(x+h) = 2x^2 + 4xh + 2h^2 + 4x + 4h - 3 f(x+h)=2x2+4xh+2h2+4x+4h−3

STEP 17

STEP 18

Divide by h h h.f(x+h)−f(x)h=4xh+2h2+4hh=4x+2h+4 \frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 4h}{h} = 4x + 2h + 4 hf(x+h)−f(x)​=h4xh+2h2+4h​=4x+2h+4

STEP 19

Take the limit as h h h approaches 0.lim⁡h→0(4x+2h+4)=4x+4 \lim_{h \to 0} (4x + 2h + 4) = 4x + 4 h→0lim​(4x+2h+4)=4x+4The derivative of y=2x2+4x−3 y = 2x^2 + 4x - 3 y=2x2+4x−3 is dydx=4x+4 \frac{dy}{dx} = 4x + 4 dxdy​=4x+4.

STEP 20

For part (c), we need to use integration by substitution to evaluate ∫01(3x2+2)4 dx \int_{0}^{1} (3x^2 + 2)^4 \, dx ∫01​(3x2+2)4dx.Let u=3x2+2 u = 3x^2 + 2 u=3x2+2.

STEP 21

Differentiate u u u with respect to x x x.dudx=6x \frac{du}{dx} = 6x dxdu​=6xdu=6x dx du = 6x \, dx du=6xdx

STEP 22

Rewrite the integral in terms of u u u.∫01(3x2+2)4 dx=∫01u4⋅du6x \int_{0}^{1} (3x^2 + 2)^4 \, dx = \int_{0}^{1} u^4 \cdot \frac{du}{6x} ∫01​(3x2+2)4dx=∫01​u4⋅6xdu​

STEP 23

Since u=3x2+2 u = 3x^2 + 2 u=3x2+2, when x=0 x = 0 x=0, u=2 u = 2 u=2, and when x=1 x = 1 x=1, u=5 u = 5 u=5.∫25u4⋅du6x \int_{2}^{5} u^4 \cdot \frac{du}{6x} ∫25​u4⋅6xdu​

STEP 24

Since du=6x dx du = 6x \, dx du=6xdx, the integral simplifies to:∫25u46 du \int_{2}^{5} \frac{u^4}{6} \, du ∫25​6u4​du

STEP 25

Factor out the constant 16 \frac{1}{6} 61​.16∫25u4 du \frac{1}{6} \int_{2}^{5} u^4 \, du 61​∫25​u4du

STEP 26

Integrate u4 u^4 u4.16[u55]25 \frac{1}{6} \left[ \frac{u^5}{5} \right]_{2}^{5} 61​[5u5​]25​

STEP 27

Evaluate the definite integral.16(555−255) \frac{1}{6} \left( \frac{5^5}{5} - \frac{2^5}{5} \right) 61​(555​−525​)16(625−325) \frac{1}{6} \left( 625 - \frac{32}{5} \right) 61​(625−532​)

STEP 28

Simplify the expression.16(625−6.4) \frac{1}{6} \left( 625 - 6.4 \right) 61​(625−6.4)16⋅618.6=103.1 \frac{1}{6} \cdot 618.6 = 103.1 61​⋅618.6=103.1The value of the integral is 103.1 103.1 103.1.

STEP 29

STEP 30

Differentiate u u u and v v v.u′=6x u' = 6x u′=6xv′=5 v' = 5 v′=5

STEP 31

STEP 32

Differentiate u u u and v v v.u′=2t u' = 2t u′=2tv′=cos⁡t v' = \cos t v′=cost

STEP 33

Apply the product rule.dydx=(2t)(sin⁡t)+(t2)(cos⁡t) \frac{dy}{dx} = (2t)(\sin t) + (t^2)(\cos t) dxdy​=(2t)(sint)+(t2)(cost)dydx=2tsin⁡t+t2cos⁡t \frac{dy}{dx} = 2t \sin t + t^2 \cos t dxdy​=2tsint+t2cost

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To find the gradient of the curve at $x = 2$ , we first need to find the derivative of the function $y = x^3 - 3x - 5$ .
$\frac{dy}{dx} = \frac{d}{dx}(x^3 - 3x - 5)$

Differentiate the function term by term.
$\frac{dy}{dx} = 3x^2 - 3$

Evaluate the derivative at $x = 2$ .
$\left. \frac{dy}{dx} \right|_{x=2} = 3(2)^2 - 3$

Calculate the value.
$\left. \frac{dy}{dx} \right|_{x=2} = 3(4) - 3 = 12 - 3 = 9$
The gradient of the curve at $x = 2$ is 9.

