Question
The slope of the tangent to the curve x=t2+3t−8,y=2t2−2t−5 at the point t = 2 is
Answer: Option C
:
C
We have,
dxdt=2t+3anddydt=4t−2dydx=dy/dtdx/dt=4t−22t+3
Thus, slope of the tangent to the curve at the point t = 2 is
[dydx]t−2=4(2)−22(2)+3=67
Thus, slope of the tangent to the curve at the point t = 2 is
Hence (c) is the correct answer
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:
C
We have,
dxdt=2t+3anddydt=4t−2dydx=dy/dtdx/dt=4t−22t+3
Thus, slope of the tangent to the curve at the point t = 2 is
[dydx]t−2=4(2)−22(2)+3=67
Thus, slope of the tangent to the curve at the point t = 2 is
Hence (c) is the correct answer
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