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Mathematics Question on Applications of Derivatives

The slope of the tangent to the curve x=t2+3t−8,y=2t2−5x=t^2+3t-8,y=2t^2-5 at the point (2,−1)(2, −1) is

A

227\frac{22}{7}

B

67\frac{6}{7}

C

76\frac{7}{6}

D

−67\frac{-6}{7}

Answer

67\frac{6}{7}

Explanation

Solution

The correct answer is B:67\frac{6}{7}
The given curve is x=t2+3t−8x=t^2+3t-8 and y=2t2−2t−5y=2t^2-2t-5.
∴dxdt=2t+3∴ \frac{dx}{dt}=2t+3 and dydt=4t−2\frac{dy}{dt}=4t-2
∴dydx=dydt.dtdx=4t−22t+3∴ \frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{4t-2}{2t+3}
The given point is (2,−1)(2, −1)
At x=2x = 2, we have:
t2+3t−8=2t^2+3t-8=2
t2+3t−10=0t^2+3t-10=0
(t−2)(t+5)=0(t-2)(t+5)=0
t=2 or t=−5t=2\, or\, t=-5
At y=−1y=-1, we have:
2t2−2t−5=−12t^2-2t-5=-1
⇒2t2−2t−4=0⇒ 2t^2-2t-4=0
⇒2(t2−t−2)=0⇒ 2(t^2-t-2)=0
⇒(t−2)(t+1)=0⇒ (t-2)(t+1)=0
⇒t=2 or t=1⇒ t=2\, or\, t=1
The common value of t is 2.
Hence, the slope of the tangent to the given curve at point (2, −1) is
dydx]t=2=4(2)−22(2)+3=8−24+3=67\frac{dy}{dx}\bigg]_{t=2}=\frac{4(2)-2}{2(2)+3}=\frac{8-2}{4+3}=\frac{6}{7}