To find the coordinates of the maximum and minimum turning points, we need to find the critical points by setting the derivative equal to zero.
$3x^2 - 3 = 0$

Solve for $x$ .
$3x^2 = 3$
$x^2 = 1$
$x = \pm 1$

The coordinates of the turning points are $(1, -7)$ and $(-1, -3)$ .

To determine whether these points are maxima or minima, we need to find the second derivative of the function.
$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 3)$

Differentiate the first derivative.
$\frac{d^2y}{dx^2} = 6x$

Since $\frac{d^2y}{dx^2}$ is positive at $x = 1$ , the point $(1, -7)$ is a minimum. Since $\frac{d^2y}{dx^2}$ is negative at $x = -1$ , the point $(-1, -3)$ is a maximum.

For part (b), we need to find the derivative of $y = 2x^2 + 4x - 3$ from first principles.
The definition of the derivative from first principles is:
$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Substitute $y = 2x^2 + 4x - 3$ into the definition.
$f(x+h) = 2(x+h)^2 + 4(x+h) - 3$

Expand $f(x+h)$ .
$f(x+h) = 2(x^2 + 2xh + h^2) + 4x + 4h - 3$
$f(x+h) = 2x^2 + 4xh + 2h^2 + 4x + 4h - 3$

Divide by $h$ .
$\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 4h}{h} = 4x + 2h + 4$

Take the limit as $h$ approaches 0.
$\lim_{h \to 0} (4x + 2h + 4) = 4x + 4$
The derivative of $y = 2x^2 + 4x - 3$ is $\frac{dy}{dx} = 4x + 4$ .

For part (c), we need to use integration by substitution to evaluate $\int_{0}^{1} (3x^2 + 2)^4 \, dx$ .
Let $u = 3x^2 + 2$ .

Differentiate $u$ with respect to $x$ .
$\frac{du}{dx} = 6x$
$du = 6x \, dx$

Rewrite the integral in terms of $u$ .
$\int_{0}^{1} (3x^2 + 2)^4 \, dx = \int_{0}^{1} u^4 \cdot \frac{du}{6x}$

Since $u = 3x^2 + 2$ , when $x = 0$ , $u = 2$ , and when $x = 1$ , $u = 5$ .
$\int_{2}^{5} u^4 \cdot \frac{du}{6x}$

Since $du = 6x \, dx$ , the integral simplifies to:
$\int_{2}^{5} \frac{u^4}{6} \, du$

Factor out the constant $\frac{1}{6}$ .
$\frac{1}{6} \int_{2}^{5} u^4 \, du$

Integrate $u^4$ .
$\frac{1}{6} \left[ \frac{u^5}{5} \right]_{2}^{5}$

Evaluate the definite integral.
$\frac{1}{6} \left( \frac{5^5}{5} - \frac{2^5}{5} \right)$
$\frac{1}{6} \left( 625 - \frac{32}{5} \right)$

Simplify the expression.
$\frac{1}{6} \left( 625 - 6.4 \right)$
$\frac{1}{6} \cdot 618.6 = 103.1$
The value of the integral is $103.1$ .

Differentiate $u$ and $v$ .
$u' = 6x$
$v' = 5$

Differentiate $u$ and $v$ .
$u' = 2t$
$v' = \cos t$

Apply the product rule.
$\frac{dy}{dx} = (2t)(\sin t) + (t^2)(\cos t)$
$\frac{dy}{dx} = 2t \sin t + t^2 \cos t$