The gradient of a curve which passes through P(2, 7) is given by dy/dx = 15/sqrt(7x + 2). a) Find the…

Question

The gradient of a curve which passes through P(2, 7) is given by dy/dx = 15/sqrt(7x + 2). a) Find the equation of the curve. b) Find the equation of the normal to the curve at P.

ScanSolve AI Tutor · Accepted Answer

Question 10: The gradient of a curve which passes through P(2, 7) is given by dy/dx = (15)/(√(7x + 2)). a) Find the equation of the curve. Step 1: Integrate the given gradient function to find the equation of the curve. y = ∫ (15)/(√(7x + 2)) dx Rewrite the integrand using exponent notation: y = ∫ 15(7x + 2)^(−1/2) dx; Step 2: Apply the integration rule ∫ (ax+b)ⁿ dx = ((ax+b)ⁿ⁺¹)/(a(n+1)) + C. Here, a=7, n=−1/2. So, n+1 = 1/2. y = 15 · ((7x + 2)^(1/2))/(7 · (1/2)) + C; y = 15 · 2/7 (7x + 2)^(1/2) + C; y = 30/7 √(7x + 2) + C; Step 3: Use the given point P(2, 7) to find the value of the constant C. Substitute x=2 and y=7 into the equation. 7 = 30/7 √(7(2) + 2) + C; 7 = 30/7 √(14 + 2) + C; 7 = 30/7 √(16) + C; 7 = 30/7 · 4 + C; 7 = 120/7 + C; C = 7 − 120/7; C = 49/7 − 120/7; C = −71/7 Step 4: Substitute the value of C back into the curve's equation. y = 30/7 √(7x + 2) − 71/7 b) Find the equation of the normal to the curve at P. Step 1: Find the gradient of the tangent to the curve at point P(2, 7) by substituting x=2 into the given dy/dx. m_t = dy/dx |_x=2 = (15)/(√(7(2) + 2)); m_t = (15)/(√(14 + 2)); m_t = (15)/(√(16)); m_t = 15/4 Step 2: Find the gradient of the normal, m_n. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent's gradient. m_n = −1/m_t; m_n = −(1)/(15/4); m_n = −4/15; Step 3: Use the point-slope form of a linear equation, y − y₁ = m(x − x₁), with point P(2, 7) and m_n = −4/15. y − 7 = −4/15(x − 2); Step 4: Multiply both sides by 15 to eliminate the fraction and rearrange the equation into the general form Ax + By + C = 0. 15(y − 7) = −4(x − 2); 15y − 105 = −4x + 8; 4x + 15y − 105 − 8 = 0; 4x + 15y − 113 = 0

The gradient of a curve which passes through P(2, 7) is given by dy/dx = 15/sqrt(7x + 2). a) Find the equation of the curve. b) Find the equation of the normal to the curve at P.

Answers

Step-by-step solution

Part 10

Part a· 4 steps
$y = \frac{30}{7} \sqrt{7x + 2} - \frac{71}{7}$

Integrate the given gradient function to find the equation of the curve.

Apply the integration rule $\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$ .

Use the given point $P(2, 7)$ to find the value of the constant $C$ .

Substitute the value of $C$ back into the curve's equation.

Part b· 4 steps
$4x + 15y - 113 = 0$

Find the gradient of the tangent to the curve at point $P(2, 7)$ by substituting $x=2$ into the given $\frac{dy}{dx}$ .

Find the gradient of the normal, $m_n$ .

Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , with point $P(2, 7)$ and $m_n = -\frac{4}{15}$ .

Multiply both sides by 15 to eliminate the fraction and rearrange the equation into the general form $Ax + By + C = 0$ .

Try one yourself

Still stuck on this one?

More Mathematics Questions

The gradient of a curve which passes through P(2, 7) is given by dy/dx = 15/sqrt(7x + 2). a) Find the equation of the curve. b) Find the equation of the normal to the curve at P.

Answers

Step-by-step solution

Part 10

Part a· 4 stepsy=3077x+2−717y = \frac{30}{7} \sqrt{7x + 2} - \frac{71}{7}y=730​7x+2​−771​

Integrate the given gradient function to find the equation of the curve.

Apply the integration rule ∫(ax+b)n dx=(ax+b)n+1a(n+1)+C\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C∫(ax+b)ndx=a(n+1)(ax+b)n+1​+C.

Use the given point P(2,7)P(2, 7)P(2,7) to find the value of the constant CCC.

Substitute the value of CCC back into the curve's equation.

Part b· 4 steps4x+15y−113=04x + 15y - 113 = 04x+15y−113=0

Find the gradient of the tangent to the curve at point P(2,7)P(2, 7)P(2,7) by substituting x=2x=2x=2 into the given dydx\frac{dy}{dx}dxdy​.

Find the gradient of the normal, mnm_nmn​.

Use the point-slope form of a linear equation, y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​), with point P(2,7)P(2, 7)P(2,7) and mn=−415m_n = -\frac{4}{15}mn​=−154​.

Multiply both sides by 15 to eliminate the fraction and rearrange the equation into the general form Ax+By+C=0Ax + By + C = 0Ax+By+C=0.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Part a· 4 steps
$y = \frac{30}{7} \sqrt{7x + 2} - \frac{71}{7}$

Apply the integration rule $\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$ .

Use the given point $P(2, 7)$ to find the value of the constant $C$ .

Substitute the value of $C$ back into the curve's equation.

Part b· 4 steps
$4x + 15y - 113 = 0$

Find the gradient of the tangent to the curve at point $P(2, 7)$ by substituting $x=2$ into the given $\frac{dy}{dx}$ .

Find the gradient of the normal, $m_n$ .

Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , with point $P(2, 7)$ and $m_n = -\frac{4}{15}$ .

Multiply both sides by 15 to eliminate the fraction and rearrange the equation into the general form $Ax + By + C = 0$ .